Question

The armature of an ac generator is a circular coil with 50 turns and radius 2.30 cm. When the armature rotates at 350.0 rev/min, the amplitude of the emf in the coil is 20.0 V. What is the magnitude of the magnetic field (assumed to be uniform)

Answer 1

Answer:

B = 6.57 T

Explanation:

The amplitude of emf of an AC generator coil can be given by the following formula:

$EMF = N\omega AB$

where,

EMF = amplitude of emf = 20 V

N = No. of turns = 50

ω = angular speed= (350 rev/min)(2π rad/1 rev)(1 min/60 s) = 36.65 rad/s

A = Cros-sectional area of coil = πr² = π(0.023 m)² = 1.66 x 10⁻³ m²

Therefore,

$20\ V = (50)(36.65\ rad/s)(1.66\ x\ 10^{-3}\ m^2)B\\\\B = \frac{20\ V}{(50)(36.65\ rad/s)(1.66\ x\ 10^{-3}\ m^2)}$

B = 6.57 T