Question

What is the peak emf generated by rotating a a980-turn, 11cm 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 7 ms? g

Answer 1

Answer:

The peak emf generated by the coil is 418.3 mV

Explanation:

Given;

number of turns, N = 980 turns

diameter, d = 11 cm = 0.11 m

magnetic field, B = 5 x 10⁻⁵ T

time, t = 7 ms = 7 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 = 980 x (π x 0.11²/4) x 5 x 10⁻⁵ x (2π/0.007)

V₀ = 980 x 0.00951 x  5 x 10⁻⁵ x 897.71

V₀ = 0.4183 V = 418.3 mV

The peak emf generated by the coil is 418.3 mV

Answer 2

Answer:

E = 426mV

Explanation:

Given N = 980

B = 5×10-⁵ T

R = 11cm = 0.11m

Δt = 7ms = 7×10-³s

Area A = π×D²/4 = π×0.11²/4 = 0.0095m²

E = - N×ΔBAωSinθ = NBA×2π/t

The field and the Area are not changing but the angle between the area vector changes from 0 to 90° and Sinθ changes from 0 to 1. Therefore the flux changes from zero to maximum.

So E = - 980×5.10×10-⁵×0.0095×2π/(7×10-³)

E = 426×10-³V= 426mV