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All questions of Electromagnetic Induction for NEET Exam
A flux of 1m Wb passes through a strip having an area A = 0.02 m2. The plane of the strip is at an angle of 60º to the direction of a uniform field B. The value of B is
- a)0.1 T
- b)0.058 T
- c)4.0 mT
- d)none of the above
Correct answer is option 'B'. Can you explain this answer?
A flux of 1m Wb passes through a strip having an area A = 0.02 m2. The plane of the strip is at an angle of 60º to the direction of a uniform field B. The value of B is
a)
0.1 T
b)
0.058 T
c)
4.0 mT
d)
none of the above
|
Mira Joshi answered |
formula of flux is = BAcosθ
θ is angle between A and B
θ=30∘
ϕ=1×10−3wb
A=0.02 m2
angle between screen and B is 60∘
B = ϕ/acos30 = 10-3/ 2 x 10-2 x cos√3/2 = 0.058T
In the arrangement shown in given figure current from A to B is increasing in magnitude. Induced current in the loop will
- a)have clockwise direction
- b)have anticlockwise direction
- c)be zero
- d)oscillate between clockwise and anticlockwise
Correct answer is option 'A'. Can you explain this answer?
In the arrangement shown in given figure current from A to B is increasing in magnitude. Induced current in the loop will
a)
have clockwise direction
b)
have anticlockwise direction
c)
be zero
d)
oscillate between clockwise and anticlockwise
|
New Words answered |
The direction of the induced current is as shown in the figure, according to Lenz’s law which states that the indeed current flows always in such a direction as to oppose the change which is giving rise to it.
A conducting square loop of side I and resistance R moves in its plane with a uniform velocity v perpendicular to one of its sides. A uniform and constant magnetic field B exists along the perpendicular to the plane of the loop in fig. The current induced in the loop is
- a)RvB
- b)zero
- c)vBL/R
- d)vBL
Correct answer is option 'B'. Can you explain this answer?
A conducting square loop of side I and resistance R moves in its plane with a uniform velocity v perpendicular to one of its sides. A uniform and constant magnetic field B exists along the perpendicular to the plane of the loop in fig. The current induced in the loop is
a)
RvB
b)
zero
c)
vBL/R
d)
vBL
|
Sivappriya answered |
As the coil is neither moving inside the magnetic flux nor moving outside the magnetic flux the change is magnetic fulx is zero . therefore emf is zero hencecurrent induced in the coil is zero the thing is the coil moves in region of constant and uniform magnetic field ,so as said above current induced is zero option 2 is correct.
A long metal bar of 30 cm length is aligned along a north south line and moves eastward at a speed of 10 ms-1. A uniform magnetic field of 4.0 T points vertically downwards. If the south end of the bar has a potential of 0V, the induced potential at the north end of the bar is- a) + 12 V
- b)-12 V
- c)0 V
- d)Cannot be determined since there is not closed circuit
Correct answer is option 'A'. Can you explain this answer?
A long metal bar of 30 cm length is aligned along a north south line and moves eastward at a speed of 10 ms-1. A uniform magnetic field of 4.0 T points vertically downwards. If the south end of the bar has a potential of 0V, the induced potential at the north end of the bar is
a)
+ 12 V
b)
-12 V
c)
0 V
d)
Cannot be determined since there is not closed circuit
|
Anjali Sharma answered |
Induced emf =Blv=12V. It is induced in the northward direction by right hand rule (emf=
)
therefore if the south end of the pole has potential of 0V, the north end will have a potential of 12V.
)therefore if the south end of the pole has potential of 0V, the north end will have a potential of 12V.
A square coil ABCD is placed in x-y plane with its centre at origin. A long straight wire, passing through origin, carries a current in negative z-direction. Current in this wire increases with time.The induced current in the coil is
- a)clockwise
- b)anticlockwise
- c)zero
- d)alternating
Correct answer is option 'C'. Can you explain this answer?
A square coil ABCD is placed in x-y plane with its centre at origin. A long straight wire, passing through origin, carries a current in negative z-direction. Current in this wire increases with time.The induced current in the coil is
a)
clockwise
b)
anticlockwise
c)
zero
d)
alternating
|
Geetika Shah answered |
As magnetic field lines due to current carrying wire is along the surface and hence,
ϕ=B.A=BAcos90O=0
ϕ=B.A=BAcos90O=0
A conducting loop of radius R is present in a uniform magnetic field B perpendicular the plane of the ring. If radius R varies as a function of time `t', as R = R0 + t. The e.m.f induced in the loop is
a)2p (R0 + t) B clockwiseb)p(R0 + t)B clockwisec)2p(R0 + t)B anticlockwised)zeroCorrect answer is option 'C'. Can you explain this answer?
|
Lavanya Menon answered |
ε=− dϕ/dt
=−β(dA/dt)
=−β(dA/dt)
⇒∣ε∣=βπd(R0+t)2
=2πβ(R0+t)
Since inward flux is increasing the EMF will be induced counter clockwise.
=2πβ(R0+t)
Since inward flux is increasing the EMF will be induced counter clockwise.
Two infinitely long conducting parallel rails are connected through a capacitor C as shown in the figure. A conductor of length l is moved with constant speed v0. Which of the following graph truly depicts the variation of current through the conductor with time ?
