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?

15 H

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2.5 H

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3.5 H

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0.5 H

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Solution

The correct option is A $15 H$
Given : $μ_{r} = 6000$ $N = 2000$ turns $l = 0.3 m$
Cross-section area of solenoid $A = 1.5 \times 10^{- 4} m^{2}$
Self inductance $L = \frac{μ_{r} μ_{o} N^{2} A}{l}$ where $μ = μ_{r} μ_{o}$
$∴$ $L = \frac{6000 \times 4 π \times 10^{- 7} \times (2000)^{2} \times 1.5 \times 10^{- 4}}{0.3} = 15 H$

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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?

The correct option is A 15HGiven : μr=6000 N=2000 turns l=0.3m Cross-section area of solenoid A=1.5×10−4m2Self inductance L=μrμoN2Al where μ=μrμo∴ L=6000×4π×10−7×(2000)2×1.5×10−40.3=15H

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?