Magnitude and direction of induced current The loop in the picture below has dim

Question

Magnitude and direction of induced current

The loop in the picture below has dimensions L = 23.0 cm and W = 12.0 cm. It is being pulled out of a region containing a magnetic field of magnitude 0.3 T at a speed v = 9.0 m/s. If the loop has a total resistance of 0.6 Ohm, determine magnitude and direction of the induced current. Explain how you determined the direction of the induced current. ANSWER: 0.54 A, counter-clockwise. If you get a different answer please don't mind posting it, thank you. Calculate the force required to pull the loop out of the magnetic field at the constant speed indicated in part (a). ANSWER: 0.0194 N If you get a different answer please don't mind posting it, thank you.

Explanation / Answer

a) Induced emf in the loop = B*dA/dt

= B*d(L*w)/dt

= B*w*dL/dt

= B*W*v

induced current, I = induced emf/R

= B*W*v/R

= 0.3*0.12*9/0.6

= 0.54 A

direction : counter-clockwise

b) magnetic force acting on the loop, FB = B*I*w

= 0.3*0.54*0.12

= 0.0194 N

so, required external force to pull the loop with constant speed = FB

= 0.0194 N

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