A rod of mass m and resistance R slides smoothly over two parallel

Question:

A rod of mass m and resistance R slides smoothly over two parallel perfectly conducting wires kept sloping at an angle θ with respect to the horizontal. The

circuit is closed through a perfect conductor at the top. There is a constant magnetic field B along the vertical direction. If the rod is initially at rest, find the

velocity of the rod as a function of time.

Solution:

The angle between B and PQ = 90o

dϕ = B.dA

dϕ = B v d cos θ

-ε = B v d cos θ

I = -Bvd/R cos θ

Solving the above equation using Newton’s second law, we get, v as

v = α g sin θ [1 – e t/ꞷ]

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