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Two infinitely long conducting parallel rails are connected through a capacitor C as shown in the figure. A conductor of length l is moved with constant speed v0. Which of the following graph truly depicts the variation of current through the conductor with time ?
a)
b)
c)
d)
|
|
Anaya Patel answered |
By Faraday's Law of induction,
ε=− dϕ/dt
=−Bl (dx/dt) =−Blv0
This emf should induce the movement of charges creating a current. But due to the attached capacitor, all charges are conserved.
Thus I= dq/dt =0
The correct option is C.
ε=− dϕ/dt
=−Bl (dx/dt) =−Blv0
This emf should induce the movement of charges creating a current. But due to the attached capacitor, all charges are conserved.
Thus I= dq/dt =0
The correct option is C.
Whenever the magnetic flux (
) linked with a coil changes the emf is induced in the circuit, This emf lasts- a)as long as the change in flux continues.
- b)for a long time
- c)for a short time
- d)forever
Correct answer is option 'A'. Can you explain this answer?
Whenever the magnetic flux () linked with a coil changes the emf is induced in the circuit, This emf lasts
) linked with a coil changes the emf is induced in the circuit, This emf lastsa)
as long as the change in flux continues.
b)
for a long time
c)
for a short time
d)
forever
|
Idamangala Merlin answered |
Answer a it is faraday 1st law
A magnet is moved towards the coil (a) quickly and (b) slowly, and then the work done is- a)does not depend on the motion of the magnet.
- b)smaller in case (a)
- c)equal in both cases
- d)larger in case (a)
Correct answer is option 'D'. Can you explain this answer?
A magnet is moved towards the coil (a) quickly and (b) slowly, and then the work done is
a)
does not depend on the motion of the magnet.
b)
smaller in case (a)
c)
equal in both cases
d)
larger in case (a)
|
Jayant Mishra answered |
A magnet is moved towards the coil (a) quickly and (b) slowly, and then the work done is This is because when the magnet is moved quickly, opposing emf induced in the coil will be more.
Two identical conductors P and Q are placed on two frictionless fixed conducting rails R and S in a uniform magnetic field directed into the plane. If P is moved in the direction shown in figure with a constant speed, then rod Q
- a)will be attracted towards P
- b)will be repelled away from P
- c)will remain stationary
- d)may be repelled or attracted towards P
Correct answer is option 'A'. Can you explain this answer?
Two identical conductors P and Q are placed on two frictionless fixed conducting rails R and S in a uniform magnetic field directed into the plane. If P is moved in the direction shown in figure with a constant speed, then rod Q
a)
will be attracted towards P
b)
will be repelled away from P
c)
will remain stationary
d)
may be repelled or attracted towards P
|
Neha Sharma answered |
As the conductor P moves away from Q, the area of the loop enclosed by the conductors and the rails increases. This in turn increases the flux through the loop.
EMF will be induced in such a way that the change in flux will be resisted. The induced current will cause Q to move towards P thereby reducing the area and thus the flux back.
EMF will be induced in such a way that the change in flux will be resisted. The induced current will cause Q to move towards P thereby reducing the area and thus the flux back.
A small conducting rod of length l, moves with a uniform velocity v in a uniform magnetic field B as shown in fig-
- a)Then the end X of the rod becomes positively charged
- b)the end Y of the rod becomes positively charged
- c)the entire rod is unevely charged
- d)the rod becomes hot due to joule heating
Correct answer is option 'B'. Can you explain this answer?
A small conducting rod of length l, moves with a uniform velocity v in a uniform magnetic field B as shown in fig-
a)
Then the end X of the rod becomes positively charged
b)
the end Y of the rod becomes positively charged
c)
the entire rod is unevely charged
d)
the rod becomes hot due to joule heating
|
Dr Manju Sen answered |
The rod is moving towards the right in a field directed into the page.
Now, if we apply Fleming's right hand rule, then the direction of induced current will be from end X to end Y.
But, according to Lenz's law the emf induced in the rod will be such that it opposes the motion of the rod.
Hence, the actual emf induced will be from end Y to end X. So, the current will also flow from end Y to end X.
Now, using the convention of current end Y should be positive and end X should be negative.
So, correct answer is option b
Now, if we apply Fleming's right hand rule, then the direction of induced current will be from end X to end Y.
But, according to Lenz's law the emf induced in the rod will be such that it opposes the motion of the rod.
Hence, the actual emf induced will be from end Y to end X. So, the current will also flow from end Y to end X.
Now, using the convention of current end Y should be positive and end X should be negative.
So, correct answer is option b
The instantaneous flux associated with a closed circuit of 10Ω resistance is indicated by the following reaction f = 6t2 _ 5t + 1, then the value in amperes of the induced current at t = 0.25 sec will be- a)1.2
- b)0.8
- c)6
- d)0.2
Correct answer is option 'D'. Can you explain this answer?
The instantaneous flux associated with a closed circuit of 10Ω resistance is indicated by the following reaction f = 6t2 _ 5t + 1, then the value in amperes of the induced current at t = 0.25 sec will be
a)
1.2
b)
0.8
c)
6
d)
0.2
|
Vivek Rana answered |
ϕ=6t2−5t+1
e= dϕ/dt =12t−5
at t=0.25sec
∣e∣=(12/4−5)=∣3−5∣=2
i=e/R=2/10=0.2A
e= dϕ/dt =12t−5
at t=0.25sec
∣e∣=(12/4−5)=∣3−5∣=2
i=e/R=2/10=0.2A
A coil of area 10 cm² of 20 turns is placed in uniform magnetic field of 104 gauss. The normal to the plane of the coil makes an angle of 60° with the magnetic field. The flux in maxwell through the coil is
- a)105
- b)106
- c)103
- d)104
Correct answer is option 'B'. Can you explain this answer?
