All Exams > NEET > 30-Day Revision Course for NEET > All Questions
All questions of Magnetism and Matter for NEET Exam
Which combination of magnetic field lines and poles shows two magnets repelling each other?- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Which combination of magnetic field lines and poles shows two magnets repelling each other?
a)
b)
c)
d)
 | Dr Manju Sen answered |
- The discovery that one particular pole of a magnet orients northward, whereas the other pole orients southward allowed people to identify the north and south poles of any magnet.
- It was then noticed that the north poles of two different magnets repel each other, and likewise for the south poles. Conversely, the north pole of one magnet attracts the south pole of other magnets.
- This situation is analogous to that of electric charge, where like charges repel and unlike charges attract. In magnets, we simply replace the charge with a pole: Like poles repel and unlike poles attract.
A bar magnet of magnetic moment 1.5 J/T lies parallel to the direction of a uniform magnetic field of 0.22 T. What is the torque acting on it?- a)-0.3J
- b)0.23J
- c)0.0J
- d)0.3J
Correct answer is option 'B'. Can you explain this answer?
A bar magnet of magnetic moment 1.5 J/T lies parallel to the direction of a uniform magnetic field of 0.22 T. What is the torque acting on it?
a)
-0.3J
b)
0.23J
c)
0.0J
d)
0.3J
 | Preeti Iyer answered |
Answer :-
Solution :- torque when 
=180°
=180°
= 1.5 × 0.22 × sin 180°
= 0.33 × 0 = 0 Nm
When the switch is closed a magnetic field is produced by the coil. Which answer shows the shape of the field?- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
When the switch is closed a magnetic field is produced by the coil. Which answer shows the shape of the field?
a)
b)
c)
d)
 | Anand Kumar answered |
Ya it's B ....
Emerges from one side and enter from another side...
Emerges from one side and enter from another side...
A toroid wound with 60.0 turns/m of wire carries a current of 5.00 A. The torus is iron, which has a magnetic permeability of μm=5000μ0 under the given conditions. H and B inside the iron are- a)340AT/m,2.88T
- b)340AT/m,1.88T
- c)380AT/m,1.98T
- d)300AT/m,1.88T
Correct answer is option 'D'. Can you explain this answer?
A toroid wound with 60.0 turns/m of wire carries a current of 5.00 A. The torus is iron, which has a magnetic permeability of μm=5000μ0 under the given conditions. H and B inside the iron are
a)
340AT/m,2.88T
b)
340AT/m,1.88T
c)
380AT/m,1.98T
d)
300AT/m,1.88T
 | Neha Sharma answered |
B=B0+Bm
= μ0nI+ μmnI
=nI(μ+ μm)
=60x5(5000μ0+ μ0)
B=300(5001μ0)
B=1.88T
H=B/μm
=300x5001μ0/5000μ
=300AT
= μ0nI+ μmnI
=nI(μ+ μm)
=60x5(5000μ0+ μ0)
B=300(5001μ0)
B=1.88T
H=B/μm
=300x5001μ0/5000μ
=300AT
Which of the following statements about earth's magnetism is correct- a)The earth behaves as a magnet with the magnetic field pointing approximately from the geographic south-west to the north-east.
- b)The earth behaves as a magnet with the magnetic field pointing approximately from the geographic south-east to the north-west.
- c)The earth behaves as a magnet with the magnetic field pointing approximately from the geographic east to the west.
- d)The earth behaves as a magnet with the magnetic field pointing approximately from the geographic south to the north.
Correct answer is option 'D'. Can you explain this answer?
Which of the following statements about earth's magnetism is correct
a)
The earth behaves as a magnet with the magnetic field pointing approximately from the geographic south-west to the north-east.
b)
The earth behaves as a magnet with the magnetic field pointing approximately from the geographic south-east to the north-west.
c)
The earth behaves as a magnet with the magnetic field pointing approximately from the geographic east to the west.
d)
The earth behaves as a magnet with the magnetic field pointing approximately from the geographic south to the north.
 | Sarthak Bansal answered |
According to recent researches the magnetic field of earth is considered due to a large bar magnet situated in earth's core.
It is considered that the north pole of this large magnet is situated at the geographical south of earth and vice versa and as the magnetic field due to a bar magnet is from north pole to south pole of the maget thus the earth's magnetic field is considered from geographical south to geographical north which are respectively north and south poles of the bar magnet.
It is considered that the north pole of this large magnet is situated at the geographical south of earth and vice versa and as the magnetic field due to a bar magnet is from north pole to south pole of the maget thus the earth's magnetic field is considered from geographical south to geographical north which are respectively north and south poles of the bar magnet.
A short bar magnet has a magnetic moment of 0.48 J/T .Magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on the axis has a direction and magnitude of .- a)1.06 G along S-N direction.
- b)0.96 G along S-N direction.
- c)0.86 G along S-N direction.
- d)0.76 G along S-N direction.
Correct answer is option 'B'. Can you explain this answer?
A short bar magnet has a magnetic moment of 0.48 J/T .Magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on the axis has a direction and magnitude of .
a)
1.06 G along S-N direction.
b)
0.96 G along S-N direction.
c)
0.86 G along S-N direction.
d)
0.76 G along S-N direction.
 | Tanuja Kapoor answered |
Here M=0.48JT−1,d=10cm=0.1m
On axis,
B= (μ0/4π)(2M/d3)=10−7×(2×0.48/0.13)=0.96×10−4T
On axis,
B= (μ0/4π)(2M/d3)=10−7×(2×0.48/0.13)=0.96×10−4T
In the magnetic meridian of a certain place, the horizontal component of the earth’s magnetic field is 0.26G and the dip angle is 600. What is the magnetic field of the earth at this location- a)0.65G
- b)0.62G
- c)0.58G
- d)0.52G
Correct answer is option 'D'. Can you explain this answer?
In the magnetic meridian of a certain place, the horizontal component of the earth’s magnetic field is 0.26G and the dip angle is 600. What is the magnetic field of the earth at this location
a)
0.65G
b)
0.62G
c)
0.58G
d)
0.52G
 | Om Desai answered |
The earth's magnetic field is Be and its horizontal and vertical components are He and Hv
cosθ= He/Be
∴cos60o= (0.26×10−4/ Be )T
⇒Be=(0.26×10−4)/ (½)=0.52×10−4T=0.52G
cosθ= He/Be
∴cos60o= (0.26×10−4/ Be )T
⇒Be=(0.26×10−4)/ (½)=0.52×10−4T=0.52G
Correct unit of Bohr magneton is- a)T
- b)T/J
- c)J/T
- d)J
Correct answer is option 'C'. Can you explain this answer?
Correct unit of Bohr magneton is
a)
T
b)
T/J
c)
J/T
d)
J
 | Sushil Kumar answered |
The Bohr magneton μB is a physical constant and the natural unit for expressing the magnetic moment of an electron caused by either its orbital or spin angular momentum.
μB= eℏ/2me
where e is the elementary charge, ℏ is the reduced Planck's constant, me is the electron rest mass.
The value of Bohr magneton in SI units is 9.27400968(20)×10−24JT−1
μB= eℏ/2me
where e is the elementary charge, ℏ is the reduced Planck's constant, me is the electron rest mass.
