All questions of Inducing Currents for Grade 9 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 Weber in closed circuit of resistance 15 ohm varies with time t sec as
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 circular coil of area 200 cm2 and 25 turns rotates about its vertical diameter with an angular speed of 20 ms-1 in a uniform horizontal magnetic field of magnitude 0.05 T. The maximum voltage induced in the coil is- a)0.5 V
- b)1.5 V
- c)2.5 V
- d)2.0 V
Correct answer is option 'A'. Can you explain this answer?
A circular coil of area 200 cm2 and 25 turns rotates about its vertical diameter with an angular speed of 20 ms-1 in a uniform horizontal magnetic field of magnitude 0.05 T. The maximum voltage induced in the coil is
a)
0.5 V
b)
1.5 V
c)
2.5 V
d)
2.0 V
|
Hansa Sharma answered |
It is induced EMF of periodic EMI, So formula is:
E= NBAw
Here w is angular speed.
E= NBAw
Here w is angular speed.
So, E = 25×0.05×200×20/10000 =0.5V
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.
Eddy currents have negative effects. Because they produce- a)Harmful radiation
- b)Heating and damping
- c)Damping only
- d)Heating only
Correct answer is option 'B'. Can you explain this answer?
Eddy currents have negative effects. Because they produce
a)
Harmful radiation
b)
Heating and damping
c)
Damping only
d)
Heating only
|
Mamali . answered |
When a conductive material is subjected to a time-varying magnetic flux, eddy currents are generated in the conductor. These eddy currents circulate inside the con- ductor generating a magnetic field of opposite polarity as the applied magnetic field. The interaction of the two magnetic fields causes a force that resists the change in magnetic flux. However, due to the internal resistance of the conductive material, the eddy currents will be dissipated into heat and the force will die out. As the eddy currents are dissipated, energy is removed from the system, thus producing a damp- ing effect.
In the dead beat galvanometer, the coil is wound on a frame made of:- a)good conductor
- b)non magnetic material
- c)bad conductor
- d)magnetic material
Correct answer is option 'A'. Can you explain this answer?
In the dead beat galvanometer, the coil is wound on a frame made of:
a)
good conductor
b)
non magnetic material
c)
bad conductor
d)
magnetic material
|
Nikita Singh answered |
Aluminum is used as a coil which is a good conductor.
Dead beat galvanometer:
Dead beat galvanometer:
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.
Which of the following will not increase the size and effect of eddy current?- a)Low resistivity materials
- b)Strong magnetic field
- c)Thicker material
- d)Thinner material
Correct answer is option 'D'. Can you explain this answer?
Which of the following will not increase the size and effect of eddy current?
a)
Low resistivity materials
b)
Strong magnetic field
c)
Thicker material
d)
Thinner material
|
|
Hansa Sharma answered |
Stronger magnetic field, thicker material and low resistivity material will increase the size and effect of eddy current whereas thinner material will reduce the effect of eddy currents.
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
In which of the following cases with a bar magnet and the coil no induced emf is induced.- a)When there is no relative motion between the magnet and the coil.
- b)When the coil is moved towards or away from the magnet.
- c)When the magnet is taken away from the coil.
- d)When the magnet is inserted into the coil.
Correct answer is option 'A'. Can you explain this answer?
In which of the following cases with a bar magnet and the coil no induced emf is induced.
a)
When there is no relative motion between the magnet and the coil.
b)
When the coil is moved towards or away from the magnet.
c)
When the magnet is taken away from the coil.
d)
When the magnet is inserted into the coil.
|
|
Lavanya Menon answered |
This is because induced emf directly depends on relative motion between the magnet and coil, but if there is no relative motion between them, there will be no emf induced.
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.
Identify the type of commercial motor which works as a consequence of eddy currents.- a)Compressors
- b)Induction motors
- c)Turbines
- d)Hydropowered motors
Correct answer is option 'B'. Can you explain this answer?
Identify the type of commercial motor which works as a consequence of eddy currents.
a)
Compressors
b)
Induction motors
c)
Turbines
d)
Hydropowered motors
|
Varun Kapoor answered |
A rotating magnetic field is produced employing two single-phase currents. A metallic rotor placed inside the rotating magnetic field starts rotating due to large eddy currents produced in it. These motors are commonly used in fans.
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.
Eddy currents do not cause:- a)sparking
- b)heating
- c)loss of energy
- d)damping
Correct answer is option 'A'. Can you explain this answer?
