The coefficient of self-inductance of a solenoid is 0.18 mH. If a core of soft iron of relative permeability 900 is inserted, then the coefficient of self-inductance will become nearly

Question

The coefficient of self-inductance of a solenoid is 0.18 mH. If a core of soft iron of relative permeability 900 is inserted, then the coefficient of self-inductance will become nearly (A) 5.4 mH (B) 162 mH (C) 0.006 mH (D) 0.0002 mH

Tardigrade Admin · Accepted Answer

L propto μ ⇒ (L2/L1)=(μ/μ0) ( n and i are same) ⇒ L2=μr L1=900 × 0.18=162 mH

Q. The coefficient of self-inductance of a solenoid is $0.18 m H$ . If a core of soft iron of relative permeability $900$ is inserted, then the coefficient of self-inductance will become nearly

5.4 mH

162 mH

0.006 mH

0.0002 mH

$L \propto μ \Rightarrow \frac{L _{2}}{L _{1}} = \frac{μ}{μ _{0}}$
( $n$ and $i$ are same)
$\Rightarrow L_{2} = μ_{r} L_{1} = 900 \times 0.18 = 162 m H$

Q. The coefficient of self-inductance of a solenoid is 0.18mH. If a core of soft iron of relative permeability 900 is inserted, then the coefficient of self-inductance will become nearly