A 96-turn square coil of side 20.0 cm rotates about a vertical axis at ? = 1.72

Question

A 96-turn square coil of side 20.0 cm rotates about a vertical axis at ? = 1.72 x 101 rev/min as indicated in the figure below. The horizontal component of Eart's magnetic field at the coll's location is equal to 2.00 10 ? T 20.0 cm 20.0 cm m emf induced in I by this field. (b) What is the orientation of the coil with respect to the magnetic field when the maximum emf occurs? O The plane of the coil is perpendicular to the magnetic field. The plane of the coil is oriented 450 with respect to the magnetic field. O The plane of the coil is parallel to the magnetic field.

Explanation / Answer

Part A

EMF is given by

EMF = N*A*B*w*sin wt

emf will be max when sin wt = 1

EMF)max = N*A*B*w

A = area = 20 cm*20 cm = 0.2*0.2 m^2

w = 1.72*10^3 rev/min = 1720 rev/min

w = 1720*2*pi/60 = 180.12 rad/sec

Using given values:

EMF)max = 96*0.2*0.2*2*10^-5*180.12

EMF = 13.83*10^-3 V = 13.83 mV

Part B

EMF will be max when

sin wt = 1

wt = 90 deg = angle between plane of the coil and magnetic field

Correct option is A

The plane of the coil is perpendicular to the magnetic field.

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