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Question

A uniform magnetic field exists in the region given by B=3^i5^k. A rod of length 5 m placed along yaxis is moved along xaxis with constant speed 1 ms1. Then induced emf in the rod is

A
Zero
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B
25 V
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C
5 V
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D
10 V
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Solution

The correct option is B 25 V
Force acting on the charges in the moving rod
F=q(v×B)

Now in the question,
F=qvBz (only the perpendicular component of magnetic field will produce a force)

The electrons in the rod will experience the force in the direction as shown. As a result, they will accumuate at one end of the rod.


The motion of the electrons will stop only when the force due to the electric field developed across the rod balances the force due to magnetic field.
At equilibrium,
qE=qvBz
E=vBz(1)

Then induced emf across the rod ϵ=V1V2=El=vBzl
(l is length of the rod)

ϵ=(1 m/s)×(5 T)×(5 m)=25 V

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