A constant magnetic field passes through a single rectangular loop whose dimensi

Question

A constant magnetic field passes through a single rectangular loop whose dimensions are 0.35 m 0.55 m. The magnetic field has a magnitude of 2.1 T and is inclined at an angle of 75° with respect to the normal to the plane of the loop.

(a) If the magnetic field decreases to zero in a time of 0.45 s, what is the magnitude of the average emf induced in the loop?
V

(b) If the magnetic field remains constant at its initial value of 2.1 T, what is the magnitude of the rate A / t at which the area should change so that the average emf has the same magnitude as in part (a)?
m2/s

Explanation / Answer

| emf | = d(flux)/dt

Since flux = (magnetic field)*(area), differentiate to get:

d(flux)/dt = (magnetic field)*d(area)/dt + (area)*d(magnetic field)/dt

In part A, the area was constant, so d(area)/dt = 0, and the equation for d(flux)/dt becomes:
d(flux)/dt = (area)*d(magnetic field)/dt

Plug in the following:
area = (0.35 m)(0.55 m)cos(75 deg)
d(magnetic field)/dt = (2.1 T)/(0.45 s)

This gives:
| emf | = d(flux)/dt = (area)*d(magnetic field)/dt = [(0.35 m)(0.55 m)cos(75 deg)][(2.1 T)/(0.45 s)] = 0.2325 V

In part B, the magnetic field is constant. So, d(magnetic field)/dt = 0, and the equation for d(flux)/dt becomes:
d(flux)/dt = (magnetic field)*d(area)/dt

Plug in the following:
magnetic field = 2.1 T
d(flux)/dt = 0.2325 V

This gives:
d(flux)/dt = (magnetic field)*d(area)/dt
d(area)/dt = [d(flux)/dt]/[magnetic field] = [0.2325 V]/[2.1 T] = 0.1107 m^2 / sec

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