The value of Bohr magneton in SI units is 9.27400968(20)×10−24JT−1
A cube-shaped permanent magnet is made of a ferromagnetic material with a magnetization M of about The side length is 2 cm. Magnetic dipole moment of the magnet is.- a)5Am2
- b)3Am2
- c)4Am2
- d)6Am2
Correct answer is option 'C'. Can you explain this answer?
A cube-shaped permanent magnet is made of a ferromagnetic material with a magnetization M of about The side length is 2 cm. Magnetic dipole moment of the magnet is.
a)
5Am2
b)
3Am2
c)
4Am2
d)
6Am2
 | Infinity Academy answered |
Correct Answer :- C
Explanation : Magnetic Moment = Magnetization × Volume
Side of cube = 20 cm
= 0.2 m
= 500 × 0.2 × 0.2 × 0.2
= 4Am2
A short bar magnet of magnetic moment 5.25×10−2JT−1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45 with earth’s field on its normal bisector . Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.- a)5.5 cm
- b)7.0 cm
- c)6.0 cm
- d)5 cm
Correct answer is option 'D'. Can you explain this answer?
A short bar magnet of magnetic moment 5.25×10−2JT−1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45 with earth’s field on its normal bisector . Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
a)
5.5 cm
b)
7.0 cm
c)
6.0 cm
d)
5 cm
| | Nandini Iyer answered |
According to the question, the resultant field is inclined at 45o with earth's magnetic field.
tanθ=BH/B
θ=45o
tanθ=1=BH/B
BH=B= μoM/4πr3
Given, B= 0.42×10−4T
M=5.25×10−2J/T
Therefore,0.42+10−4=10−7×[(5.25×10−2)/r3]
r3=12.5×10−5=125×10−6
r=5×10−2m = 5 cm
tanθ=BH/B
θ=45o
tanθ=1=BH/B
BH=B= μoM/4πr3
Given, B= 0.42×10−4T
M=5.25×10−2J/T
Therefore,0.42+10−4=10−7×[(5.25×10−2)/r3]
r3=12.5×10−5=125×10−6
r=5×10−2m = 5 cm
A cube-shaped permanent magnet is made of a ferromagnetic material with a magnetization M of about The side length is 2 cm. Magnetic field due to the magnet at a point 10 cm from the magnet along its axis is- a)0.003 T
- b)0.002 T
- c)0.004 T
- d)0.001 T
Correct answer is option 'D'. Can you explain this answer?
A cube-shaped permanent magnet is made of a ferromagnetic material with a magnetization M of about The side length is 2 cm. Magnetic field due to the magnet at a point 10 cm from the magnet along its axis is
a)
0.003 T
b)
0.002 T
c)
0.004 T
d)
0.001 T
 | Riya Banerjee answered |
Correct Answer :- d
Explanation : μtotal = MV
= (8*105)(2*10-2)
= 6A m2
Magnetic field on the axis of a current loop with magnetic material μtotal is:
B = μ0 μtotal / (2π(x2 + a2)1/2
B = (4π*10-7)(6) / [2π(0.1)]
= 1 * 10-3 T
= 0.001 T
For diamagnetic materials- a)additional field caused by atoms is always parallel in direction to that of the external field
- b)additional field caused by atoms is always perpendicular in direction to that of the external field
- c)additional field caused by atoms is always in same direction as that of the external field
- d)additional field caused by atoms is always opposite in direction to that of the external field
Correct answer is option 'D'. Can you explain this answer?
For diamagnetic materials
a)
additional field caused by atoms is always parallel in direction to that of the external field
b)
additional field caused by atoms is always perpendicular in direction to that of the external field
c)
additional field caused by atoms is always in same direction as that of the external field
d)
additional field caused by atoms is always opposite in direction to that of the external field
 | Ahtesham Rashid answered |
It is because when we make a group of electrons inside the material it is called domain. So when we apply external magnetic field on the material. Those domains start to align themselves in the direction of magnetic field. BUT NOT ALL DOMAINS ALIGN THEMSELVES IN THAT DIRECTION. So option D is correct.
If a magnet is suspended over a container of liquid air, it attracts droplets to its poles. The droplets contain only liquid oxygen and no nitrogen because- a)oxygen is ferromagnetic whereas nitrogen is diamagnetic
- b)oxygen is ferrimagnetic whereas nitrogen is diamagnetic
- c)oxygen is diamagnetic whereas nitrogen is paramagnetic
- d)oxygen is paramagnetic whereas nitrogen is diamagnetic
Correct answer is option 'D'. Can you explain this answer?
If a magnet is suspended over a container of liquid air, it attracts droplets to its poles. The droplets contain only liquid oxygen and no nitrogen because
a)
oxygen is ferromagnetic whereas nitrogen is diamagnetic
b)
oxygen is ferrimagnetic whereas nitrogen is diamagnetic
c)
oxygen is diamagnetic whereas nitrogen is paramagnetic
d)
oxygen is paramagnetic whereas nitrogen is diamagnetic
| | Anaya Patel answered |
If a magnet is suspended over a container of liquid air, it attracts droplets to its poles. The droplets contain only liquid oxygen; even though nitrogen is the primary constituent of air, it is not attracted to the magnet. Explain what this tells you about the magnetic susceptibilities of oxygen and nitrogen, and explain why a magnet in ordinary, room-temperature air doesn’t attract molecules of oxygen gas to its poles.
Which one of the Maxwell’s laws leads to the conclusion that there are no magnetic field loops that are not closed?- a)Faraday’s law
- b)Gauss’ law for magnetism
- c)Gauss’ law for electricity
- d)Ampere-Maxwell law
Correct answer is option 'B'. Can you explain this answer?
Which one of the Maxwell’s laws leads to the conclusion that there are no magnetic field loops that are not closed?
a)
Faraday’s law
b)
Gauss’ law for magnetism
c)
Gauss’ law for electricity
d)
Ampere-Maxwell law
 | Jyoti Kapoor answered |
In physics, Gauss's law for magnetism is one of the four Maxwell's equations that underlie classical electrodynamics. It states that the magnetic field B has divergence equal to zero,in other words, that it is a solenoidal vector field. It is equivalent to the statement that magnetic monopoles do not exist.Rather than "magnetic charges", the basic entity for magnetism is the magnetic dipole. (If monopoles were ever found, the law would have to be modified, as elaborated below.)
Gauss's law for magnetism can be written in two forms, a differential form and an integral form. These forms are equivalent due to the divergence theorem.
The name "Gauss's law for magnetism"is not universally used. The law is also called "Absence of free magnetic poles";one reference even explicitly says the law has "no name".It is also referred to as the "transversality requirement"because for plane waves it requires that the polarization be transverse to the direction of propagation.
A toroidal solenoid with 500 turns is wound on a ring with a mean radius of 2.90 cm. Find the current in the winding that is required to set up a magnetic field of 0.350 T in the ring if the ring is made of annealed iron Km=1400- a)82.5mA
- b)72.5mA
- c)79.5mA
- d)69.5mA
Correct answer is option 'B'. Can you explain this answer?
