Question

The loop is in a magnetic field 0.22 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A = 0.285 m2.Suppose the radius of the elastic loop increases at a constant rate, dr/dt = 3.45 cm/s .

A)

Determine the emf induced in the loop at t = 0 and at t = 1.00 s .

Express your answer using two sifnificant figures

E(0) =		mV

B)

Express your answer using two sifnificant figures.

E(1.00) =		mV

Answer 1

  the induced emf occurs because of changing flux, and is described by

emf = - N * rate of change of flux

where N is the number of coils of the loop and flux = B * A cos(theta)

B = mag field
A=area of loop
theta = angle between mag field and normal to loop; here, this angle is zero so cos(theta) = 1 and flux = B * A

the only changing variable is the area of the loop, so we have

induced emf = d/dt ( BA)

the area of the loop = pi r^2

but r is a function of t, so dA/dt = 2 pi r dr/dt

we are told that dr/dt = 0.0345m/s, therefore

dA/dt = 2 pi r * 0.0345m/s

or induced emf = -0.22T x 2 pi r * 0.0345m

when t=0, we are told that A = pi r^2 = 0.285m^2 so that r = 0.3011m

substitute numbers and solve for emf;

when t = 1s, the radius will have increased by 0.0345m/s x 1s = 0.0345m so will be 0.3011m + 0.0345m = 0.3356m, and the emf is then

emf = - B dA/dt = - 0.22T * 2 pi* 0.3356m * 0.0345m/s =0.016000v