Question

A 114-turn square coil of side 20.0 cm rotates about a vertical axis at ω = 1.20 103 rev/min as indicated in the figure below. The horizontal component of Earth's magnetic field at the coil's location is equal to 2.00 10-5 T.
(a) Calculate the maximum emf induced in the coil by this field. 1.824 mV (wrong)
(b) What is the orientation of the coil with respect to the magnetic field when the maximum emf occurs? The plane of the coil is perpendicular to the magnetic field.
The plane of the coil is oriented 45° with respect to the magnetic field.
The plane of the coil is parallel to the magnetic field. (correct)

Answer 1

PART (a):

Area of the coil = a² = (0.2 m)² = 0.04 m²

Angular speed = 1.2 x 10³ rev/min = 1.2 x 10³ x (2π/60) rad/s

Therefore, Maximum emf induced in the coil is given by:

E = NABω

E = (114) x (0.04 m²) x (2 x 10^-5 T) x (1.2 x 10³ x (2π/60) rad/s)

E = 0.001146 V

E = 11.46 mV ----------- (answer)

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PART (b):

Correct option is:

• The plane of the coil is parallel to the magnetic field.