Question

A simple AC generator consists of a single loop of wire rotating 50 times per second within a magnetic field. The loop is a rectangle 5cm by 12 cm. The field strength is 0.21 T. What is the peak output emf?

Answer 1

Answer:

39.56 x 10⁻² V

Explanation:

Peak emf in a rotating coil

= n BAω

Where n is no of turns of coil , B is magnetic field , A is area of coil and ω is angular velocity of rotation of coil

Here n = 1

B = .21 T

A = 5 X 12 X 10⁻⁴ = 60 X 10⁻⁴

ω = 2π X 50 = 100π

Emf = 1 x .21 x 60 x 10⁻⁴ x 100 x 3.14

=  39.56 x 10⁻² V