A toroidal solenoid with 500 turns is wound on a ring with a mean radius of 2.90 cm. Find the current in the winding that is required to set up a magnetic field of 0.350 T in the ring if the ring is made of annealed iron Km=1400
a)
82.5mA
b)
72.5mA
c)
79.5mA
d)
69.5mA
 | Geetika Shah answered |
Given,
Number of turns, N = 500 turns
Radius of solenoid, r = 2.9
Relative permeability of annealed iron of Km=1400
Permeability of free space, μ0 =4π
The magnetic field, B=0.350 T
Therefore,
μ/μ0= μr
μ= μr x μ0
B= μrμ0NI/2πR
0.350=1400x4x 3.14x500xI/2xπx2.90
I=72.mA
Number of turns, N = 500 turns
Radius of solenoid, r = 2.9
Relative permeability of annealed iron of Km=1400
Permeability of free space, μ0 =4π
The magnetic field, B=0.350 T
Therefore,
μ/μ0= μr
μ= μr x μ0
B= μrμ0NI/2πR
0.350=1400x4x 3.14x500xI/2xπx2.90
I=72.mA
For paramagnetic materials- a)orbital and spin magnetic moments of the electrons are of the order of bohr magneton
- b)orbital magnetic moments of the electrons are zero
- c)orbital and spin magnetic moments of the electrons are less than zero
- d)spin magnetic moments of the electrons are zero
Correct answer is option 'A'. Can you explain this answer?
For paramagnetic materials
a)
orbital and spin magnetic moments of the electrons are of the order of bohr magneton
b)
orbital magnetic moments of the electrons are zero
c)
orbital and spin magnetic moments of the electrons are less than zero
d)
spin magnetic moments of the electrons are zero
| | Shubham Jain answered |
Paramagnetic materials are those materials that are weakly attracted by a magnetic field. They have unpaired electrons in their outermost shell, which causes a net magnetic moment due to the spin and orbital motion of the electrons.
Orbital and Spin Magnetic Moments:
- Orbital magnetic moment: It is the magnetic moment of an electron due to its motion around the nucleus.
- Spin magnetic moment: It is the magnetic moment of an electron due to its intrinsic spin.
Order of Bohr Magneton:
- Bohr magneton is a unit of magnetic moment, which is equal to approximately 9.27 x 10^-24 joules per tesla.
- The orbital and spin magnetic moments of electrons in paramagnetic materials are of the order of Bohr magneton.
Explanation of Option A:
- Option A states that the orbital and spin magnetic moments of electrons in paramagnetic materials are of the order of Bohr magneton.
- This is correct because the unpaired electrons in paramagnetic materials have a net magnetic moment due to their spin and orbital motion, which is of the order of Bohr magneton.
- This magnetic moment causes the paramagnetic material to be weakly attracted to a magnetic field.
Other Options:
- Option B states that the orbital magnetic moments of electrons in paramagnetic materials are zero, which is incorrect because the unpaired electrons have an orbital magnetic moment.
- Option C states that the orbital and spin magnetic moments of electrons in paramagnetic materials are less than zero, which is incorrect because magnetic moments cannot be negative.
- Option D states that the spin magnetic moments of electrons in paramagnetic materials are zero, which is incorrect because the unpaired electrons have a spin magnetic moment.
Orbital and Spin Magnetic Moments:
- Orbital magnetic moment: It is the magnetic moment of an electron due to its motion around the nucleus.
- Spin magnetic moment: It is the magnetic moment of an electron due to its intrinsic spin.
Order of Bohr Magneton:
- Bohr magneton is a unit of magnetic moment, which is equal to approximately 9.27 x 10^-24 joules per tesla.
- The orbital and spin magnetic moments of electrons in paramagnetic materials are of the order of Bohr magneton.
Explanation of Option A:
- Option A states that the orbital and spin magnetic moments of electrons in paramagnetic materials are of the order of Bohr magneton.
- This is correct because the unpaired electrons in paramagnetic materials have a net magnetic moment due to their spin and orbital motion, which is of the order of Bohr magneton.
- This magnetic moment causes the paramagnetic material to be weakly attracted to a magnetic field.
Other Options:
- Option B states that the orbital magnetic moments of electrons in paramagnetic materials are zero, which is incorrect because the unpaired electrons have an orbital magnetic moment.
- Option C states that the orbital and spin magnetic moments of electrons in paramagnetic materials are less than zero, which is incorrect because magnetic moments cannot be negative.
- Option D states that the spin magnetic moments of electrons in paramagnetic materials are zero, which is incorrect because the unpaired electrons have a spin magnetic moment.
How can a magnetic field be produced?- a)Using a permanent magnet
- b)Electric current
- c)Using a temporary magnet
- d)Using a permanent magnet or electric current
Correct answer is option 'D'. Can you explain this answer?
How can a magnetic field be produced?
a)
Using a permanent magnet
b)
Electric current
c)
Using a temporary magnet
d)
Using a permanent magnet or electric current
 | Sanaya Kumar answered |
Production of Magnetic Field
Magnetic field can be produced in various ways. The correct answer is option 'D', which states that a magnetic field can be produced using a permanent magnet or electric current. Let's discuss both of these methods in detail.
Using a Permanent Magnet
A permanent magnet is a magnet that retains its magnetic properties even in the absence of an external magnetic field. The magnetic field produced by a permanent magnet is due to the alignment of its atomic dipoles. The magnetic field produced by a permanent magnet is static and does not change in strength or direction over time.
Using an Electric Current
An electric current is a flow of electric charge through a conductor. When an electric current flows through a conductor, it produces a magnetic field around the conductor. The strength and direction of the magnetic field depend on the strength and direction of the current flowing through the conductor. The magnetic field produced by an electric current is dynamic and can change in strength and direction over time.
Conclusion
In conclusion, a magnetic field can be produced using a permanent magnet or electric current. While the magnetic field produced by a permanent magnet is static, the magnetic field produced by an electric current is dynamic and can change in strength and direction over time.
Magnetic field can be produced in various ways. The correct answer is option 'D', which states that a magnetic field can be produced using a permanent magnet or electric current. Let's discuss both of these methods in detail.
Using a Permanent Magnet
A permanent magnet is a magnet that retains its magnetic properties even in the absence of an external magnetic field. The magnetic field produced by a permanent magnet is due to the alignment of its atomic dipoles. The magnetic field produced by a permanent magnet is static and does not change in strength or direction over time.
Using an Electric Current
An electric current is a flow of electric charge through a conductor. When an electric current flows through a conductor, it produces a magnetic field around the conductor. The strength and direction of the magnetic field depend on the strength and direction of the current flowing through the conductor. The magnetic field produced by an electric current is dynamic and can change in strength and direction over time.
Conclusion
In conclusion, a magnetic field can be produced using a permanent magnet or electric current. While the magnetic field produced by a permanent magnet is static, the magnetic field produced by an electric current is dynamic and can change in strength and direction over time.
A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null–point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)- a)0.64 G in the direction of earth’s field.
- b)0.62 G in the direction of earth’s field.
- c)0.54 G in the direction of earth’s field.
