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

Step-by-step 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=(\mu_o N^2A)/(l)`

A is the area of the solenoid

`L=(4\pi * 10^(-7)* (590)^2* \pi * (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.
`(dI)/(dt)=3\ A/s`

EMF in the solenoid is given by :

`E=-L(dI)/(dt)`

`E=-6.83* 3`

E = -20.49 volt

Hence, this is the required solution.