Question

Please Answer! Thank you!

1. The coil of a generator has a radius of 0.13 m. When this coil is unwound, the wire from which it is made has a length of 6.00 m. The magnetic field of the generator is 0.36 T, and the coil rotates at an angular speed of 34 rad/s. What is the peak emf of this generator?

2. A conducting coil of 2200 turns is connected to a galvanometer, and the total resistance of the circuit is 50 Ω. The area of each turn is 5.75 × 10^-4 m². This coil is moved from a region where the magnetic field is zero into a region where it is nonzero, the normal to the coil being kept parallel to the magnetic field. The amount of charge that is induced
to flow around the circuit is measured to be 7.5 × 10^-3 C. Find the magnitude of the magnetic field.

Answer 1

2.)

use the equation emf = N(Φ/t)

Since emf is voltage you can rewrite V = N(Φ/t)

Since V=IR by Ohms law you get IR = N(Φ/t)

Then, since I = (Q/t) you get (Q/t)R = N(Φ/t)

Thus your t's cancel and you get QR = N(Φ)

Rewrite this as QR = N(BA)

Now plug in you numbers:

(7.5e-3)(50) = 2200[B(5.75e-4)]

Solve for B

B = 0.296 T ~0.3T .....Answer.