- d)0.58 G in the direction of earth’s field.
Correct answer is option 'C'. Can you explain this answer?
A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null–point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)
a)
0.64 G in the direction of earth’s field.
b)
0.62 G in the direction of earth’s field.
c)
0.54 G in the direction of earth’s field.
d)
0.58 G in the direction of earth’s field.
 | Janhavi Kaur answered |
To find the total magnetic field on the normal bisector of the magnet at the same distance as the null point, we need to consider the contributions from both the magnet and the Earth's magnetic field.
1. Magnetic Field due to the Magnet:
The null points on the axis of the magnet indicate that the field due to the magnet is equal and opposite to the horizontal component of the Earth's magnetic field. Therefore, the horizontal component of the magnet's field at the null point is equal to the Earth's magnetic field.
Given that the null point is at a distance of 14 cm from the center of the magnet, we can use the formula for the magnetic field along the axis of a short bar magnet:
B = (μ₀/4π) * (2M/(d³))
Where:
B is the magnetic field
μ₀ is the permeability of free space (4π x 10^-7 Tm/A)
M is the magnetic moment of the magnet
d is the distance from the center of the magnet
Since the null point is at 14 cm from the center, the distance from the center to the null point is 7 cm (0.07 m). The magnetic field due to the magnet at this point is equal to the Earth's magnetic field, which is given as 0.36 G (1 G = 10^-4 T).
2. Total Magnetic Field:
The total magnetic field on the normal bisector at the same distance from the center of the magnet can be found by vector addition of the Earth's magnetic field and the magnetic field due to the magnet.
Since the angle of dip is zero, the Earth's magnetic field is entirely horizontal. Therefore, the total magnetic field will be the vector sum of the Earth's magnetic field and the magnetic field due to the magnet.
The total magnetic field is given by:
B_total = √(B_magnet² + B_earth² + 2B_magnetB_earthcosθ)
Where:
B_magnet is the magnetic field due to the magnet
B_earth is the Earth's magnetic field
θ is the angle between the two fields (which is 180° since they are opposite in direction)
Plugging in the values:
B_total = √((0.36 G)² + (0.36 G)² + 2(0.36 G)(0.36 G)cos180°)
B_total = √(0.1296 G² + 0.1296 G² - 2(0.1296 G²))
B_total = √(0.2592 G² - 0.2592 G²)
B_total = √0 G²
B_total = 0 G
Therefore, the total magnetic field on the normal bisector of the magnet at the same distance as the null point is 0 G. None of the given options (a, b, c, d) match the correct answer.
1. Magnetic Field due to the Magnet:
The null points on the axis of the magnet indicate that the field due to the magnet is equal and opposite to the horizontal component of the Earth's magnetic field. Therefore, the horizontal component of the magnet's field at the null point is equal to the Earth's magnetic field.
Given that the null point is at a distance of 14 cm from the center of the magnet, we can use the formula for the magnetic field along the axis of a short bar magnet:
B = (μ₀/4π) * (2M/(d³))
Where:
B is the magnetic field
μ₀ is the permeability of free space (4π x 10^-7 Tm/A)
M is the magnetic moment of the magnet
d is the distance from the center of the magnet
Since the null point is at 14 cm from the center, the distance from the center to the null point is 7 cm (0.07 m). The magnetic field due to the magnet at this point is equal to the Earth's magnetic field, which is given as 0.36 G (1 G = 10^-4 T).
2. Total Magnetic Field:
The total magnetic field on the normal bisector at the same distance from the center of the magnet can be found by vector addition of the Earth's magnetic field and the magnetic field due to the magnet.
Since the angle of dip is zero, the Earth's magnetic field is entirely horizontal. Therefore, the total magnetic field will be the vector sum of the Earth's magnetic field and the magnetic field due to the magnet.
The total magnetic field is given by:
B_total = √(B_magnet² + B_earth² + 2B_magnetB_earthcosθ)
Where:
B_magnet is the magnetic field due to the magnet
B_earth is the Earth's magnetic field
θ is the angle between the two fields (which is 180° since they are opposite in direction)
Plugging in the values:
B_total = √((0.36 G)² + (0.36 G)² + 2(0.36 G)(0.36 G)cos180°)
B_total = √(0.1296 G² + 0.1296 G² - 2(0.1296 G²))
B_total = √(0.2592 G² - 0.2592 G²)
B_total = √0 G²
B_total = 0 G
Therefore, the total magnetic field on the normal bisector of the magnet at the same distance as the null point is 0 G. None of the given options (a, b, c, d) match the correct answer.
Which of the following statements about magnetic field lines true?- a)The magnetic field lines of a magnet (or a solenoid) form continuous closed loops
- b)The smaller the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field B.
- c)The perpendicular to the field line at a given point gives direction of magnetic field
- d)The larger the number of field lines crossing per unit area, the weaker is the magnitude of the magnetic field B.
Correct answer is option 'A'. Can you explain this answer?
Which of the following statements about magnetic field lines true?
a)
The magnetic field lines of a magnet (or a solenoid) form continuous closed loops
b)
The smaller the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field B.
c)
The perpendicular to the field line at a given point gives direction of magnetic field
d)
The larger the number of field lines crossing per unit area, the weaker is the magnitude of the magnetic field B.
 | Krishna Iyer answered |
Magnetic poles exist in pairs. It is not possible to isolate a north pole or a south pole. Magnetic field lines start from the north pole and go to the south pole and return to the north pole. They form continuous closed loops unlike electric lines of force which do not as an electric monopole, a single charge does exist.
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 220 with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.- a)0.48G
- b)0.28G
- c)0.58G
- d)0.38G
Correct answer is option 'D'. Can you explain this answer?
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 220 with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.
a)
0.48G
b)
0.28G
c)
0.58G
d)
0.38G
 | Hansa Sharma answered |
We know the horizontal component of earth magnetic field H=R cos θ
Where θ=angle of dip=220 and the value of H=0.35G
So, putting the values respectively R= H/cosθ=0.35/cos22=0.35/0.927=0.377=0.38G
Where θ=angle of dip=220 and the value of H=0.35G
So, putting the values respectively R= H/cosθ=0.35/cos22=0.35/0.927=0.377=0.38G
In which case of comparing solenoid and bar magnet there is no exact similarity?- a)There is a current entering and a current leaving a solenoid
- b)soenoid can be broken into two weaker solenoids
- c)flux lines enter one end of a solenoid
- d)moving a small compass needle in the neighbourhood of a solenoid enables tracing the flux lines
Correct answer is option 'A'. Can you explain this answer?
In which case of comparing solenoid and bar magnet there is no exact similarity?
a)
There is a current entering and a current leaving a solenoid
b)
soenoid can be broken into two weaker solenoids
c)
flux lines enter one end of a solenoid
d)
moving a small compass needle in the neighbourhood of a solenoid enables tracing the flux lines
| | Kiran Khanna answered |
Comparing Solenoid and Bar Magnet
Introduction:
Solenoid and bar magnet are two types of magnets that have different properties. However, there are some similarities between them. In this question, we need to identify the case where there is no exact similarity between solenoid and bar magnet.
