A square, 45-turn coil 10.0 cm on a side with a resistance of 0.820 is placed be

Question

A square, 45-turn coil 10.0 cm on a side with a resistance of 0.820 is placed between the poles of a large electromagnet. The electromagnet produces a constant, uniform magnetic field of 0.600 T directed into the page. As suggested by the figure below the field drops sharply to zero at the edges of the magnet. The coil moves to the right at a constant velocity of 2.00 cm/s.

1) Determine the magnitude of the force on the left segment of the coil while the coil is leaving the field. (Express your answer to two significant figures.)

Explanation / Answer

Apply the expression , E = Blv, to determine the induced emf.

Now, while leaving the field right branch will have no induced EMF.

So, EMF = Blv on the left branch only.

therefore, I = E/R

=> Blv*n/R = (0.6*0.1*0.2*45)/0.82 = 0.66 A

So, the requisite force, F = B*I*l = 0.60*0.66*0.10 = 0.0396 N

Answer in two significant digits = 0.040 N

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