The angle formed by B and PQ is 90
dϕ = B.dA
dϕ = B v d cos θ
-ε = B v d cos θ
I = -Bvd/R cos θ
Using Newton’s second law to solve the preceding problem, we get v as
v = α g sin θ [1 – e t/ꞷ]
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.
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.