Question:

If a solenoid has \(500\) turns, length \(0.5\,\text{m}\) and cross-sectional area \(4\times10^{-4}\,\text{m}^2\), then its self-inductance is

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For a long solenoid, \[ \boxed{ L=\frac{\mu_0N^2A}{l} } \] where \(N\) is the number of turns, \(A\) is the cross-sectional area and \(l\) is the length.
Updated On: Jul 14, 2026
  • \(0.25\,\text{mH}\)
  • \(0.63\,\text{mH}\)
  • \(1.26\,\text{mH}\)
  • \(2.5\,\text{mH}\)
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The Correct Option is C

Solution and Explanation

Step 1: Use the formula for self-inductance of a solenoid. \[ L=\frac{\mu_0N^2A}{l}, \] where \[ \mu_0=4\pi\times10^{-7}\,\text{H/m}, \] \[ N=500,\qquad A=4\times10^{-4}\,\text{m}^2,\qquad l=0.5\,\text{m}. \]

Step 2:
Substitute the values. \[ L= \frac{4\pi\times10^{-7}\times(500)^2\times4\times10^{-4}}{0.5} \] \[ =1.26\times10^{-3}\,\text{H} =1.26\,\text{mH}. \] Hence, \[ \boxed{1.26\,\text{mH}} \] Therefore, \[ \boxed{(C)} \] is the correct answer.
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