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A rectangular conductor LMNO is placed...

A rectangular conductor LMNO is placed in a uniform magnetic field of 0.5 T. The field is directed perpendicular to the plane of the conductor. When

the arm MN of length of 20 cm is moved towards left with a velocity of `10ms^(-1)`, calculate the emf induced in the arm. Given the resistance of the arm to the `5 Omega` ( assuming that other arms are of negligible resistance) find the value of the current in the arm.

Text Solution

Verified by Experts

Let ON be at some point x.
The emf induced in the loop `= epsilon`
`epsilon= (-d phi)/(dt) = (-d (Blx))/(dt)= Blv`
`= 0.5xx0.2xx10= 1V`
`:.` Current in the arm,
`I = (epsilon)/(R) = 1/5 = 0.2 A`.
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