Question

The self inductance of an iron core solenoid 3.14m long, cross sectional area 10^-3 m2 and having 500 turns
a)0.01mH
b)0.02mH
c)0.03mH
d)0.04mH

Answer 1

Dear Tulsi,

◆ Answer -

Self inductance = 10^-4 H

● Explaination -

# Given -

n = 500 turns

A = 10^-3 m^2

l = 3.14 m

# Solution -

Self inductance of solenoid coil is calculated by formula -

L = μ0.n².A / l

L = 4π×10^-7 × (500)² × 10^-3 / 3.14

L = 4 × 10^-7 × 25×10^4 × 10^-3

L = 10^-4 H

L = 0.1 H

Therefore, self inductance of the solenoid coil will be 10^-4 H.

Thanks dear...

Answer 2

Explanation:

option A is the correct answer