If self inductance of air core solenoid increases from 0.04 mH to 16 mH on introducing an iron core into it, then what is relative magnetic permeability of the core?

Question

If self inductance of air core solenoid increases from 0.04 mH to 16 mH on introducing an iron core into it, then what is relative magnetic permeability of the core? (A) 400 (B) 500 (C) 600 (D) 700

Tardigrade Admin · Accepted Answer

As L0=(μ0 N2 A/ℓ), L=(μ N2 A/ℓ) (L propto μ) μ r =(μ/μ0) μ r =( L / L 0)=(16/0.04)=400

Q. If self inductance of air core solenoid increases from $0.04 m H$ to $16 m H$ on introducing an iron core into it, then what is relative magnetic permeability of the core?

400

500

600

700

As $L_{0} = \frac{μ _{0} N ^{2} A}{ℓ}$ ,
$L = \frac{μ N ^{2} A}{ℓ} (L \propto μ)$
$μ_{r} = \frac{μ}{μ _{0}}$
$μ_{r} = \frac{L}{L _{0}} = \frac{16}{0.04} = 400$

Q. If self inductance of air core solenoid increases from 0.04mH to 16mH on introducing an iron core into it, then what is relative magnetic permeability of the core?

400

500

600

700

As L0​=ℓμ0​N2A​, L=ℓμN2A​(L∝μ) μr​=μ0​μ​ μr​=L0​L​=0.0416​=400

Q. If self inductance of air core solenoid increases from $0.04 m H$ to $16 m H$ on introducing an iron core into it, then what is relative magnetic permeability of the core?

As $L_{0} = \frac{μ _{0} N ^{2} A}{ℓ}$ ,
$L = \frac{μ N ^{2} A}{ℓ} (L \propto μ)$
$μ_{r} = \frac{μ}{μ _{0}}$
$μ_{r} = \frac{L}{L _{0}} = \frac{16}{0.04} = 400$