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

`84.8* 10^(-6) V`

Step-by-step explanation:

We are given that

Radius,r=1.5 mm=
`1.5* 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* 10^(-3) s`

`1 ms=10^(-3)s`

We know that average induced emf

`E=(d\phi)/(dt)=-(d(BA))/(dt)=A(dB)/(dt)=A((B-B_0))/(dt)`

Substitute the values

`E_(avg)=\pi (1.5* 10^(-3))^2* (1.5-0)/(125* 10^(-3))`

`E_(avg)=84.8* 10^(-6) V`

Answer 2

The average induced emf around the border of the circular region is
`8.48* 10^(-5)\ V`.

Step-by-step 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=(-d\phi)/(dt)\\\\\epsilon=A(-d(B'-B))/(dt)\\\\\epsilon=\pi (1.5* 10^(-3))^2* (1.5)/(0.125)\\\\\epsilon=8.48* 10^(-5)\ V`

So, the average induced emf around the border of the circular region is
`8.48* 10^(-5)\ V`.