Question

An electric generator contains a coil of 99 turns of wire, each forming a rectangular loop 73.9 cm by 34.9 cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 2.96 T and initially perpendicular to the coil's plane. What is in volts the maximum value of the emf produced when the loop is spun at 1200 rev/min about an axis perpendicular to the magnetic field?

Answer 1

Answer:

12078.46 V

Explanation:

Applying,

E₀ = BANω.................... Equation 1

Where: E₀ = maximum emf, B = magnetic Field, A = Area of the coil, N = Number of turns of the coil. ω = angular velocity

Given: N = 99 turns, B = 2.96 T, ω = 1200 rev/min = (1200+0.10472) = 125.664 rad/s

A = L×W, where L= Length = 93.9 cm = 0.939 m, W = width = 34.9 cm = 0.349 m

A = (0.939×0.349) = 0.328 m²

Substitute into equation 1

E₀ = 99(2.96)(0.328)(125.664)

E₀ = 12078.46 V

Hence the maximum value of the emf produced = 12078.46 V

Answer 2

Answer:

The maximum emf induced in the loop is 9498.268 V

Explanation:

Given;

number of turns of coil, N = 99 turns

area of the rectangular loop, A = 0.739 m x 0.349 m = 0.2579 m²

magnetic field strength, B = 2.96 T

angular speed of the loop, ω = 1200 rev/min

angular speed of the loop, ω (rad/s) = (2π x 1200) / 60 = 125.68 rad/s

The maximum value of the emf produced is calculated using the formula below;

ξ = NABω

Substitute the given values and calculate the maximum emf induced;

ξ = (99)(0.2579)(2.96)(125.68)

ξ = 9498.268 volts

Therefore, the maximum emf induced in the loop is 9498.268 V