Question

Consider a generator that rotates its 200-turn, 0.19 m diameter coil at 3450 rpm in a 0.55 T field. What is the induced EMF (electromotive force) in the coil?

a) 25.7 V
b) 36.5 V
c) 45.6 V
d) 52.3 V

Answer 1

The induced EMF in the generator coil can be calculated using Faraday's law, and by converting the given values into the appropriate units. Using the formula with the provided values results in an induced EMF of approximately 36.5 V.

Step-by-step explanation:

To calculate the induced EMF in a generator coil, we use Faraday's law of electromagnetic induction. The formula for EMF (ε) is given by ε = NABωsin(θ), where:

• N is the number of turns in the coil,
• A is the area of the coil,
• B is the magnetic field strength,
• ω is the angular velocity in radians per second, and
• θ is the angle between the magnetic field and the normal to the coil.

Given:

• N = 200 turns
• Diameter of the coil, d = 0.19 m, so the radius r = d/2 = 0.095 m
• B = 0.55 T
• Rotational speed = 3450 rpm (revolutions per minute)

To find the angular velocity (ω), we convert rpm to rad/s:

ω = (3450 rpm)(2π rad/rev)(1 min/60 s) = 361.03 rad/s

The area of the coil (A) is πr2 = π(0.095 m)2.

Assuming the coil rotates in such a way that at the peak, the angle θ is 90 degrees (or π/2 radians), sin(θ) = 1. We can now calculate the peak EMF:

ε = NABωsin(θ)

ε = (200)(π(0.095 m)2)(0.55 T)(361.03 rad/s)(1)

ε ≈ 36.5 V

Therefore, the correct answer is (b) 36.5 V.