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?

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

Explanation:

Given that,

Electric field

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

Where, x = distance

n = number of turns per unit length

E = electric field

r = radius

Put the value into the formula

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

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