Question:
A solenoid has the length \(1\ \text{m}\) and the area of cross section \(0.02\ \text{m}^2\). If the number of turns in the solenoid is \(5000\), then the self inductance of the solenoid is
A solenoid has the length \(1\ \text{m}\) and the area of cross section \(0.02\ \text{m}^2\). If the number of turns in the solenoid is \(5000\), then the self inductance of the solenoid is
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For a long air-core solenoid,
\[
L=\frac{\mu_0N^2A}{l}.
\]
Self inductance is directly proportional to the square of the number of turns and the cross-sectional area, and inversely proportional to the length of the solenoid.
Updated On: Jun 26, 2026
- \(0.2\pi\ \text{henry}\)
- \(0.4\pi\ \text{henry}\)
- \(0.02\pi\ \text{henry}\)
- \(0.04\pi\ \text{henry}\)
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The Correct Option is A
Solution and Explanation
Step 1: Use the formula for self inductance of a solenoid.
The self inductance of a long solenoid is given by \[ L=\frac{\mu_0N^2A}{l} \] where \[ \mu_0=4\pi\times10^{-7}\ \text{H m}^{-1}, \] \[ N=5000, \] \[ A=0.02\ \text{m}^2, \] \[ l=1\ \text{m}. \]
Step 2: Substitute the given values.
\[ L = \frac{(4\pi\times10^{-7})(5000)^2(0.02)}{1} \] Since \[ (5000)^2 = 25\times10^6, \] we get \[ L = (4\pi\times10^{-7}) (25\times10^6) (0.02) \]
Step 3: Simplify the numerical part.
\[ 25\times0.02 = 0.5 \] Therefore, \[ L = 4\pi\times10^{-7}\times0.5\times10^6 \] \[ L = 2\pi\times10^{-1} \] \[ L = 0.2\pi\ \text{H} \]
Step 4: Final conclusion.
Hence, the self inductance of the solenoid is \[ \boxed{0.2\pi\ \text{henry}} \]
The self inductance of a long solenoid is given by \[ L=\frac{\mu_0N^2A}{l} \] where \[ \mu_0=4\pi\times10^{-7}\ \text{H m}^{-1}, \] \[ N=5000, \] \[ A=0.02\ \text{m}^2, \] \[ l=1\ \text{m}. \]
Step 2: Substitute the given values.
\[ L = \frac{(4\pi\times10^{-7})(5000)^2(0.02)}{1} \] Since \[ (5000)^2 = 25\times10^6, \] we get \[ L = (4\pi\times10^{-7}) (25\times10^6) (0.02) \]
Step 3: Simplify the numerical part.
\[ 25\times0.02 = 0.5 \] Therefore, \[ L = 4\pi\times10^{-7}\times0.5\times10^6 \] \[ L = 2\pi\times10^{-1} \] \[ L = 0.2\pi\ \text{H} \]
Step 4: Final conclusion.
Hence, the self inductance of the solenoid is \[ \boxed{0.2\pi\ \text{henry}} \]
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