Question

A wire has length 1 m, radius 2mm and
relative permeability = 20000. Find its self inductance.
(1) 2 mH
(2) 1mH
(3) 3 mH
(4) 4 mH​

Answer 1

Answer:

40H.

Explanation:

The self inductance of a wire will be having a formulae of self inductance, L which is = μ(permeability)*n(number of turns per unit length)V(A*l which is volume or area * length).

The self inductance only depends on the physical quantities, the number of turns per unit length of the wire will be 1 since it is a single wire. So on substituting the values will get that the self inductance L= μnAl or 20000*1*2*10^-3 which will be 40H.