A coil of area 10 cm² of 20 turns is placed in uniform magnetic field of 104 gauss. The normal to the plane of the coil makes an angle of 60° with the magnetic field. The flux in maxwell through the coil is
a)
105
b)
106
c)
103
d)
104
|
Hansa Sharma answered |
The number of turns in a long solenoid is 500. The area of cross-section of solenoid is 2 × 10-3 m2. If the value of magnetic induction, on passing a current of 2 amp, through it is 5 × 10-3 tesla, the magnitude of magnetic flux connected with it in webers will be- a)5 × 10-3
- b)10-2
- c)10-5
- d)2.5
Correct answer is option 'A'. Can you explain this answer?
The number of turns in a long solenoid is 500. The area of cross-section of solenoid is 2 × 10-3 m2. If the value of magnetic induction, on passing a current of 2 amp, through it is 5 × 10-3 tesla, the magnitude of magnetic flux connected with it in webers will be
a)
5 × 10-3
b)
10-2
c)
10-5
d)
2.5
|
Preeti Iyer answered |
Φ=BAN
=5x10-3x2x10-3x500
=5x10-3x1000x10-3
=5x10-3
=5x10-3x2x10-3x500
=5x10-3x1000x10-3
=5x10-3
An air-cored solenoid with length 30 cm, area of cross-section 25 cm2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10−3s. Average back emf induced across the ends of the open switch in the circuit is- a)6.5 V .
- b)4.6 V
- c)4.5 V
- d)5 V3.
Correct answer is option 'A'. Can you explain this answer?
An air-cored solenoid with length 30 cm, area of cross-section 25 cm2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10−3s. Average back emf induced across the ends of the open switch in the circuit is
a)
6.5 V .
b)
4.6 V
c)
4.5 V
d)
5 V3.
|
Arka Das answered |
-4 seconds. Find the average induced emf in the solenoid during this time.
We can use Faraday's law of induction to find the average induced emf in the solenoid:
emf = -N (ΔΦ/Δt)
where N is the number of turns, ΔΦ is the change in magnetic flux, and Δt is the time interval.
Since the current is suddenly switched off, the magnetic flux through the solenoid changes from its maximum value to zero. The maximum value of magnetic flux through the solenoid is given by:
Φ = B A
where B is the magnetic field strength and A is the area of cross-section. Since the solenoid is air-cored, the magnetic field is given by:
B = μ0 N I / L
where μ0 is the permeability of free space, I is the current, and L is the length of the solenoid. Substituting the given values, we get:
B = (4π × 10^-7) × 500 × 2.5 / 0.3 = 1.05 T
Therefore, the maximum magnetic flux through the solenoid is:
Φ = 1.05 × 25 × 10^-4 = 2.625 × 10^-5 Wb
During the brief time of 10^-4 seconds, the change in magnetic flux is equal to the maximum flux, since the current is switched off suddenly. Therefore, ΔΦ = 2.625 × 10^-5 Wb.
Substituting the given values in the formula for emf, we get:
emf = -500 (2.625 × 10^-5) / (10^-4) = -1.3125 V
Therefore, the average induced emf in the solenoid during the brief time of 10^-4 seconds is 1.3125 V. Note that the negative sign indicates that the induced emf opposes the change in current.
We can use Faraday's law of induction to find the average induced emf in the solenoid:
emf = -N (ΔΦ/Δt)
where N is the number of turns, ΔΦ is the change in magnetic flux, and Δt is the time interval.
Since the current is suddenly switched off, the magnetic flux through the solenoid changes from its maximum value to zero. The maximum value of magnetic flux through the solenoid is given by:
Φ = B A
where B is the magnetic field strength and A is the area of cross-section. Since the solenoid is air-cored, the magnetic field is given by:
B = μ0 N I / L
where μ0 is the permeability of free space, I is the current, and L is the length of the solenoid. Substituting the given values, we get:
B = (4π × 10^-7) × 500 × 2.5 / 0.3 = 1.05 T
Therefore, the maximum magnetic flux through the solenoid is:
Φ = 1.05 × 25 × 10^-4 = 2.625 × 10^-5 Wb
During the brief time of 10^-4 seconds, the change in magnetic flux is equal to the maximum flux, since the current is switched off suddenly. Therefore, ΔΦ = 2.625 × 10^-5 Wb.
Substituting the given values in the formula for emf, we get:
emf = -500 (2.625 × 10^-5) / (10^-4) = -1.3125 V
Therefore, the average induced emf in the solenoid during the brief time of 10^-4 seconds is 1.3125 V. Note that the negative sign indicates that the induced emf opposes the change in current.
The dimensions of permeability of free space can be given by- a)[MLT-2A-2]
- b)[MLA-2]
- c)[ML-3T2A2]
- d)[MLA-1]
Correct answer is option 'A'. Can you explain this answer?
The dimensions of permeability of free space can be given by
a)
[MLT-2A-2]
b)
[MLA-2]
c)
[ML-3T2A2]
d)
[MLA-1]
|
|
Lavanya Menon answered |
In SI units, permeability is measured in Henries per meter H/m or Hm−1.
Henry has the dimensions of [ML2T−2A−2].
Dimensions for magnetic permeability will be [ML2T−2A−2]/[L]=[MLT−2A−2]
Henry has the dimensions of [ML2T−2A−2].
