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A small circular loop of area 2 cm2 is placed in the plane of, and concentric with, a large circular loop of radius 1 m. The current in the large loop is changed at constant rate from 200A to -200A ( a change in direction) in a time of 1 s, starting at t=0. What is magnitude of the magnetic field at the center of the small loop due to the current in the large loop at (a) t=0, (b) t=0.5 s, and (c) t=1 s? (d) from t=0 to t=1 s, is reversed? Because the inner loop is small, assume is uniform over its area. (e) What emf is induced in the small loop at t=0.5 s?

A small circular loop of area 2 cm2 is placed in the plane of, and concentric with, a large circular loop of radius 1 m. The current in the large loop is changed at constant rate from 200A to -200A ( a change in direction) in a time of 1 s, starting at t=0. What is magnitude of the magnetic field at the center of the small loop due to the current in the large loop at (a) t=0, (b) t=0.5 s, and (c) t=1 s? (d) from t=0 to t=1 s, is reversed? Because the inner loop is small, assume is uniform over its area. (e) What emf is induced in the small loop at t=0.5 s?

Grade:12

1 Answers

Aman Bansal
592 Points
8 years ago

Dear Sandeep,

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
$\vert\epsilon\vert = 5.04 \times 10^{-8} V$

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