Electromagnetic Induction Test

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Electromagnetic Induction Test - 37

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Question 1
1 / -0

The two rails of a railway track, insulated from each other and the ground, are connected to millivoltmeter. What is the reading of the millivoltmeter when a train passes at a speed of 180 km/hr along the track, given that the vertical component of earth's magnetic field is $$0.2\times 10^{-4}wb/m^2$$ and rails are separated by 1 metre

A
$$10^{-1} volt$$

B
$$10 \ mV$$

C
$$1 \ volt$$

D
$$1 \ mV$$

Solution

$$\varepsilon =Blv$$
$$=(0.2\times 10^{-4})(1)(180\times 5/18)$$
$$=10^{-3}V$$
$$=1mV$$
Question 2
1 / -0

A metal rod moves at a constant velocity in a direction perpendicular to its length. A constant uniform magnetic field exists in space in a direction perpendicular to the rod as well its velocity. Select correct statements (s) from the following

A
the entire rod is at the same potential

B
there is an electric field in the rod

C
the electric potential is highest at the centre

D
the electric potential is lowest at its centre and increases towards its ends

Solution

The moving rod develops an electric field = Bvl
and the EMF induced is such that it opposes the change of magnetic field.
Due to shifting of electrons, one end of the rod becomes positive and the other end negative. This developes a electric field in the rod.
Question 3
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A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth's magnetic field is $$0.2\times 10^{-4}T$$, then the e.m.f developed between the two ends of the conductor is

A
5 mV

B
$$50\mu V$$

C
$$5\mu V$$

D
50 mV

Solution

$$I=2m, w=5 rad/s,$$

$$B=0.2\times 10^{-4}T$$

$$\varepsilon=\dfrac {B\omega l}{2}$$

$$=\dfrac {0.2\times 10^{-4}\times 5\times 1}{2}$$

$$=50\mu V$$
Question 4
1 / -0

The flux linked with a coil at any instant 't' is given by $$\phi=10t^2-50t+250$$. The induced emf at $$t=3s$$ is

A
-190 V

B
-10 V

C
10 V

D
190 V

Solution

Farady's law states that time varying magnetic flux can induce an e.m.f.
E = Electric field induced

$$E = - \cfrac{d \phi}{dt}$$

$$E(t) = -\cfrac{d (10t^2 -50t+250)}{dt} = -(20t-50) V$$

$$E(t=3 s) = -(20(3)-50)V = -10 V$$
Question 5
1 / -0

The dimensions of self inductance are

A
$$\left[ ML{ T }^{ -2 }{ A }^{ -2 } \right] $$

B
$$\left[ M{ L }^{ 2 }{ T }^{ -1 }{ A }^{ -2 } \right] $$

C
$$\left[ M{ L }^{ 2 }{ T }^{ -2 }{ A }^{ -2 } \right] $$

D
$$\left[ M{ L }^{ 2 }{ T }^{ -1 }{ A }^{ - 1} \right] $$

Solution

$$U=\cfrac { 1 }{ 2 } L{ i }^{ 2 }\quad $$

$$\therefore \quad \left[ L \right] =\left[ \cfrac { U }{ { i }^{ 2 } } \right] =\left[ \cfrac { M{ L }^{ 2 }{ T }^{ -2 } }{ { A }^{ 2 } } \right] $$

$$=\left[ M{ L }^{ 2 }{ T }^{ -2 }{ A }^{ -2 } \right] $$
Question 6
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A conducting rod is rotating about one end in a plane perpendicular to a uniform magnetic field with constant angular velocity. The correct graph between the induced emf($$e$$) across the rod and time ($$t$$) is

A

B

C

D

Solution

$$EMF\quad induced\quad is\quad \frac { 1 }{ 2 } B\omega { l }^{ 2 },\quad which\quad is\quad constant\quad with\quad time.$$
Question 7
1 / -0

A solenoid that is highly wound with wire of diameter $$0.10cm$$ has a cross-sectional area $$0.90{cm}^{2}$$ and is $$40cm$$ long. Calculate the self inductance of the solenoid.

A
$$4.5 \times 10^{-5} H$$

B
$$7.5 \times 10^{-5} H$$

C
$$10 \times 10^{-5} H$$

D
$$12.5 \times 10^{-5} H$$

Solution

Self-inductance in solenoid, $$L=\cfrac { { \mu }_{ 0 }{ N }^{ 2 }S }{ l } $$

$$and \quad N=\cfrac { l }{ d } $$

$$\Rightarrow \quad L=\cfrac { { \mu }_{ 0 }lS }{ { d }^{ 2 } } =\cfrac { \left( 4\pi \times { 10 }^{ -7 } \right) \left( 0.4 \right) \left( 0.9\times { 10 }^{ -4 } \right) }{ { \left( 0.1\times { 10 }^{ -2 } \right) }^{ 2 } } \quad =4.5\times { 10 }^{ -5 }H$$
Question 8
1 / -0

A solenoid that is highly wound with wire of diameter $$0.10cm$$ has a cross-sectional area $$0.90{cm}^{2}$$ and is $$40cm$$ long. If the current through the solenoid decreases uniformly from $$10A$$ to $$0A$$ in $$0.10s$$, what is the emf induced between the ends of the solenoid?

A
$$4.5 \times 10^{-4} V$$

B
$$5.5 \times 10^{-3} V$$

C
$$5.0 \times 10^{-8} V$$

D
$$4.5 \times 10^{-3} V$$

Solution

Induced emf, $$\left| e \right| =\left| L.\cfrac { \Delta i }{ \Delta t } \right| $$

$$=\cfrac { \left( 4.5\times { 10 }^{ -5 } \right) \left( 10 \right) }{ 0.1 } $$

$$=4.5\times { 10 }^{ -3 }V$$
Question 9
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When a loop moves towards a stationary magnet with speed $$v$$, the induced emf in the loop is $$E$$. If the magnet also moves away from the loop with the same speed, then the emf induced in the loop is

A
$$E$$

B
$$2E$$

C
$$\cfrac{E}{2}$$

D
zero

Solution

Correct option: D

Explanation for correct option:

$$\bullet$$Faraday's Law states that an electric field is induced by changing magnetic flux.
$$\bullet$$If both move with the same velocity in the same direction, the relative velocity is zero, hence there is no change of magnetic flux across the loop with time.
Hence no EMF is induced.
Question 10
1 / -0

A conducting straight wire $$PQ$$ of length $$l$$ is fixed along a diameter of a non-conducting ring as shown in the figure. The ring is given a pure rolling motion on a horizontal surface such that its centre of mass has a veleocity $$v$$. There exists a uniform horizontal magnetic field $$B$$ in horizontal direction perpendicular to the plane of ring. The magnitude of induced emf in the wire $$PQ$$ at the position shown in the figure will be :

A
$$Bvl$$

B
$$2Bvl$$

C
$$3Bvl/2$$

D
zero

Solution

$$\omega=\cfrac{v}{R}$$
$$=\cfrac{v}{l/2}=\cfrac{2v}{l}$$
$$e=\cfrac{B\omega {l}^{2}}{2}$$
$$=\cfrac { B\left( \cfrac { 2v }{ l } \right) { l }^{ 2 } }{ 2 } =Bvl$$