An air-cored coil has a self-inductance of 0.1 H. A soft-iron core of relative permeability 1000 is introduced and the number of turns is reduced to 1 / 10 text th . The value of self-inductance now is

Question

An air-cored coil has a self-inductance of 0.1 H. A soft-iron core of relative permeability 1000 is introduced and the number of turns is reduced to 1 / 10 text th . The value of self-inductance now is (A) 0.1 H (B) 1 mH (C) 1 H (D) 10 mH

Tardigrade Admin · Accepted Answer

L0=(μ0 π N2 ⋅ r/2)=0.1 H L=(μ0 μr π((N/10))2 ⋅ r/2) As μr=1000 L=(1000/102)((μ0 π N2 ⋅ r/2))=10 L0=10 × 0.1 H =1 H

Q. An air-cored coil has a self-inductance of $0.1 H$ . A soft-iron core of relative permeability 1000 is introduced and the number of turns is reduced to $1/1 0^{th}$ . The value of self-inductance now is

$0.1 H$

$1 m H$

$1 H$

$10 m H$

$L_{0} = \frac{μ _{0} π N ^{2} \cdot r}{2} = 0.1 H$
$L = \frac{μ _{0} μ _{r} π ( \frac{N}{10} ) ^{2} \cdot r}{2}$
As $μ_{r} = 1000$
$L = \frac{1000}{1 0 ^{2}} (\frac{μ _{0} π N ^{2} \cdot r}{2}) = 10 L_{0} = 10 \times 0.1 H = 1 H$

Q. An air-cored coil has a self-inductance of 0.1H. A soft-iron core of relative permeability 1000 is introduced and the number of turns is reduced to 1/10th . The value of self-inductance now is

0.1H

1mH

1H

10mH