Answer:
The correct answer is option 'A' - "There is a current entering and a current leaving a solenoid". Let's understand why this is the correct answer.
Solenoid:
A solenoid is a coil of wire that produces a magnetic field when an electric current is passed through it. The magnetic field produced by a solenoid is similar to that of a bar magnet. However, there are some differences.
Bar Magnet:
A bar magnet is a permanent magnet that has a north pole and a south pole. The magnetic field of a bar magnet is uniform and is strongest at the poles.
Comparison:
Now let's compare the two magnets based on the given options.
a) There is a current entering and a current leaving a solenoid:
This is a property unique to solenoids. When an electric current is passed through a solenoid, there is a current entering at one end and a current leaving at the other end. This is because a solenoid is a coil of wire that is wrapped around a core. As a result, the magnetic field produced by a solenoid is concentrated inside the coil and is weaker outside.
b) Solenoid can be broken into two weaker solenoids:
This is a property that is similar to a bar magnet. A bar magnet can be broken into two weaker magnets, each with its own north and south pole. Similarly, a solenoid can be broken into two weaker solenoids, each with its own magnetic field.
c) Flux lines enter one end of a solenoid:
This is a property that is similar to a bar magnet. The magnetic field of a bar magnet is uniform and the flux lines enter at one end and exit at the other end. Similarly, the magnetic field of a solenoid is strongest at the ends and the flux lines enter at one end and exit at the other end.
d) Moving a small compass needle in the neighbourhood of a solenoid enables tracing the flux lines:
This is a property that is similar to a bar magnet. The magnetic field of a bar magnet can be visualized using iron filings or a small compass needle. Similarly, the magnetic field of a solenoid can be visualized using a small compass needle.
Conclusion:
Based on the above comparison, we can see that option 'A' is the correct answer as it is the only option that is unique to solenoids and does not have a similar property in bar magnets.
Introduction:
Solenoid and bar magnet are two types of magnets that have different properties. However, there are some similarities between them. In this question, we need to identify the case where there is no exact similarity between solenoid and bar magnet.
Answer:
The correct answer is option 'A' - "There is a current entering and a current leaving a solenoid". Let's understand why this is the correct answer.
Solenoid:
A solenoid is a coil of wire that produces a magnetic field when an electric current is passed through it. The magnetic field produced by a solenoid is similar to that of a bar magnet. However, there are some differences.
Bar Magnet:
A bar magnet is a permanent magnet that has a north pole and a south pole. The magnetic field of a bar magnet is uniform and is strongest at the poles.
Comparison:
Now let's compare the two magnets based on the given options.
a) There is a current entering and a current leaving a solenoid:
This is a property unique to solenoids. When an electric current is passed through a solenoid, there is a current entering at one end and a current leaving at the other end. This is because a solenoid is a coil of wire that is wrapped around a core. As a result, the magnetic field produced by a solenoid is concentrated inside the coil and is weaker outside.
b) Solenoid can be broken into two weaker solenoids:
This is a property that is similar to a bar magnet. A bar magnet can be broken into two weaker magnets, each with its own north and south pole. Similarly, a solenoid can be broken into two weaker solenoids, each with its own magnetic field.
c) Flux lines enter one end of a solenoid:
This is a property that is similar to a bar magnet. The magnetic field of a bar magnet is uniform and the flux lines enter at one end and exit at the other end. Similarly, the magnetic field of a solenoid is strongest at the ends and the flux lines enter at one end and exit at the other end.
d) Moving a small compass needle in the neighbourhood of a solenoid enables tracing the flux lines:
This is a property that is similar to a bar magnet. The magnetic field of a bar magnet can be visualized using iron filings or a small compass needle. Similarly, the magnetic field of a solenoid can be visualized using a small compass needle.
Conclusion:
Based on the above comparison, we can see that option 'A' is the correct answer as it is the only option that is unique to solenoids and does not have a similar property in bar magnets.
Magnetic field strength due to a short bar magnet on its axial line at a distance x is B. What is its value at the same distance on the equatorial line?- a)B/2
- b)B
- c)2B
- d)4B
Correct answer is option 'A'. Can you explain this answer?
Magnetic field strength due to a short bar magnet on its axial line at a distance x is B. What is its value at the same distance on the equatorial line?
a)
B/2
b)
B
c)
2B
d)
4B
 | Vivek Rana answered |
The magnetic field at any axial point is given by, B = 2μo/ 4πx3
Similarly, the field at any equatorial point is given by, B = μo/ 4πx3
Thus, the field at any equatorial point is half of what it is at an axial point.
Similarly, the field at any equatorial point is given by, B = μo/ 4πx3
Thus, the field at any equatorial point is half of what it is at an axial point.
In Ferromagnetic materials- a)atomic magnetic moments are random in strong external fields
- b)strong interactions between atomic magnetic moments cause them to line up anti parallel to each other in external field
- c)weak interactions between atomic magnetic moments cause them to line up anti parallel to each other in external field
- d)strong interactions between atomic magnetic moments cause them to line up parallel to each other in regions called magnetic domains even if there is no external field
Correct answer is option 'D'. Can you explain this answer?
In Ferromagnetic materials
a)
atomic magnetic moments are random in strong external fields
b)
strong interactions between atomic magnetic moments cause them to line up anti parallel to each other in external field
c)
weak interactions between atomic magnetic moments cause them to line up anti parallel to each other in external field
d)
strong interactions between atomic magnetic moments cause them to line up parallel to each other in regions called magnetic domains even if there is no external field
 | Rajesh Chauhan answered |
A ferromagnetic substance contains permanent atomic magnetic dipoles that are spontaneously oriented parallel to one another even in the absence of an external field. The magnetic repulsion between two dipoles aligned side by side with their moments in the same direction makes it difficult to understand the phenomenon of ferromagnetism. It is known that within a ferromagnetic material, there is a spontaneous alignment of atoms in large clusters. A new type of interaction, a quantum mechanical effect known as the exchange interaction, is involved. A highly simplified description of how the exchange interaction aligns electrons in ferromagnetic materials is given here.
A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 450 with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east. Determine the horizontal component of the earth’s magnetic field at the location- a)0.34G
- b)0.49G
- c)0.29G
- d)0.39G
Correct answer is option 'D'. Can you explain this answer?
A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 450 with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east. Determine the horizontal component of the earth’s magnetic field at the location
a)
0.34G
b)
0.49G
c)
0.29G
d)
0.39G
| | Shruti Sarkar answered |
The needle will point along W-E if the result of the earth’s magnetic field and magnetic field due to the coil have a resultant in the W-E direction. This happens if B cos 45o=Earth's field
⟹ (μ0nI/2r) cos45o=(4π×10−7×30×0.35)/(2×0.12) ×(1/√2)=0.39G=Earth's magnetic field.
⟹ (μ0nI/2r) cos45o=(4π×10−7×30×0.35)/(2×0.12) ×(1/√2)=0.39G=Earth's magnetic field.
