Atechnician 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?

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 :

A is the area of the solenoid

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.

EMF in the solenoid is given by :

E = -20.49 volt

Hence, this is the required solution.

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2

because growing process had stoped after 4-5 year