A solenoid 30 cm long is made by winding 2000 loops of wire on an iron rod whose cross-section is 1.5 cm2 .If the relative permeability of the iron is 6000. What is the self-inductance of the solenoid?

Question

A solenoid 30 cm long is made by winding 2000 loops of wire on an iron rod whose cross-section is 1.5 cm2 .If the relative permeability of the iron is 6000. What is the self-inductance of the solenoid? (A) 1.5 H (B) 2.5 H (C) 3.5 H (D) 0.5 H

Tardigrade Admin · Accepted Answer

Self-inductance of the solenoid L =(μr ⋅ μ0 N2 A/I) =(600 × 4 π × 10-7 ×(2000)2 ×(1.5) × 10-4/0.3) =1.5 H

Q. A solenoid $30 c m$ long is made by winding $2000$ loops of wire on an iron rod whose cross-section is $1.5 c m^{2}$ .If the relative permeability of the iron is $6000$ . What is the self-inductance of the solenoid?

1.5 H

2.5 H

3.5 H

0.5 H

Self-inductance of the solenoid
$L = \frac{μ _{r} \cdot μ _{0} N ^{2} A}{I}$
$= \frac{600 \times 4 π \times 1 0 ^{- 7} \times ( 2000 ) ^{2} \times ( 1.5 ) \times 1 0 ^{- 4}}{0.3}$
$= 1.5 H$

Q. A solenoid 30cm long is made by winding 2000 loops of wire on an iron rod whose cross-section is 1.5cm2 .If the relative permeability of the iron is 6000. What is the self-inductance of the solenoid?