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

Answer:

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

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

$E = \dfrac{x}{2}\mu_{0}n\dfrac{dI}{dt}$

Where, x = distance

n = number of turns per unit length

E = electric field

r = radius

Put the value into the formula

$4.0\times10^{-6}=\dfrac{2.0\times10^{-2}}{2}\times4\pi\times10^{-7}\times800\times\dfrac{dI}{dt}$

$\dfrac{dI}{dt}=\dfrac{4.0\times10^{-6}\times2}{2.0\times10^{-2}\times4\pi\times10^{-7}\times800}$

$\dfrac{dI}{dt}=0.397\ A/s$

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