Dimensions for magnetic permeability will be [ML2T−2A−2]/[L]=[MLT−2A−2]
In an iron cored coil the iron core is removed so that the coil becomes an air cored coil. The inductance of the coil will- a)increase
- b)decrease
- c)remain the same
- d)initially increase and then decrease
Correct answer is option 'B'. Can you explain this answer?
In an iron cored coil the iron core is removed so that the coil becomes an air cored coil. The inductance of the coil will
a)
increase
b)
decrease
c)
remain the same
d)
initially increase and then decrease
|
|
Nandini Iyer answered |
In an iron cored coil the iron core is removed so that the coil becomes an air cored coil. The inductance of the coil will decrease.
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s-1 in a uniform horizontal magnetic field of magnitude 3×10-2 T. If the coil resistance is 10Ω, maximum emf induced in the coil,maximum value of current in the coil and average power loss due to Joule heating are- a)0.603 V, 0.0303 A,0.018 W
- b)0.603 V, 0.0603 A,0.038 W
- c)0.603 V, 0.0603 A,0.018 W .
- d)0.303 V, 0.0603 A,0.018 W
Correct answer is option 'C'. Can you explain this answer?
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s-1 in a uniform horizontal magnetic field of magnitude 3×10-2 T. If the coil resistance is 10Ω, maximum emf induced in the coil,maximum value of current in the coil and average power loss due to Joule heating are
a)
0.603 V, 0.0303 A,0.018 W
b)
0.603 V, 0.0603 A,0.038 W
c)
0.603 V, 0.0603 A,0.018 W .
d)
0.303 V, 0.0603 A,0.018 W
|
Bhavana Banerjee answered |
e = NBAω = 20 x 3 x 10-2 x (3.14 x 0.08 x 0.08) x 50 = 0.603C
average power loss due to joule heating
Consider the situation shown in fig. The resistanceless wire AB is slid on the fixed rails with a constant velocity. If the wire AB is replaced by a resistanceless semicircular wire, the magnitude of the induced current will
- a)increase
- b)remain the same
- c)decrease
- d)increase or decrease depending on whether the semicircle bulges towards the resistance or away from it
Correct answer is option 'B'. Can you explain this answer?
Consider the situation shown in fig. The resistanceless wire AB is slid on the fixed rails with a constant velocity. If the wire AB is replaced by a resistanceless semicircular wire, the magnitude of the induced current will
a)
increase
b)
remain the same
c)
decrease
d)
increase or decrease depending on whether the semicircle bulges towards the resistance or away from it
|
Om Rana answered |
As we know I = B L V. But L would be a vector quantity here(If you have doubt why L would be vector quantity, see the derivation of induced current). So we have to take L effective. But L effective here will be diameter of the semicircle as it is perpendicular to both B and V vectors. As the diameter of the circle is same as the length of the wire we used first So there will be no change in the equation of induced current. So it remains same.
In the given circuit the maximum deflection in the galvanometer occurs when
- a)magnet is rotated inside the coil
- b)magnet is stationary at the center of the coil.
- c)numbers of turns in the coil is reduced.
- d)magnet is pushed into the coil.
Correct answer is option 'D'. Can you explain this answer?
In the given circuit the maximum deflection in the galvanometer occurs when
a)
magnet is rotated inside the coil
b)
magnet is stationary at the center of the coil.
c)
numbers of turns in the coil is reduced.
d)
magnet is pushed into the coil.
|
sankalp raj answered |
This question based on lenz law. as the bar magnet is moving , there is a change in magnetic flux.This will give polarity in the coil. Hence the current developed in the coil. so the option d is right.
A constant current I is maintained in a solenoid. Which of the following quantities will not increase if an iron rod is inserted in the solenoid along its axis?- a)Magnetic field at the centre.
- b)Self-inductance of the solenoid.
- c)Magnetic flux linked with the solenoid.
- d)Rate of joule heating.
Correct answer is option 'D'. Can you explain this answer?
A constant current I is maintained in a solenoid. Which of the following quantities will not increase if an iron rod is inserted in the solenoid along its axis?
a)
Magnetic field at the centre.
b)
Self-inductance of the solenoid.
c)
Magnetic flux linked with the solenoid.
d)
Rate of joule heating.
|
Fahad Chauhan answered |
Magnetic field is directly proportional to permiability and when iron rod is inserted the relative permeability changes and thus magnetic field , self inductance is proportional to flux which is proportional to magnetic field and thus follow above explanation, and remaining last option rate of heating =I²R and both these factors doesn't get affected by iron rod , and hence this is correct option.
electromagnetic induction i.e currents can be induced in coils (Select the best)- a)Only if the coil moves
- b)Only if the coil moves and magnet also moves in the same direction
- c)if relative motion of coil and magnet is present.
