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What will happen to the inductance L of a solenoid when the number of turns and the length are doubled keeping the area of cross – section same?

Option: 1

\frac{L}{2}


Option: 2

L


Option: 3

2L


Option: 4

4L


Answers (1)

best_answer

In case of a solenoid as \mathrm{B}=\mu_0 \mathrm{nI}

\phi=\mathrm{B}(\mathrm{n} / \mathrm{S})=\mu_0 \mathrm{n}^2 l \mathrm{SI} and hence

\mathrm{L}=\frac{\phi}{\mathrm{I}}=\mu_0 \mathrm{n}^2 l \mathrm{~S}=\mu_0 \frac{\mathrm{N}^2}{l} \mathrm{~S} \quad\left(\text { as } \mathrm{n}=\frac{\mathrm{N}}{l}\right)

When N and l are doubled, then

\mathrm{L}^{\prime}=\mu_0 \frac{(2 \mathrm{~N})^2}{2 l} \mathrm{~S}=2 \mu_0 \frac{\mathrm{N}^2}{l} \mathrm{~S}=2 \mathrm{~L}

i.e., inductance of the solenoid will be doubled.

Posted by

Rakesh

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