Question

A technician wraps wire around a tube of length 33 cm having a diameter of 8.1 cm. When the windings are evenly spread over the full length of the tube, the result is a solenoid containing 590 turns of wire. (a) Find the self-inductance of this solenoid. (b) If the current in this solenoid increases at the rate of 3 A/s, what is the self-induced emf in the solenoid?

Answer 1

Explanation:

It is given that,

Length of the tube, l = 33 cm = 0.33 m

Diameter of the tube, d = 8.1 cm = 0.081 m

Number of turns, N = 590

(a) We need to find the self- inductance of the solenoid. The formula for the self inductance is given by :

$L=\dfrac{\mu_o N^2A}{l}$

A is the area of the solenoid

$L=\dfrac{4\pi \times 10^{-7}\times (590)^2\times \pi \times (0.0405)^2}{0.33}$

L = 0.00683 H

or

L = 6.83 mH

So, the self inductance of the solenoid is 6.83 mH.

(b) The current in this solenoid increases at the rate of 3 A/s. $\dfrac{dI}{dt}=3\ A/s$

EMF in the solenoid is given by :

$E=-L\dfrac{dI}{dt}$

$E=-6.83\times 3$

E = -20.49 volt

Hence, this is the required solution.