Problem 3

 
3) A small circular loop of area 2.00 cm² is placed in the plane of, and concentric with, a large circular loop of radius 1.00 m. The current in the large loop is changed uniformly from 200 A to -200 A in a time 1.00 s, beginning at t=0 s, as shown in the figure. (a) What is the magnetic field at the center of the small circular loop due to the current in the large loop at t=0, t=0.5 s, and t=1.00 s (12.5 pts)? (b) What emf is induced in the small loop at t=0.5 s (12.5 pts)? (Since the inner loop is small, assume the field $\vec{B}$ due to the outer loop is uniform over the area of the smaller loop.)  

Answer:
a)
The equation for the current as a function of time is given by
I = I₀ (1 - 2t)
The magnetic field at the center of a current loop is given by
$B = \frac{\mu_{0}I}{2R}$
At t = 0, the current is 200 A, so the magnetic field is given by
$B = \frac{\mu_{0}(200 A)}{2R}$
$B = 1.26 \times 10^{-4} T$
At t = .5, the current is 0 A, so the magnetic field is zero. At t = 1, the current is -200 A, so the magnetic field is given by
$B = -1.26 \times 10^{-4} T$

b)
The magnetic flux through the loop is given by
$\Phi_{m} = B A \cos \theta$
Here, we take $\theta = 0$.
So, the flux is given by
$\Phi_{m} = \frac{\mu_{0}IA}{2R}$
The area of the loop is given by
$A = 2.00 cm^{2} ~=~ 2 \times 10^{-4} m^{2}$
So, the change in magnetic flux over time is given by
$\frac{d\Phi_{m}}{dt} = \frac{d}{dt} \frac{A \mu_{0}}{2R} I$
Since A, R, and $\mu_{0}$ are constants in time, this becomes
$\frac{d\Phi_{m}}{dt} = \frac{A \mu_{0}}{2R} \frac{dI}{dt}$
$\frac{d\Phi_{m}}{dt} = \frac{A \mu_{0}}{2R} (-2 I_{0})$
Well, the change in magnetic flux over time is known as the EMF, so we write it as such.
$\epsilon = -\frac{A \mu_{0} I_{0}}{R}$
Plugging in the numbers, we find that
$\vert\epsilon\vert = 5.04 \times 10^{-8} V$