A coil is wrapped with 474 turns?

Question

A coil is wrapped with 474 turns of wire on
the perimeter of a circular frame (of radius
28 cm). Each turn has the same area, equal
to that of the frame. A uniform magnetic
field is directed perpendicular to the plane of
the coil. This field changes at a constant rate
from 29 mT to 69 mT in 31 ms.
What is the magnitude of the induced E in
the coil at the instant the magnetic field has a magnitude of 51 mT? Answer in units of V

Answer 1

1) you can assume that the change in fild occur uniformly over the given period of time, therefore the field will have a magnitude of 51mT at 0.55*31ms (51 is 55% of the total increase of field).
2) find the change in magnetic flux = magnetic filed (55mT, given) * area (you know the radius of the frame!).
3) use the formula induced E = -(change in magnetic flux * n. turns) / time (calculated in 1).
this formula is a combination of Faraday's law of mangetic induction and Lenz's law.

hope it helps!

Answer 2

This is a direct application of Faraday's law.

E = N * d phi / dt

E is in Volts
N is the number of turns of the coil (ideally described in this problem)
phi is the magnetic flux in any cross-sectional area of the field in Webers or Tesla m^2

Since the flux is normal to the coil, the integral to convert from Teslas (B field) to Webers is a simple multiplication by the area of the coil.

E = Area * N * dB/dt

since dB/dt is a constant the voltage is the same at 29, 51, or 69 mT The rate of change is still 1290 T/s

I get 1506 volts.

And that 'sounds' reasonable based on my experience.
.