Question

A long, thin solenoid has 390 turns per meter and a radius of 1.20 cm. The current in the solenoid is increasing at a uniform rate didt. The magnitude of the induced electric field at a point which is near the center of the solenoid and a distance of 3.48 cm from its axis is 8.00×10−6 V/m.

Calculate di/dt.

Answer 1

To solve this problem it is necessary to apply the concepts related to Faraday's law and the induced emf.

By definition the induced electromotive force is defined as

`\int E dl = -(d\phi)/(dt)`

`\int E dl = -((dB)/(dt))A`

Where,

`\phi =` Electric field

B = Magnetic Field

A = Area

At the theory the magnetic field is defined as,

`B = \mu_0 NI`

Where,

N = Number of loops

I = current

`\mu_0 =` Permeability constant

We know also that the cross sectional area, is the area from a circle, and the length is equal to the perimeter then

A = \pi r^2

l = 2\pi r

Replacing at the previous equation we have that

`E (2\pi r) = \mu_0 n ((di)/(dt))(\pi R^2)`

Where,

R = Radius of the solenoid

r = The distance from the axis

Re-arrange to find the current in function of time,

`(di)/(dt) = (Er)/(\mu_0 NR^2)`

Replacing our values we have

`(di)/(dt) = ((8.00*10^(-6))(0.0348))/((4\pi*10^(-7))(390)(1.2*10^-2)^2)`

`(di)/(dt) = 3.94487A/s`