- d)Only if the magnet moves
Correct answer is option 'C'. Can you explain this answer?
electromagnetic induction i.e currents can be induced in coils (Select the best)
a)
Only if the coil moves
b)
Only if the coil moves and magnet also moves in the same direction
c)
if relative motion of coil and magnet is present.
d)
Only if the magnet moves
|
Ezhil The Champ answered |
Consider a cylindrical copper coil connected serially to a galvanometer . a strong magnet with north or south pole is taken towards it. and coil is moved up and down.
when ever there is a relative motion between coil and the magnet the galvanometer shows deflection . indicating flow of induced current.
the deflection is momentary . it last so long as there is relative motion between coil and magnet.
the direction of induced current changes if magnet or coil is moved towards or away frm it
the deflection is more when the relative motion is faster or less when it is slow.
t
when ever there is a relative motion between coil and the magnet the galvanometer shows deflection . indicating flow of induced current.
the deflection is momentary . it last so long as there is relative motion between coil and magnet.
the direction of induced current changes if magnet or coil is moved towards or away frm it
the deflection is more when the relative motion is faster or less when it is slow.
t
A thin wire of length 2m is perpendicular to the xy plane. It is moved with velocity 
through a region of magnetic induction 
Wb/m2. Then potential difference induced between the ends of the wire- a)2 volts
- b)4 volts
- c) 0 volts
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
A thin wire of length 2m is perpendicular to the xy plane. It is moved with velocity through a region of magnetic induction Wb/m2. Then potential difference induced between the ends of the wire
through a region of magnetic induction Wb/m2. Then potential difference induced between the ends of the wirea)
2 volts
b)
4 volts
c)
0 volts
d)
None of these
|
|
Nikita Singh answered |
The magnetic flux through a stationary loop with resistance R varies during interval of time T as f = at (T _ t). The heat generated during this time neglecting the inductance of loop will be- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The magnetic flux through a stationary loop with resistance R varies during interval of time T as f = at (T _ t). The heat generated during this time neglecting the inductance of loop will be
a)
b)
c)
d)
|
|
Hansa Sharma answered |
E=dx/dt =d(a(Tt-t2))/dt=a(T-2t)
I=E/R=a(T-2t)/R
Heat liberated,
△k=I2Rdt
=(a2 /R)[T2t+(4t3/3)-2Tt2]T0
=△H=a2T3/3R
I=E/R=a(T-2t)/R
Heat liberated,
△k=I2Rdt
=(a2 /R)[T2t+(4t3/3)-2Tt2]T0
=△H=a2T3/3R
The north pole of a magnet is brought near a coil. The induced current in the coil as seen by an observer on the side of magnet will be- a)in the clockwise direction
- b)in the anticlockwise direction
- c)initially in the clockwise and then anticlockwise direction
- d)initially in the anticlockwise and then clockwise direction
Correct answer is option 'B'. Can you explain this answer?
The north pole of a magnet is brought near a coil. The induced current in the coil as seen by an observer on the side of magnet will be
a)
in the clockwise direction
b)
in the anticlockwise direction
c)
initially in the clockwise and then anticlockwise direction
d)
initially in the anticlockwise and then clockwise direction
|
|
Geetika Shah answered |
The direction of the current will be anticlockwise.
According to Lenz's law, the current in the coil will be induced in the direction that'll oppose the external magnetic field .
As the flux due to the external magnet is increasing ( as the N-pole is brought close ) , the coil will have to induce current in the anticlockwise direction to oppose this increase in flux and not in clockwise direction as that'll end up supporting the external flux.
According to Lenz's law, the current in the coil will be induced in the direction that'll oppose the external magnetic field .
As the flux due to the external magnet is increasing ( as the N-pole is brought close ) , the coil will have to induce current in the anticlockwise direction to oppose this increase in flux and not in clockwise direction as that'll end up supporting the external flux.
The total number of magnetic lines of force crossing a surface normally is termed as- a)Magnetic field.
- b)Magnetic permeability.
- c)Magnetic flux.
- d)Magnetic susceptibility.
Correct answer is option 'C'. Can you explain this answer?
The total number of magnetic lines of force crossing a surface normally is termed as
a)
Magnetic field.
b)
Magnetic permeability.
c)
Magnetic flux.
d)
Magnetic susceptibility.
|
Nabanita Pillai answered |
The measurement of the total magnetic field that passes through a given area is known as magnetic flux. It is helpful in describing the effects of the magnetic force on something occupying a given area.
If we consider a simple flat area A as our example and angle θ as the angle between the normal to the surface and a magnetic field vector, then the magnetic flux is given by the equation:
ϕ=BAcosΘ
If we consider a simple flat area A as our example and angle θ as the angle between the normal to the surface and a magnetic field vector, then the magnetic flux is given by the equation:
ϕ=BAcosΘ
Two identical coaxial circular loops carry a current i each circulating in the same direction. If the loops approach each other- a)the current in each will decrease
- b)the current in each will increase
- c)the current in each will remain the same
- d)the current in one will increase and in other will decrease
Correct answer is option 'A'. Can you explain this answer?
Two identical coaxial circular loops carry a current i each circulating in the same direction. If the loops approach each other
a)
the current in each will decrease
b)
the current in each will increase
c)
the current in each will remain the same
d)
the current in one will increase and in other will decrease
|
Tejas Desai answered |
Ans.
Option (a)
As coils approach each other, the flux linked with each coil increases. A current will be induced in each coil which will try to decrease the flux. This implies induced current in each coil will be opposite to initial current. So, current in each coil decreases as the coils approach each other.
When a magnet is being moved towards a coil, the induced emf does not depend upon- a)the number of turns of the coil
- b)the motion of the magnet
- c)the magnetic moment of the magnet
- d)the resistance of the coil
Correct answer is option 'D'. Can you explain this answer?
When a magnet is being moved towards a coil, the induced emf does not depend upon
a)
the number of turns of the coil
b)
the motion of the magnet
c)
the magnetic moment of the magnet
d)
the resistance of the coil
|
Chirag Joshi answered |
The magnitude of induced emf is independent of the resistance of the coil.
