Question

The magnetic field inside a 5.0-cm-diameter solenoid is 2.0 T and decreasing at 4.60 T/s. Part A What is the electric field strength inside the solenoid at a point on the axis

Answer 1

Complete Question

The magnetic field inside a 5.0-cm-diameter solenoid is 2.0 T and decreasing at 4.60 T/s.

Part A What is the electric field strength inside the solenoid at a point on the axis?

Part B

What is the electric field strength inside the solenoid at a point 1.50 cm from the axis?

Answer:

Part A

`E = 0 \ V/m`

Part B

`E_(15) = 0.0345 \ V/m`

Step-by-step explanation:

From the question we are told that

The diameter of the solenoid is
`d = 5.0 \ cm = 0.05 \ m`

The magnetic field is
`B = 2.0 \ T`

The rate of the change of the magnetic field is
`(dB)/(dt) = 4.60 \ T/s`

The radius of the solenoid is mathematically represented as

`R = ( d)/(2)`

substituting values

`R = ( 5.0 *10^(-2))/(2) = 0.025 \ m`

Generally the of the solenoid is mathematically represented as

`E = ( r)/(2) * |(dB)/(dt) |`

Now at the point on axis is r = 0 given that the axis is the origin so

`E = ( 0)/(2) * |(dB)/(dt) |`

`E = 0 \ V/m`

Now the electric field strength inside the solenoid at a point 1.50cm from the axis is mathematically represented as

`E_(15) = ( 15*10^(-2 ))/(2) * |4.60 |`

`E_(15) = 0.0345 \ V/m`