A 43.0-cm wire is moving to the right at a speed of 7.90 m/s across two parallel

Question

A 43.0-cm wire is moving to the right at a speed of 7.90 m/s across two parallel wire rails that are connected on the left side, as shown in the figure. The whole apparatus is immersed in a uniform magnetic field which has a magnitude of 0.750 T and is directed into the screen. What is the emf induced in the wire? induced emf causes a current to flow in the circuit formed by the moving wire and the rails. In what direction does the current flow around the circuit ? If the moving wire and the rails have a combined total resistance of 1.55 Ohm, what applied force would be required to keep the wire moving at the given velocity? Assume that there is no friction between the moving wire and the rails.

Explanation / Answer

here,
length of wire, l = 43cm = 0.43 m
velocity of wire, v = 7.90 m/s
Magnatic field, B = 0.750 into the screen
Resistance, R = 1.55 ohms

Part A:
Indced EMf, E = B*l*V*sin90
E = 0.750 * 0.43*7.90
E = 2.548 V

Part b:
From the generator hand rule.

First Fingure : Direction of field
Thumb : direction of motion
middle fingure will direction of EMF,

So, Clockwise direction

Part C:
Induced Current, I = EMF/R = 2.548/1.55
I = 1.644 A

Therefore, Force, F = I*B*L = 1.644 * 0.75 * 0.43
F = 0.53 N

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