∴e=−dϕ/dt=−N (d/dt) [(BA)/dt]
Thus, e depends upon N, B and A.
But induced current depends upon R.
∴e=−dϕ/dt=−N (d/dt) [(BA)/dt]
Thus, e depends upon N, B and A.
But induced current depends upon R.
When a bar magnet is pushed towards the coil connected in series with a galvanometer, the pointer in the galvanometer G- a)shows deflection while bar is stationary
- b)oscillates
- c)shows deflection while bar is in motion
- d)does not move
Correct answer is option 'C'. Can you explain this answer?
When a bar magnet is pushed towards the coil connected in series with a galvanometer, the pointer in the galvanometer G
a)
shows deflection while bar is stationary
b)
oscillates
c)
shows deflection while bar is in motion
d)
does not move
|
Samridhi Bajaj answered |
Explanation:When bar is in motion then flux will changed hence current will be induced.
There are two stationary coils near each other. If current in coil-1 is changing with time, relationship of current in coil-2 to it's emf is described by- a)Alternating Inductance
- b)Self Inductance
- c)Variable Inductance
- d)Mutual Inductance.
Correct answer is option 'D'. Can you explain this answer?
There are two stationary coils near each other. If current in coil-1 is changing with time, relationship of current in coil-2 to it's emf is described by
a)
Alternating Inductance
b)
Self Inductance
c)
Variable Inductance
d)
Mutual Inductance.
|
Rishika Chauhan answered |
M is mutual inductance
Eddy Currents are- a)plasma currents
- b)induced solar currents that circulate throughout a river and are swirling .
- c)gaseous currents
- d)induced currents that circulate throughout the volume of a material similar to swirling eddies in a river.
Correct answer is option 'D'. Can you explain this answer?
Eddy Currents are
a)
plasma currents
b)
induced solar currents that circulate throughout a river and are swirling .
c)
gaseous currents
d)
induced currents that circulate throughout the volume of a material similar to swirling eddies in a river.
|
Raghav Yadav answered |
Explanation:Eddy currents (also called Foucault currents) are loops of electrical current induced within conductors by a changing magnetic field in the conductor, due to Faraday's law of induction. Eddy currents flow in closed loops within conductors, in planes perpendicular to the magnetic field. By lenz's law an eddy current creates a magnetic field that opposes a change in the magnetic field that created it, and thus eddy currents react back on the source of the magnetic field.
Kamla peddles a stationary bicycle the pedals of the bicycle are attached to a 100 turn coil of area 0.10m2. The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 0.02 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil?- a)0.628
- b)0.314
- c)0.714
- d)0.554
Correct answer is option 'A'. Can you explain this answer?
Kamla peddles a stationary bicycle the pedals of the bicycle are attached to a 100 turn coil of area 0.10m2. The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 0.02 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil?
a)
0.628
b)
0.314
c)
0.714
d)
0.554
|
Ishani Patel answered |
Maximum voltage
An AC generator consists of- a)A coil that can rotate in a permanent magnetic field and is provided with slip rings.
- b)A permanent magnet rotor and a stationary coil.
- c)Two stationary coils close to each other
- d)A stationary coil for emf and a movable coil fed with AC
Correct answer is option 'A'. Can you explain this answer?
An AC generator consists of
a)
A coil that can rotate in a permanent magnetic field and is provided with slip rings.
b)
A permanent magnet rotor and a stationary coil.
c)
Two stationary coils close to each other
d)
A stationary coil for emf and a movable coil fed with AC
|
Arshiya Choudhury answered |
Explanation:AC generators operate on the same fundamental principles of electromagnetic induction.The turning of a coil in a magnetic field produces motional emfs in both sides of the coil.A slip ring is a method of making an electrical connection through a rotating assembly. Formally, it is an electric transmission device that allows energy flow between two electrical rotating parts, such as in a motor.
The magnetic field between the Horizontal poles of an electromagnet is uniform at any time, but its magnitude is increasing at the rate of 0.020T/s.The area of a horizontal conducting loop in the magnetic field is 120cm2, and the total circuit resistance, including the meter, is 5Ω. Induced emf and the induced current in the circuit are- a)0.18 mV,0.048 mA
- b)0.22 mV,0.048 mA
- c)0.20 mV,0.048 mA
- d)0.24 mV,0.048 mA
Correct answer is option 'D'. Can you explain this answer?
The magnetic field between the Horizontal poles of an electromagnet is uniform at any time, but its magnitude is increasing at the rate of 0.020T/s.The area of a horizontal conducting loop in the magnetic field is 120cm2, and the total circuit resistance, including the meter, is 5Ω. Induced emf and the induced current in the circuit are
a)
0.18 mV,0.048 mA
b)
0.22 mV,0.048 mA
c)
0.20 mV,0.048 mA
d)
0.24 mV,0.048 mA
|
|
Anand Saha answered |
Magnetic Field
- The magnetic field between the horizontal poles of an electromagnet is uniform at any time.
- The magnitude of the magnetic field is increasing at a rate of 0.020 T/s.
Conducting Loop
- The area of the horizontal conducting loop in the magnetic field is 120 cm^2.
- The total circuit resistance, including the meter, is 5 Ω.
Induced EMF
- The induced EMF can be calculated using Faraday's law of electromagnetic induction.
- Faraday's law states that the induced EMF is equal to the rate of change of magnetic flux through the loop.
- The magnetic flux through the loop can be calculated by multiplying the magnetic field strength by the area of the loop.