Eddy currents do not cause:
a)
sparking
b)
heating
c)
loss of energy
d)
damping
|
|
Arjun Singhania answered |
During braking, the metal wheels are exposed to a magnetic field from an electromagnet, generating eddy currents in the wheels. So, by Lenz's law, the magnetic field formed by the Eddy current will oppose its cause. Thus the wheel will face a force opposing the initial movement of the wheel.
The role of inductance is equivalent to:- a)energy
- b)force
- c)inertia
- d)momentum
Correct answer is option 'C'. Can you explain this answer?
The role of inductance is equivalent to:
a)
energy
b)
force
c)
inertia
d)
momentum
|
Rahul Bansal answered |
Self induction is that phenomenon in which a change in electric current in a coil produces an induced emf in the coil itself.
Now, it is also known as inertia of electricity as for if we were to change electric current through a current carrying coil it will tend to oppose any further change in its emf. This is similar to inertial behavior in mechanics where bodies in either rest or motion tend to oppose any change in their state. Here mass is the inertial property analogous to self inductance.
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 |
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
The no-load current drawn by transformer is usually what per cent of the full-load current ?- a)0.2 to 0.5 per cent
- b)2 to 5 per cent
- c)12 to 15 per cent
- d)20 to 30 per cent
Correct answer is option 'B'. Can you explain this answer?
The no-load current drawn by transformer is usually what per cent of the full-load current ?
a)
0.2 to 0.5 per cent
b)
2 to 5 per cent
c)
12 to 15 per cent
d)
20 to 30 per cent
|
Baishali Chakraborty answered |
The no load current is about 2-5% of the full load current and it accounts for the losses in a transformer. These no-load losses include core(iron/fixed) losses, which contains eddy current losses & hysteresis losses and the copper(I2*R) losses due to the no Load current.
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
Eddy currents can be induced in- a)Metal Blocks
- b)Wires
- c)Both metal blocks and wires
- d)Insulators
Correct answer is option 'C'. Can you explain this answer?
Eddy currents can be induced in
a)
Metal Blocks
b)
Wires
c)
Both metal blocks and wires
d)
Insulators
|
Anusaya Swain answered |
Eddy currents are produced by varying magnetic field in solid plates, blocks and wires.
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.
In electromagnetic induction, line integral of induced field E around a close path is __________, induced electric field is ___________.- a)Zero, Non Conservative
- b)Non zero, Conservative
- c)Zero, Conservative
- d)Non zero, Non Conservative
Correct answer is option 'D'. Can you explain this answer?
In electromagnetic induction, line integral of induced field E around a close path is __________, induced electric field is ___________.
a)
Zero, Non Conservative
b)
Non zero, Conservative
c)
Zero, Conservative
d)
Non zero, Non Conservative
|
|
Kalyan Chavan answered |
In electromagnetic induction, line integral of induced field E around a close path is non zero, and induced electric field is non conservative.
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.
The principle of electromagnetic induction is used in which of the following phenomenon- a)Generation of electricity in hydroelectricity.
- b)Generation of electricity in nuclear power plant.
- c)Generation of electricity in thermal power plant.
- d)Generation of electricity from wind energy.
Correct answer is option 'A'. Can you explain this answer?
The principle of electromagnetic induction is used in which of the following phenomenon
a)
Generation of electricity in hydroelectricity.
b)
Generation of electricity in nuclear power plant.
c)
Generation of electricity in thermal power plant.
d)
Generation of electricity from wind energy.
|
|
Animesh Nhagal answered |
Electrical generators (such as hydroelectric dams) where mechanical power is used to move a magnetic field past coils of wire to generate voltage
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.
A coil of 100 turns is pulled in 0.04 sec. between the poles of a magnet, if its flux changes from 40 x 10-6 Wb per turn to 10-5 Wb per turn, then the average emf induced in the coil is- a)0.75V
- b)7.5V
- c)0.0075V
- d)0.075V
Correct answer is option 'D'. Can you explain this answer?
A coil of 100 turns is pulled in 0.04 sec. between the poles of a magnet, if its flux changes from 40 x 10-6 Wb per turn to 10-5 Wb per turn, then the average emf induced in the coil is
a)
0.75V
b)
7.5V
c)
0.0075V
d)
0.075V
|
Shalini Basu answered |
Given:
Number of turns, N = 100
Time taken, t = 0.04 sec
Initial flux per turn, Φ1 = 40 x 10^-6 Wb
Final flux per turn, Φ2 = 10^-5 Wb
To find: Average emf induced in the coil
Formula used:
The average emf induced in a coil is given by
E = ΔΦ/Δt
where ΔΦ is the change in magnetic flux and Δt is the time taken for the change.
