Question

A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. Initially the field is zero and then changes to 1.50 T, pointing upward when viewed from above, perpendicular to the circular plane, in a time of 125 ms
what is the average induced emf around the border of the circular region?

Answer 1

Answer:

The average induced emf around the border of the circular region is $8.48\times 10^{-5}\ V$.

Explanation:

Given that,

Radius of circular region, r = 1.5 mm

Initial magnetic field, B = 0

Final magnetic field, B' = 1.5 T

The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :

$\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\dfrac{-d(B'-B)}{dt}\\\\\epsilon=\pi (1.5\times 10^{-3})^2\times \dfrac{1.5}{0.125}\\\\\epsilon=8.48\times 10^{-5}\ V$

So, the average induced emf around the border of the circular region is $8.48\times 10^{-5}\ V$.

Answer 2

Answer:

$84.8\times 10^{-6} V$

Explanation:

We are given that

Radius,r=1.5 mm=$1.5\times 10^{-3} m$

$1mm=10^{-3} m$

Initial magnetic field,$B_0=0$

Final magnetic field,$B=1.5 T$

Time,$\Delta t=$125 ms=$125\times 10^{-3} s$

$1 ms=10^{-3}s$

We know that average induced emf

$E=\frac{d\phi}{dt}=-\frac{d(BA)}{dt}=A\frac{dB}{dt}=A\frac{(B-B_0)}{dt}$

Substitute the values

$E_{avg}=\pi (1.5\times 10^{-3})^2\times \frac{1.5-0}{125\times 10^{-3}}$

$E_{avg}=84.8\times 10^{-6} V$