A rectangular loop with dimensions 0.050 m by 0.080 m is being pulled at constan

Question

A rectangular loop with dimensions 0.050 m by 0.080 m is being pulled at constant speed v = 12.0 m/s out of a region of uniform magnetic field. The magnetic field has magnitude B = 3.0 T and is directed into the page. The loop has resistance R = 0.40 ohm. What are the magnitude and direction (clockwise or-counterclockwise) of the induced current in the loop at the instant when the loop is half-way out of the field region? Ans. I = ___________ direction _________ When the loop is half-way out of the field region, what is the magnitude of the external force that must be applied to the loop to maintain its constant velocity?

Explanation / Answer

The flux through the rectangular coil is given by B = BA = Blx where l = 0.05 and for current instant x = 0.08/2 = 0.04 m

the induced emf is given by e = -dB / dt = -d/dt(Blx) = Bldx/dt = -Blv

for our case B = 3 T, l = 0.05 and v = 12 m/s

e = -3*.05*12 = -1.8 V

Induced current I = e/R = 1.8/0.4 = 4.5 A

The directionof induced current will be such that it opposes the magnetic field. Hence magnetic field by loop should be coming out of the plane. Hence current must be counter-clockwise.

b) The forces on wires which are of length 0.08 m due to magnetic field will be opposite in direction and same in direction. Hence only the force in 0.05 m wire will matter.

The force on a current carrying conductor due to magnetic field B is given by F = ILBsin(theta)

For our case I = 4.5 A, L = 0.05 m, B = 3 T, theta = 90o

F = 4.5*0.05*3*1 = 0.675 N

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