XPQY is a vertical smooth long loop having a total resistance R where PX is parallel to QY and separation between them is l. A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod CD of length L(L > l) and mass m is made to slide down from rest under the gravity as shown in figure. The terminal speed acquired by the rod is m/s. (g = acceleration due to gravity) [figure]
- mgRB2l2
- 2mgRB2L2
- 2mgRB2l2
- 8mgRB2l2
Answer
(1) mgRB2l2
Solution
Only the part of the rod between the rails (length l) is in the circuit.
At speed v: motional emf = Blv
Current I = BlvR
Magnetic force on the rod (upward, opposing the motion by Lenz's law): F = BIl = B2l2vR
At terminal speed the net force is zero: mg = B2l2vtR
vt = mgRB2l2
Answer: (1)