When iron filings are sprinkled on a sheet of glass placed over a short bar magnet then,- a)The iron filings form a pattern of suggesting that the magnet has north pole similar to the positive charge
- b)The iron filings form a random pattern of suggesting that the magnet has random pole
- c)The iron filings form a pattern suggesting that the magnet has two poles similar to the positive and negative charge of an electric dipole
- d)The iron filings form a pattern of suggesting that the magnet has south pole similar to the negative charge
Correct answer is option 'C'. Can you explain this answer?
When iron filings are sprinkled on a sheet of glass placed over a short bar magnet then,
a)
The iron filings form a pattern of suggesting that the magnet has north pole similar to the positive charge
b)
The iron filings form a random pattern of suggesting that the magnet has random pole
c)
The iron filings form a pattern suggesting that the magnet has two poles similar to the positive and negative charge of an electric dipole
d)
The iron filings form a pattern of suggesting that the magnet has south pole similar to the negative charge
 | Anjana Sen answered |
Explanation:North pole act as positive end and South pole act as negative end. The iron fillings in the presence of magnetic field gets magnetized and form a tiny magnet which then gets attracted to the poles of bar magnet. North pole of bar magnet attracts south pole of tiny magnet and vice-versa. So all the iron fillings are arranged as the magnetic field lines of bar magnet.
A current I flows along the length of an infinitely long, straight, thin walled pipe. Then- a)the magnetic field at all points inside the pipe is the same, but not zero
- b)the magnetic field is zero only on the axis of the pipe
- c)the magnetic field is different at different points inside the pipe
- d)the magnetic field at any point inside the pipe is zero
Correct answer is option 'D'. Can you explain this answer?
A current I flows along the length of an infinitely long, straight, thin walled pipe. Then
a)
the magnetic field at all points inside the pipe is the same, but not zero
b)
the magnetic field is zero only on the axis of the pipe
c)
the magnetic field is different at different points inside the pipe
d)
the magnetic field at any point inside the pipe is zero
 | Shivani Tiwari answered |
Understanding the Magnetic Field Inside a Thin-Walled Pipe
When a current I flows through an infinitely long, straight, thin-walled pipe, it's essential to analyze the magnetic field produced by this configuration.
Magnetic Field Inside the Pipe
- According to Ampère's Law, the magnetic field around a current-carrying conductor can be determined using the symmetry of the situation.
- For a thin-walled pipe, which can be visualized as a cylindrical shell carrying current, the magnetic field inside the pipe is influenced by the current flowing along its surface.
Application of Gauss's Law
- To determine the magnetic field inside, we can apply Gauss's Law for magnetism, which states that the net magnetic flux through a closed surface is zero.
- Inside a cylindrical conductor, the enclosed current is zero when considering a Gaussian surface that lies entirely within the pipe's walls.
Conclusion: Magnetic Field Behavior
- The magnetic field inside the pipe is uniform and equals zero because:
- There are equal currents in opposite directions on the thin walls of the pipe that cancel each other out.
- Thus, at any point inside the pipe, the contributions to the magnetic field from all parts of the pipe cancel each other.
Final Note
- Therefore, the correct answer is option 'D': the magnetic field at any point inside the pipe is zero.
When a current I flows through an infinitely long, straight, thin-walled pipe, it's essential to analyze the magnetic field produced by this configuration.
Magnetic Field Inside the Pipe
- According to Ampère's Law, the magnetic field around a current-carrying conductor can be determined using the symmetry of the situation.
- For a thin-walled pipe, which can be visualized as a cylindrical shell carrying current, the magnetic field inside the pipe is influenced by the current flowing along its surface.
Application of Gauss's Law
- To determine the magnetic field inside, we can apply Gauss's Law for magnetism, which states that the net magnetic flux through a closed surface is zero.
- Inside a cylindrical conductor, the enclosed current is zero when considering a Gaussian surface that lies entirely within the pipe's walls.
Conclusion: Magnetic Field Behavior
- The magnetic field inside the pipe is uniform and equals zero because:
- There are equal currents in opposite directions on the thin walls of the pipe that cancel each other out.
- Thus, at any point inside the pipe, the contributions to the magnetic field from all parts of the pipe cancel each other.
Final Note
- Therefore, the correct answer is option 'D': the magnetic field at any point inside the pipe is zero.
Diamagnetic substances are- a)those which have tendency to move from stronger to the weaker part of the external magnetic field
- b)those that are uneffected by external magnetic fields
- c)those that develop a net magnetic moment in direction of applied field
- d)those which have tendency to move from weaker to the stronger part of the external magnetic field
Correct answer is option 'A'. Can you explain this answer?
Diamagnetic substances are
a)
those which have tendency to move from stronger to the weaker part of the external magnetic field
b)
those that are uneffected by external magnetic fields
c)
those that develop a net magnetic moment in direction of applied field
d)
those which have tendency to move from weaker to the stronger part of the external magnetic field
| | Nabanita Sen answered |
When a diamagnetic material is placed in an external magnetic field the spin motion of electrons is so modified that the electrons which produce the moments in the direction of the external field show down while the electrons which produce magnetic moments in opposite directions get accelerated.
Thus, a net magnetic moment is induced in the opposite direction of the applied magnetic field. Hence the substance is magnetized opposite of the external field. Thus, it moves from stronger. Weaker parts of the magnetic.
Thus, a net magnetic moment is induced in the opposite direction of the applied magnetic field. Hence the substance is magnetized opposite of the external field. Thus, it moves from stronger. Weaker parts of the magnetic.
A bar magnet of magnetic moment 1.5 J/T lies aligned with the direction of a uniform magnetic field of 0.22 T. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment opposite to the field direction?- a)0.66J
- b)0.76J nC
- c)0.86J
- d)0.56J
Correct answer is option 'A'. Can you explain this answer?
A bar magnet of magnetic moment 1.5 J/T lies aligned with the direction of a uniform magnetic field of 0.22 T. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment opposite to the field direction?
a)
0.66J
b)
0.76J nC
c)
0.86J
d)
0.56J
| | Amar Choudhary answered |
Magnetic moment, M= 1.5 J T-1
Magnetic field strength, B= 0.22 T
W = - MB (cos θ2 - cos θ1)
= -1.5 x 0.22 (cos 180° - cos 0°)
= -0.33 (-1 - 1)
= 0.66 J
= -1.5 x 0.22 (cos 180° - cos 0°)
= -0.33 (-1 - 1)
= 0.66 J
A current of 5A exists in a square loop of side 1√2m. Then the magnitude of the magnetic field B at the centre of the square loop will be p × 10–6 T. where, value of p is ______.
[Take µ0 = 4π × 10–7 T mA–1].Correct answer is '8'. Can you explain this answer?
A current of 5A exists in a square loop of side 1√2m. Then the magnitude of the magnetic field B at the centre of the square loop will be p × 10–6 T. where, value of p is ______.
[Take µ0 = 4π × 10–7 T mA–1].
