Question

An electric field of 4.0 μV/m is induced at a point 2.0 cm from the axis of a long solenoid (radius = 3.0 cm, 800 turns/m). At what rate is the current in the solenoid changing at this instant?

Answer 1

The rate of current in the solenoid is 0.398 A/s

Step-by-step explanation:

Given that,

Electric field
`E = 4.0\ \mu V/m`

Distance = 2.0 cm

Radius = 3.0 cm

Number of turns per unit length = 800

We need to calculate the rate of current

Using formula of electric field for solenoid

`E = (x)/(2)\mu_(0)n(dI)/(dt)`

Where, x = distance

n = number of turns per unit length

E = electric field

r = radius

Put the value into the formula

`4.0*10^(-6)=(2.0*10^(-2))/(2)*4\pi*10^(-7)*800*(dI)/(dt)`

`(dI)/(dt)=(4.0*10^(-6)*2)/(2.0*10^(-2)*4\pi*10^(-7)*800)`

`(dI)/(dt)=0.397\ A/s`

Hence, The rate of current in the solenoid is 0.398 A/s.