Question

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

Answer 1

Answer:

The self inductance of wire is 0.2512 m H

Explanation:

Given as :

The length of wire = 1 m = 1000 mm

The radius of wire = r = 2 mm

The relative permeability = $\mu$ = 20000

Let The self inductance = L  mH

According to question

Self inductance = $\dfrac{\mu \times N^{2}\times A }{l}$

where  $\mu$ is permeability

N is number of wire turns  , let N = 1

A , area of wire

l is length of wire

So, Area = π × r²

Or, A = 3.14 × 4

i.e A = 12.56 mm²

So, Self inductance = $\dfrac{\mu \times N^{2}\times A }{l}$

Or, L = $\dfrac{20000 \times 1\times 12.56}{1000}$

i.e L = 20 × 12.56

So, L =251.2 mm H

or, L = 0.2512 m H

Hence, The self inductance of wire is 0.2512 m H Answer