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:
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