Calculation:
Change in magnetic flux, ΔΦ = Φ2 - Φ1
= (10^-5 - 40 x 10^-6) Wb
= 6 x 10^-6 Wb
Time taken for the change, Δt = t/N
= 0.04/100 sec
= 4 x 10^-4 sec
Average emf induced in the coil, E = ΔΦ/Δt
= (6 x 10^-6)/(4 x 10^-4) V
= 0.075 V
Therefore, the average emf induced in the coil is 0.075 V, which is option D.
Number of turns, N = 100
Time taken, t = 0.04 sec
Initial flux per turn, Φ1 = 40 x 10^-6 Wb
Final flux per turn, Φ2 = 10^-5 Wb
To find: Average emf induced in the coil
Formula used:
The average emf induced in a coil is given by
E = ΔΦ/Δt
where ΔΦ is the change in magnetic flux and Δt is the time taken for the change.
Calculation:
Change in magnetic flux, ΔΦ = Φ2 - Φ1
= (10^-5 - 40 x 10^-6) Wb
= 6 x 10^-6 Wb
Time taken for the change, Δt = t/N
= 0.04/100 sec
= 4 x 10^-4 sec
Average emf induced in the coil, E = ΔΦ/Δt
= (6 x 10^-6)/(4 x 10^-4) V
= 0.075 V
Therefore, the average emf induced in the coil is 0.075 V, which is option D.
The magnetic flux linked with a coil is changed from 1 Wb to 0.1 Wb in 0.1 second. The induced emf is- a)0.9V
- b)0.09V
- c)9.0V
- d)0.009V
Correct answer is option 'C'. Can you explain this answer?
The magnetic flux linked with a coil is changed from 1 Wb to 0.1 Wb in 0.1 second. The induced emf is
a)
0.9V
b)
0.09V
c)
9.0V
d)
0.009V
|
|
Nikita Singh answered |
Change n in magnetic flux Δϕ=0.1−1=−0.9 Weber
Time taken Δt=0.1 sec
So, emf induced in the coil E=− Δϕ/Δt
⟹ E=− (−0.9)/ 0.1=9 Volts
Hence option C is the correct answer.
Time taken Δt=0.1 sec
So, emf induced in the coil E=− Δϕ/Δt
⟹ E=− (−0.9)/ 0.1=9 Volts
Hence option C is the correct 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.
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.
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.
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
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 |
Why is the iron core of a transformer made laminated instead of being in one solid piece?- a)To reduce the magnetic field link losses
- b)To reduce the loss due to heating of coils
- c)To reduce the hysterisis losses
- d)To reduce the losses due to eddy currents.
Correct answer is option 'D'. Can you explain this answer?
Why is the iron core of a transformer made laminated instead of being in one solid piece?
a)
To reduce the magnetic field link losses
b)
To reduce the loss due to heating of coils
c)
To reduce the hysterisis losses
d)
To reduce the losses due to eddy currents.
|
|
Mira Sharma answered |
The laminated iron core prevents the formation of eddy currents and thus reduces the loss of energy.
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
Foucault Currents are also calleda)Both eddy and induced currentb)Direct Currentsc)Induced Currentd)Eddy CurrentsCorrect answer is option 'D'. Can you explain this answer?
|
|
Nandini Patel answered |
Eddy currents are the currents which are induced in a conductor whenever the amount of linked magnetic flux with the conductor changes. These were discovered by Foucault in the year 1895 and hence they are also called Foucault currents.
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.
A magnet is moved towards the coil (i) quickly, (ii) slowly, the induced emf is- a)more in (i) than in (ii) case.
- b)Smaller in (i) than in (ii) case.
- c)Same in both.
- d)Nothing can be said.
Correct answer is option 'A'. Can you explain this answer?
A magnet is moved towards the coil (i) quickly, (ii) slowly, the induced emf is
a)
more in (i) than in (ii) case.
b)
Smaller in (i) than in (ii) case.
c)
Same in both.
d)
Nothing can be said.
|
|
Rajesh Gupta answered |
When a magnet is moved towards the coil quickly, the rate of change of flux is larger than that if the magnetic field is moved slowly, thus larger emf is induced due to quick movement of the coil.
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.
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.
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