[Take µ0 = 4π × 10–7 T mA–1].
| | Infinity Academy answered |
The magnetic field at the center of a square loop of current is given by the formula
where: μ₀ = permeability of free space = 4π × 10⁻⁷ T·m/A, I = current in the loop, and, a = side length of the square loop
Calculation:
Let B be the magnetic field due to single side
then
=
∴ Bnet at centre O = 4B
= 8 × 10–6
P = 8
where: μ₀ = permeability of free space = 4π × 10⁻⁷ T·m/A, I = current in the loop, and, a = side length of the square loop
Calculation:
Let B be the magnetic field due to single side
then
=
∴ Bnet at centre O = 4B
= 8 × 10–6
P = 8
permeability of a paramagnetic material is expected to decrease with increasing temperature because- a)random thermal motion increases magnetic moment alignment
- b)electrons stop moving
- c)electrons go into forbidden gap
- d)random thermal motion reduces magnetic moment alignment
Correct answer is option 'D'. Can you explain this answer?
permeability of a paramagnetic material is expected to decrease with increasing temperature because
a)
random thermal motion increases magnetic moment alignment
b)
electrons stop moving
c)
electrons go into forbidden gap
d)
random thermal motion reduces magnetic moment alignment
| | Rajdeep Sengupta answered |
When a parametric material comes close to a magnetic field the atoms will align with the magnetic field which causes another magnetic field. Permeability is the total magnetic field divided by the original magnetic field. Increase in temperature makes it harder for the atoms to align which decreases the strength of the second field. This results in a lower total magnetic field (numerator) and thus lower permeability.
which of the following features of atomic structure determine whether an element is diamagnetic or paramagnetic- a)orbital angular momentum ,spin angular momentum
- b)atomic weight
- c)number of electrons
- d)principal quantum number
Correct answer is option 'A'. Can you explain this answer?
which of the following features of atomic structure determine whether an element is diamagnetic or paramagnetic
a)
orbital angular momentum ,spin angular momentum
b)
atomic weight
c)
number of electrons
d)
principal quantum number
| | Shubham Jain answered |
Atomic Structure and Diamagnetism/Paramagnetism
Diamagnetism and paramagnetism are the two types of magnetism that an element can exhibit. Diamagnetic elements are those that have all their electrons paired up and do not have any unpaired electrons. Paramagnetic elements, on the other hand, have one or more unpaired electrons and are attracted to a magnetic field.
The features of atomic structure that determine whether an element is diamagnetic or paramagnetic are as follows:
Orbital Angular Momentum and Spin Angular Momentum
The orbital angular momentum and spin angular momentum of an electron are two important properties of atomic structure that determine whether an element is diamagnetic or paramagnetic. These properties are related to the magnetic moment of an electron, which is a measure of the strength of its magnetic field.
Diamagnetic elements have all their electrons paired up, which means that their magnetic moments cancel out and they have no net magnetic moment. In contrast, paramagnetic elements have one or more unpaired electrons, which means that they have a net magnetic moment and are attracted to a magnetic field.
Atomic Weight
The atomic weight of an element is the average weight of all the isotopes of that element. This property does not directly determine whether an element is diamagnetic or paramagnetic, but it can indirectly affect the magnetic properties of an element. Heavier elements tend to have more unpaired electrons, which makes them more likely to be paramagnetic.
Number of Electrons
The number of electrons in an atom is another important feature of atomic structure that determines whether an element is diamagnetic or paramagnetic. Elements with an odd number of electrons are more likely to be paramagnetic because they have at least one unpaired electron. Elements with an even number of electrons can be either diamagnetic or paramagnetic, depending on whether their electrons are paired or unpaired.
Principal Quantum Number
The principal quantum number of an electron is a measure of its energy level. This property does not directly determine whether an element is diamagnetic or paramagnetic, but it can indirectly affect the magnetic properties of an element. Electrons with higher energy levels are more likely to be unpaired, which makes them more likely to be paramagnetic.
Conclusion
In summary, the orbital angular momentum and spin angular momentum of an electron are the features of atomic structure that determine whether an element is diamagnetic or paramagnetic. Other features, such as atomic weight, number of electrons, and principal quantum number, can indirectly affect the magnetic properties of an element.
Diamagnetism and paramagnetism are the two types of magnetism that an element can exhibit. Diamagnetic elements are those that have all their electrons paired up and do not have any unpaired electrons. Paramagnetic elements, on the other hand, have one or more unpaired electrons and are attracted to a magnetic field.
The features of atomic structure that determine whether an element is diamagnetic or paramagnetic are as follows:
Orbital Angular Momentum and Spin Angular Momentum
The orbital angular momentum and spin angular momentum of an electron are two important properties of atomic structure that determine whether an element is diamagnetic or paramagnetic. These properties are related to the magnetic moment of an electron, which is a measure of the strength of its magnetic field.
Diamagnetic elements have all their electrons paired up, which means that their magnetic moments cancel out and they have no net magnetic moment. In contrast, paramagnetic elements have one or more unpaired electrons, which means that they have a net magnetic moment and are attracted to a magnetic field.
Atomic Weight
The atomic weight of an element is the average weight of all the isotopes of that element. This property does not directly determine whether an element is diamagnetic or paramagnetic, but it can indirectly affect the magnetic properties of an element. Heavier elements tend to have more unpaired electrons, which makes them more likely to be paramagnetic.
Number of Electrons
The number of electrons in an atom is another important feature of atomic structure that determines whether an element is diamagnetic or paramagnetic. Elements with an odd number of electrons are more likely to be paramagnetic because they have at least one unpaired electron. Elements with an even number of electrons can be either diamagnetic or paramagnetic, depending on whether their electrons are paired or unpaired.
Principal Quantum Number
The principal quantum number of an electron is a measure of its energy level. This property does not directly determine whether an element is diamagnetic or paramagnetic, but it can indirectly affect the magnetic properties of an element. Electrons with higher energy levels are more likely to be unpaired, which makes them more likely to be paramagnetic.
Conclusion
In summary, the orbital angular momentum and spin angular momentum of an electron are the features of atomic structure that determine whether an element is diamagnetic or paramagnetic. Other features, such as atomic weight, number of electrons, and principal quantum number, can indirectly affect the magnetic properties of an element.
Magnetic material differences are explained by- a)atoms contain moving electrons, and these electrons form microscopic current loops that produce magnetic fields of their own.
- b)number of neutrons
- c)number of protonsatoms contain spinning electrons, and their spin forms microscopic current loops that produce magnetic
- d)fields
Correct answer is option 'A'. Can you explain this answer?
Magnetic material differences are explained by
a)
atoms contain moving electrons, and these electrons form microscopic current loops that produce magnetic fields of their own.
b)
number of neutrons
c)
number of protonsatoms contain spinning electrons, and their spin forms microscopic current loops that produce magnetic
d)
fields
 | Gayatri Sharma answered |
All substances show some kind of magnetic behaviour. After all, they are made up of charged particles: electrons and protons. It is the way in which electron clouds arrange themselves in atoms and how groups of these atoms behave that determines the magnetic properties of the material. The atom (or group of atoms) in effect becomes a magnetic dipole or a mini bar magnet that can align according to the magnetic field applied. The net effect of all these dipoles determines the magnetic properties of the Magnetic Materials.
An element of 0.05 m is placed at the origin as shown in figure which carries a large current of 10A. distance of 1m in perpendicular direction. The value of magnetic field is 
- a)4.5 × 10−8T
- b)5.5 × 10−8T
- c)5.0 × 10−8T
- d)7.5 × 10−8T
Correct answer is option 'C'. Can you explain this answer?
