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

$\int E dl = -\frac{d\phi}{dt}$

$\int E dl = -(\frac{dB}{dt})A$

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

$E (2\pi r) = \mu_0 n (\frac{di}{dt})(\pi R^2)$

Where,

R = Radius of the solenoid

r = The distance from the axis

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

$\frac{di}{dt} = \frac{Er}{\mu_0 NR^2}$

Replacing our values we have

$\frac{di}{dt} = \frac{(8.00*10^{-6})(0.0348)}{(4\pi*10^{-7})(390)(1.2*10^-2)^2}$

$\frac{di}{dt} = 3.94487A/s$