The Peak EMF in a Rotating Coil in a Magnetic Field

Calculating the Peak EMF

A 75-turn, 10.0 cm diameter coil rotates at an angular velocity of 8.00 rad/s in a 1.25 T field, starting with the plane of the coil parallel to the field. What is the peak emf?

Final answer: The peak emf for a 75-turn, 10.0 cm diameter coil rotating at an angular velocity of 8.00 rad/s in a 1.25 T field is approximately 1.47 volts, calculated using Faraday's law of electromagnetic induction.

Explanation: The peak emf induced in a rotating coil in a magnetic field can be calculated using Faraday's law of electromagnetic induction, which states that the induced emf is proportional to the rate of change of magnetic flux. The formula for peak emf (ε) is given as: ε = NBAωsin(ωt + δ). Here, N is the number of turns in the coil, B is the magnetic field strength, A is the area of the coil, ω is the angular velocity, and δ is the phase angle. In this case, the plane of the coil starts parallel to the field, thus phase angle (δ) is 0. At the peak, sin(ωt + δ) = 1. Given, N = 75 turns, diameter of the coil d = 10cm = 0.1m (because the area A = π*(d/2)^2), B = 1.25 T, and ω = 8 rad/s. Substituting the values into the formula, we get ε = (75)*(1.25)* π(0.1/2)^2 * 8 * 1. This yields a peak emf ε ≈ 1.47 volts.

Do you understand how to calculate the peak emf in a rotating coil in a magnetic field?

Yes, the peak emf in a rotating coil in a magnetic field can be calculated using Faraday's law of electromagnetic induction by considering the number of turns in the coil, magnetic field strength, area of the coil, angular velocity, and phase angle.