Question

Two horizontal, parallel conductors are connected to each other at the left end via an ammeter of internal resistance of r = 2 Ω. A conductor of length L = 0.5 m, resistance R = 8 Ω and mass m = 5 kg can slide (no friction) along the horizontal rails. The whole system is inside a uniform magnetic field of magnitude 2 T and direction perpendicular to the plane formed by the two parallel conductors (i.e., pointing into the page). At time t = 0 s, the sliding conductor has a velocity of magnitude 12 m/s and direction parallel to the rails and away from the ammeter (i.e., to the right.) At that instant an external force F is exerted that has the same direction as the velocity. The sliding conductor then moves with a constant acceleration of 2 m/s2, which has the same direction as the force F. Find a) the current as a function of time and plot the graph of I vs. t, b) the charge that goes through the ammeter during the first 5 seconds, c) the rate at which the current is increasing, and d) the magnitude of F at t = 5 s.

Answer 1

At t=0, the conductor has velocity 12m/s and constant acceleration 2m/s2. Then at any instant of time t, the velocity of conductor is v= 12+2t.

At that instant induced emf e= BLv = 12+2t, (B=2T and L=0.5m)

Then induced current at the same instant is i= emf / net resistance . Net resistance is 10 ohm

a) The current through ammeter i = 1.2+0.2t

The graph is a straight line with positive slope and positive y-intercept.

b) q = i dt. Hence the charge q = 1.2 t + 0.1 t2 . In first 5s, the charge that passes through ammeter is 8.5 C.

c) The rate of increase of current is di/dt = 0.2 A/s

d) F - BiL = ma => F = BiL + ma = (1.2 + 0.2t) + 10 => F = 11.2+0.2t

The external force acting on the rod at t=5 is 12.2N