- The rate of change of magnetic flux is equal to the rate of change of the magnetic field strength multiplied by the area of the loop.
- Substituting the given values, the rate of change of magnetic flux is (0.020 T/s) * (120 cm^2) = 2.4 T cm^2/s.
- To convert the magnetic flux to Weber (Wb), divide by 10,000: 2.4 T cm^2/s / 10,000 = 0.00024 Wb/s.
- Therefore, the induced EMF is 0.00024 V/s or 0.24 mV/s.
Induced Current
- The induced current can be calculated using Ohm's law, which states that the current is equal to the voltage divided by the resistance.
- Substituting the given values, the induced current is (0.24 mV/s) / (5 Ω) = 0.048 mA.
Conclusion
- The induced EMF in the circuit is 0.24 mV and the induced current is 0.048 mA.
- Therefore, the correct answer is option 'D', 0.24 mV, 0.048 mA.
- The magnetic field between the horizontal poles of an electromagnet is uniform at any time.
- The magnitude of the magnetic field is increasing at a rate of 0.020 T/s.
Conducting Loop
- The area of the horizontal conducting loop in the magnetic field is 120 cm^2.
- The total circuit resistance, including the meter, is 5 Ω.
Induced EMF
- The induced EMF can be calculated using Faraday's law of electromagnetic induction.
- Faraday's law states that the induced EMF is equal to the rate of change of magnetic flux through the loop.
- The magnetic flux through the loop can be calculated by multiplying the magnetic field strength by the area of the loop.
- The rate of change of magnetic flux is equal to the rate of change of the magnetic field strength multiplied by the area of the loop.
- Substituting the given values, the rate of change of magnetic flux is (0.020 T/s) * (120 cm^2) = 2.4 T cm^2/s.
- To convert the magnetic flux to Weber (Wb), divide by 10,000: 2.4 T cm^2/s / 10,000 = 0.00024 Wb/s.
- Therefore, the induced EMF is 0.00024 V/s or 0.24 mV/s.
Induced Current
- The induced current can be calculated using Ohm's law, which states that the current is equal to the voltage divided by the resistance.
- Substituting the given values, the induced current is (0.24 mV/s) / (5 Ω) = 0.048 mA.
Conclusion
- The induced EMF in the circuit is 0.24 mV and the induced current is 0.048 mA.
- Therefore, the correct answer is option 'D', 0.24 mV, 0.048 mA.
A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s−1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Emf developed between the centre and the ring is- a)120 V
- b)100 V .
- c)200 V
- d)150 V
Correct answer is option 'B'. Can you explain this answer?
A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s−1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Emf developed between the centre and the ring is
a)
120 V
b)
100 V .
c)
200 V
d)
150 V
|
Nishtha Bose answered |
On a cylindrical rod two coils are wound one above the other. What is the coefficient of mutual inductance if the inductance of each coil os 0.1H?- a)0.15H
- b)0.05H
- c)0.20H
- d)0.10H
Correct answer is option 'D'. Can you explain this answer?
On a cylindrical rod two coils are wound one above the other. What is the coefficient of mutual inductance if the inductance of each coil os 0.1H?
a)
0.15H
b)
0.05H
c)
0.20H
d)
0.10H
|
Rashi Bose answered |
As one coil is wound over the other so that coupling is tight i.e. k=1
Can you explain the answer of this question below:Predict the polarity of the capacitor in the situation described by fig.
- A:B’ will be positive with respect to plate ‘A’
- B:A’ will be positive with respect to plate ‘B’ .
- C:Not enough information
- D:No current induced
The answer is b.
Predict the polarity of the capacitor in the situation described by fig.
A:
B’ will be positive with respect to plate ‘A’
B:
A’ will be positive with respect to plate ‘B’ .
C:
Not enough information
D:
No current induced
|
|
Advait Ghosh answered |
Explanation:A will become positive with respect to B because current indused is in clockwise direction
A closed planar wire loop of area A and arbitrary shape is placed in a uniform magnetic field of magnitude B, with its plane perpendicular to magnetic field. The resistance of the wire loop is R. The loop is now turned upside down by 180° so that its plane again becomes perpendicular to the magnetic field. The total charge that must have flowen through the wire ring in the process is- a)< AB/R
- b)= AB/R
- c)= 2AB/R
- d)None
Correct answer is option 'C'. Can you explain this answer?
A closed planar wire loop of area A and arbitrary shape is placed in a uniform magnetic field of magnitude B, with its plane perpendicular to magnetic field. The resistance of the wire loop is R. The loop is now turned upside down by 180° so that its plane again becomes perpendicular to the magnetic field. The total charge that must have flowen through the wire ring in the process is
a)
< AB/R
b)
= AB/R
c)
= 2AB/R
d)
None
|
Rajesh Chatterjee answered |
Induced emf= d(flux)/dt=R[dq/dt]
⇒Δq= Δ(flux)/R=2BA/R
⇒Δq= Δ(flux)/R=2BA/R
A cylindrical bar magnet is held along the axis of a circular conducting loop. Which one of the following movements of the bar magnet will not induce a current in the loop?- a)The magnet is rotated about its own axis.
- b)The magnet is moved along its length.
- c)The magnet is rotated about an axis passing through its mid point and perpendicular to its length.
- d)The magnet is rotated about an axis through one end and perpendicular to its length.
Correct answer is option 'A'. Can you explain this answer?
