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An air-cored coil has a self-inductance ...

An air-cored coil has a self-inductance of 0.1 H. A soft iron core of relative permeability 1000 is 1/10th. The value of self-inductance now becomes

A

1 mH

B

10 mH

C

1H

D

10 H

Text Solution

Verified by Experts

The correct Answer is:
C
From `L=(mu_(0)N^(2)A)/(l)=(mu_(0)u_(r)N^(2)A)/(l)`
when, `mu_(r)=1000 and `N becomes `(1)/(10)`
`therefore `L becomes `1000xx((1)/(10))^(2)=10` times
i.e. `L=10xx0.1=1H`
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MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-ELECTROMAGNETIC INDUCTION-Example 1
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