Question

What is the peak emf generated by rotating a 940-turn, 24 cm diameter coil in the Earth’s 5·10−5 T magnetic field, given the plane of the coil is originally perpendicular to the Earth’s field and is rotated to be parallel to the field in 5 ms?

Answer 1

Answer:

The peak emf generated by the coil is 2.67 V

Explanation:

Given;

number of turns, N = 940 turns

diameter, d = 24 cm = 0.24 m

magnetic field, B = 5 x 10⁻⁵ T

time, t = 5 ms = 5 x 10⁻³ s

peak emf, V₀ = ?

V₀ = NABω

Where;

N is the number of turns

A is the area

B is the magnetic field strength

ω is the angular velocity

V₀ = NABω and ω = 2πf = 2π/t

V₀ = NAB2π/t

A = πd²/4

V₀ = N x (πd²/4) x B x (2π/t)

V₀ = 940 x (π x 0.24²/4) x 5 x 10⁻⁵ x (2π/0.005)

V₀ = 940 x 0.04524 x  5 x 10⁻⁵ x 1256.8

V₀ = 2.6723 V = 2.67 V

The peak emf generated by the coil is 2.67 V