A cylindrical bar magnet is held along the axis of a circular conducting loop. Which one of the following movements of the bar magnet will not induce a current in the loop?
a)
The magnet is rotated about its own axis.
b)
The magnet is moved along its length.
c)
The magnet is rotated about an axis passing through its mid point and perpendicular to its length.
d)
The magnet is rotated about an axis through one end and perpendicular to its length.
|
Anshika Menon answered |
A cylindrical bar magnet is kept along the axis of a circular coil. If the magnet is rotated about its axis, then there is no change in magnetic flux, there is no emf induced in the coil, no current will be induced in the coil.
) linked with a coil is related to the number (N) of turns of the coil as
A metal sheet is placed in a variable magnetic field which is increasing from zero to maximum. Induced current flows in the directions as shown in figure. The direction of magnetic field will be -
- a)normal to the paper, inwards
- b)normal to the paper, outwards.
- c)from east to west
- d)from north to south
Correct answer is option 'B'. Can you explain this answer?
A metal sheet is placed in a variable magnetic field which is increasing from zero to maximum. Induced current flows in the directions as shown in figure. The direction of magnetic field will be -
a)
normal to the paper, inwards
b)
normal to the paper, outwards.
c)
from east to west
d)
from north to south
|
|
Nandini Iyer answered |
The direction of these circular magnetic lines is dependent upon the direction of current. The density of the induced magnetic field is directly proportional to the magnitude of the current. Direction of the circular magnetic field lines can be given by Maxwell's right hand grip rule or Right handed cork screw rule.
A field of 5 x 104/p ampere-turns/metre acts at right angles to a coil of 50 turns of area 10-2 m2. The coil is removed from the field in 0.1 second. Then the induced emf in the coil is- a)0.1 V
- b)80 KV
- c)7.96 V
- d)none of the above
Correct answer is option 'A'. Can you explain this answer?
A field of 5 x 104/p ampere-turns/metre acts at right angles to a coil of 50 turns of area 10-2 m2. The coil is removed from the field in 0.1 second. Then the induced emf in the coil is
a)
0.1 V
b)
80 KV
c)
7.96 V
d)
none of the above
|
Partho Singh answered |
A metallic rod of 1 m length is rotated with a frequency of 50 rev/s about an axis passing through the centre point O. Other end of the metallic rod slides on a Metallic ring . A constant and uniform magnetic field of 2 T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring?
- a)220 V
- b)314 V .
- c)240 V
- d)300 V
Correct answer is option 'B'. Can you explain this answer?
A metallic rod of 1 m length is rotated with a frequency of 50 rev/s about an axis passing through the centre point O. Other end of the metallic rod slides on a Metallic ring . A constant and uniform magnetic field of 2 T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring?
a)
220 V
b)
314 V .
c)
240 V
d)
300 V
|
|
Nisha Kulkarni answered |
l = 1 m
= 314V
The magnetic field between the Horizontal poles of an electromagnet in is uniform at any time, but its magnitude is increasing at the rate of 0.020 T/s.The area of a loop made of perfect insulating material is 120cm2.This loop is kept horizontally in the magnetic field. Induced emf and the induced current in the circuit are- a)0.54 mV,zero
- b)0.84 mV,zero
- c)0.24 mV,zero
- d)0.14 mV,zero
Correct answer is option 'C'. Can you explain this answer?
The magnetic field between the Horizontal poles of an electromagnet in is uniform at any time, but its magnitude is increasing at the rate of 0.020 T/s.The area of a loop made of perfect insulating material is 120cm2.This loop is kept horizontally in the magnetic field. Induced emf and the induced current in the circuit are
a)
0.54 mV,zero
b)
0.84 mV,zero
c)
0.24 mV,zero
d)
0.14 mV,zero
|
|
Samridhi Bajaj answered |
e = Rate of change of magnetic field x area
=0.02×120×10−4=0.24mV
Damping in galvanometers is based oUpdate Questionn- a)eddy currents generating currents in the core opposing the motion and bringing the coil to rest quickly.
- b)dash pot opposing the motion of the galvanometer needle
- c)needle movement being opposed by rubber touching the needle lightly
- d)needle movement being opposed by oil
Correct answer is option 'A'. Can you explain this answer?
Damping in galvanometers is based oUpdate Questionn
a)
eddy currents generating currents in the core opposing the motion and bringing the coil to rest quickly.
b)
dash pot opposing the motion of the galvanometer needle
c)
needle movement being opposed by rubber touching the needle lightly
d)
needle movement being opposed by oil
|
|
Nisha Pillai answered |
Explanation : The coil of galvanometer is wound on a light metal frame. When the coil and frame rotate in the field of the permanent magnet, the eddy current set up in the frame oppose the motion so that the coil returns to zero quickly.
If the polarity of induced emf is such that it tends to produce a current which aids the change in magnetic flux that produced it, which conservation law is violated?- a)conservation of momentum
- b)conservation of charge
- c)conservation of energy.
- d)conservation of mass
Correct answer is option 'C'. Can you explain this answer?
If the polarity of induced emf is such that it tends to produce a current which aids the change in magnetic flux that produced it, which conservation law is violated?
a)
conservation of momentum
b)
conservation of charge
c)
conservation of energy.
d)
conservation of mass
|
Gauri Kaur answered |
Explanation : In this case we would be obtaining electrical energy continuously without doing any work.
Chapter doubts & questions for Electromagnetic Induction - 4 Months Preparation for NEET 2025 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 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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