An element of 0.05 m is placed at the origin as shown in figure which carries a large current of 10A. distance of 1m in perpendicular direction. The value of magnetic field is
a)
4.5 × 10−8T
b)
5.5 × 10−8T
c)
5.0 × 10−8T
d)
7.5 × 10−8T
 | Dr. Mohit Rajpoot answered |
The magnetic field due to a current element can be calculated using the Biot-Savart law.
The Biot-Savart law states that the magnetic field (dB) at a point due to a current element (Idl) is given by:
dB = (μ₀/4π) * (Idl × r̂) / r²
Where:
- μ₀ = 4π × 10⁻⁷ Tm/A (permeability of free space)
- I = current (in Amperes)
- dl = length of the current element (in meters)
- r̂ = unit vector in the direction from the current element to the point where the field is being calculated
- r = distance from the current element to the point (in meters)
Calculation:
Given values:
I = 10 A
dl = 0.05 m
r = 1 m
Using the Biot-Savart law:
⇒ dB = (μ₀/4π) * (I * dl × r̂) / r²
Since the angle between dl and r̂ is 90 degrees:
⇒ dB = (4π × 10⁻⁷ / 4π) * (10 * 0.05) / 1²
⇒ dB = 10⁻⁷ * 0.5
⇒ dB = 5 × 10⁻⁸ T
∴ The value of the magnetic field is 5.0 × 10⁻⁸ T.
A rectangular coil of 400 turns and 10−2m2 area, carrying a current of 0.5 A, is placed in a uniform magnetic field of 1 T such that the plane of the coil makes an angle of 60∘ with the direction of the magnetic field. The initial moment of force acting on the coil in Nm is
- a)√3
- b)1/√3
- c)1
- d)√3/2
Correct answer is option 'C'. Can you explain this answer??
A rectangular coil of 400 turns and 10−2m2 area, carrying a current of 0.5 A, is placed in a uniform magnetic field of 1 T such that the plane of the coil makes an angle of 60∘ with the direction of the magnetic field. The initial moment of force acting on the coil in Nm is
a)
√3
b)
1/√3
c)
1
d)
√3/2
| | Divya Khanna answered |
Understanding the Moment of Force on the Coil
To find the moment of force (torque) acting on a rectangular coil in a magnetic field, we use the formula:
Torque (τ) = n * B * I * A * sin(θ)
Where:
- n = number of turns in the coil (400 turns)
- B = magnetic field strength (1 T)
- I = current through the coil (0.5 A)
- A = area of the coil (10^-2 m²)
- θ = angle between the magnetic field and the normal to the coil's plane (60 degrees)
Calculating Each Component
- Number of Turns (n): 400
- Magnetic Field (B): 1 T
- Current (I): 0.5 A
- Area (A): 10^-2 m²
- Angle (θ): 60 degrees
Calculating the Torque
1. Calculate sin(θ):
- sin(60) = √3 / 2
2. Plugging in the values:
- τ = 400 * 1 * 0.5 * (10^-2) * (√3 / 2)
3. Simplifying the calculation:
- τ = 400 * 0.5 * (10^-2) * (√3 / 2)
- τ = 200 * (10^-2) * (√3 / 2)
- τ = 100 * (10^-2) * √3
- τ = 1 * √3
Final Result
The moment of force acting on the coil is τ = √3 Nm. Therefore, the correct answer is option 'C', which is 1 Nm (since we are looking for the numerical value).
This calculated torque indicates the effectiveness of the coil in the magnetic field, demonstrating its ability to experience rotational motion due to the magnetic interaction.
To find the moment of force (torque) acting on a rectangular coil in a magnetic field, we use the formula:
Torque (τ) = n * B * I * A * sin(θ)
Where:
- n = number of turns in the coil (400 turns)
- B = magnetic field strength (1 T)
- I = current through the coil (0.5 A)
- A = area of the coil (10^-2 m²)
- θ = angle between the magnetic field and the normal to the coil's plane (60 degrees)
Calculating Each Component
- Number of Turns (n): 400
- Magnetic Field (B): 1 T
- Current (I): 0.5 A
- Area (A): 10^-2 m²
- Angle (θ): 60 degrees
Calculating the Torque
1. Calculate sin(θ):
- sin(60) = √3 / 2
2. Plugging in the values:
- τ = 400 * 1 * 0.5 * (10^-2) * (√3 / 2)
3. Simplifying the calculation:
- τ = 400 * 0.5 * (10^-2) * (√3 / 2)
- τ = 200 * (10^-2) * (√3 / 2)
- τ = 100 * (10^-2) * √3
- τ = 1 * √3
Final Result
The moment of force acting on the coil is τ = √3 Nm. Therefore, the correct answer is option 'C', which is 1 Nm (since we are looking for the numerical value).
This calculated torque indicates the effectiveness of the coil in the magnetic field, demonstrating its ability to experience rotational motion due to the magnetic interaction.
Which of the following statements about bar magnets is correct ?- a)We can isolate the north and south pole of a bar magnet.
- b)North Pole atrracts North Pole.
- c)South Pole atrracts South Pole.
- d)We cannot isolate the north and south pole of a bar magnet.
Correct answer is option 'D'. Can you explain this answer?
Which of the following statements about bar magnets is correct ?
a)
We can isolate the north and south pole of a bar magnet.
b)
North Pole atrracts North Pole.
c)
South Pole atrracts South Pole.
d)
We cannot isolate the north and south pole of a bar magnet.
 | Srestha Choudhury answered |
Explanation:As magnetic monopole does not exist. If we split the bar magnet into two pieces each part will have its own north and south pole.
Far axial magnetic field of a solenoid as well as bar magnet vary- a)directly as R squared R = axial distance to the magnet/solenoid
- b)inversely as R squared R = axial distance to the magnet/solenoid
- c)directly as R cubed R = axial distance to the magnet/solenoid
- d)inversely as R cubed R = axial distance to the magnet/solenoid
Correct answer is option 'D'. Can you explain this answer?
Far axial magnetic field of a solenoid as well as bar magnet vary
a)
directly as R squared R = axial distance to the magnet/solenoid
b)
inversely as R squared R = axial distance to the magnet/solenoid
c)
directly as R cubed R = axial distance to the magnet/solenoid
d)
inversely as R cubed R = axial distance to the magnet/solenoid
| | Tanvi Bose answered |
For a short Bar Magnet, the magnetic induction at a point on the axis at a distance r from centre is given by the formula
B= (μ0/4π) (2M/r3)
⇒ B= K/ r3
⇒ B∝1/r3
The correct answer is option D.
B= (μ0/4π) (2M/r3)
⇒ B= K/ r3
⇒ B∝1/r3
The correct answer is option D.
Chapter doubts & questions for Magnetism and Matter - 30-Day Revision Course for NEET 2026 is part of NEET exam preparation. The chapters have been prepared according to the NEET exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for NEET 2026 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Magnetism and Matter - 30-Day Revision Course for NEET in English & Hindi are available as part of NEET exam. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
30-Day Revision Course for NEET85 videos|301 docs|171 tests |
