In the figure, a metal wire of mass m = 21.9 mg can slide with negligible fricti

Question

In the figure, a metal wire of mass m = 21.9 mg can slide with negligible friction on two horizontal parallel rails separated by distance d = 3.04 cm. The track lies in a vertical uniform magnetic field of magnitude 69.5 mT. At time t = 0 s, device G is connected to the rails, producing a constant current i = 8.68 mA in the wire and rails (even as the wire moves). At t= 56.8 ms, what are the wire's (a) speed and (b) direction of motion?

In the figure, a metal wire of mass m = 21.9 mg can slide with negligible friction on two horizontal parallel rails separated by distance d = 3.04 cm. The track lies in a vertical uniform magnetic field of magnitude 69.5 mT. At time t = 0 s, device G is connected to the rails, producing a constant current i = 8.68 mA in the wire and rails (even as the wire moves). At t= 56.8 ms, what are the wire's (a) speed and (b) direction of motion?

Explanation / Answer

a)

here , force acting on the rod is

Fm = BIL

Fm = 0.0695 * 0.00868 * 0.034

Fm = 2.05 *10^-5 N

Now, using second law of motion

a = 2.05 *10^-5/21.9 *10^-6

a = 0.937 m/s^2

at t = 0.0568 s

speed = 0.0568 * 0.937

speed = 0.0552 m/s

b)

direction of motion is right if current is anticlockwise

direction is left if current is clockwise

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