Question

The area of a 110-turn coil oriented with its plane perpendicular to a 0.21 T magnetic field is 7.0×10^(-2) m^2. Find the average induced emf in this coil if the magnetic field reverses its direction in 0.11 s.

a) 0.033 V
b) 0.21 V
c) 0.231 V
d) 0.071 V

Answer 1

The average induced emf in a coil of 110 turns, when the magnetic field of 0.21 T reverses over an area of 7.0×10⁻² m² in 0.11 s, is calculated to be 0.033 V using Faraday's Law of Induction.

Step-by-step explanation:

The question is related to the phenomenon of electromagnetic induction and Faraday's Law of Induction in Physics. We are given a coil consisting of 110 turns, and we need to calculate the average induced emf when the magnetic field reverses its direction in 0.11 s. According to Faraday's Law, the induced emf (ε) in a coil with N turns is given by ε = -N * dΨ/dt, where Ψ is the magnetic flux. The magnetic flux Ψ is equal to the product of the magnetic field B and the area A over which it is applied, i.e., Ψ = B * A. When the magnetic field reverses, the change in flux (ΔΨ) is 2 * B * A (since it goes from B to -B, the total change is B - (-B) = 2B). Plugging in the given numbers and calculating:

Ψ_initial = B * A = 0.21 T * 7.0×10⁻² m²
Ψ_final = - (Ψ_initial) (because the field reverses)
ΔΨ = Ψ_final - Ψ_initial = -2 * Ψ_initial
ΔΨ = -2 * 0.21 T * 7.0×10⁻² m²
dt = 0.11 s

The average induced emf is calculated as:

ε_average = -N * ΔΨ/dt
ε_average = -110 * (-2 * 0.21 * 7.0×10⁻²) / 0.11
ε_average = 0.033 V

Therefore, the average induced emf is 0.033 V, matching option (a).