Question

A parallel-plate capacitor with circular plates of radius 35 mm is being discharged by a current of 3.5 A. At what radius (a) inside and (b) outside the capacitor gap is the magnitude of the induced magnetic field equal to 80% of its maximum value? (c) What is that maximum value?

Answer 1

Step-by-step explanation:

Given that,

Radius R= 35 mm

Current = 3.5 A

The magnetic field B is proportional to the r.

Magnetic field is maximum ,when r = R

(a). We need to calculate the radius inside the capacitor

`(0.80B_(max))/(B_(max))=(r_(1))/(R)`

`r_(1)=0.80 R`

`r_(1)=0.80*35*10^(-3)`

`r_(1)=0.028\ m`

`r_(1)=28\ mm`

The radius inside the capacitor is 28 mm.

(b). We need to calculate the radius outside the capacitor

`(0.80B_(max))/(B_(max))=((r_2)/(R))^(-1)`

`r_(2)=(35)/(0.80)`

`r_(2)=43.75\ mm`

The radius outside the capacitor is 43.75 mm.

(c). We need to calculate the maximum value

Using formula of magnetic field

`B=(\mu_(0)I)/(2\pi R)`

Put the value into the formula

`B=(4\pi*10^(-7)*3.5)/(2*3.14*35*10^(-3))`

`B=2.00*10^(-5)\ T`

The maximum value of magnetic field is
`2.00*10^(-5)\ T`.

Hence, This is the required solution.