A uniform magnetic field exists in the region given by →B=3^i−5^k. A rod of length 5m placed along y−axis is moved along x−axis with constant speed 1ms−1. Then induced emf in the rod is
A
Zero
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B
25V
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C
5V
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D
10V
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Solution
The correct option is B25V 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=V1−V2=El=vBzl
(l is length of the rod)