Question

Calculate the self inductance of a solenoid having 500 turns about a cylindrical core of 2 cm radius in which µr = 50 for 0 < rho < 0.5 cm.

Answer 1

Answer:

The self-inductance is given as,

L=

l

μ

0

N

2

A

=

50×10

−2

4π×10

−7

×(500)

2

×π×(2×10

−2

)

2

=7.896×10

−4

H

Thus, the self-inductance of an air solenoid is 7.896×10

−4

H.

Explanation:

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