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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25 H

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35 H

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5 H

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Solution

The correct option is C $15 H$
Given : $μ_{r} = 6000$ , $l = 30 c m = 0.3 m$ , $N = 2000$
Cross section of solenoid $A = 1.5 c m^{2} = 1.5 \times 10^{- 4} m^{2}$

As we know, Self-inductance of the solenoid
$L = \frac{μ_{r} \cdot μ_{0} N^{2} A}{l}$
$= \frac{6000 \times 4 π \times 10^{- 7} \times {(2000)}^{2} \times (1.5) \times 10^{- 4}}{0.3}$
$= 15 H$

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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 C 15HGiven : μr=6000, l=30cm=0.3m, N=2000 Cross section of solenoid A=1.5cm2=1.5×10−4m2As we know, Self-inductance of the solenoidL=μr⋅μ0N2Al=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?