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]