MHT CET · Physics · Electromagnetic Induction
A long solenoid has 1500 turns. When a current of \(3.5 \mathrm{~A}\) flows through it, the magnetic flux linked with each turn of solenoid is \(2.8 \times 10^{-3}\) weber. The self-inductance of solenoid is
- A \(1.2 \mathrm{H}\)
- B \(2.4 \mathrm{H}\)
- C \(3.6 \mathrm{H}\)
- D \(6 \mathrm{H}\)
Answer & Solution
Correct Answer
(A) \(1.2 \mathrm{H}\)
Step-by-step Solution
Detailed explanation
Flux linked with each turn of the solenoid is, \(\phi=2.8 \times 10^{-3} \mathrm{~Wb}\)
\(\therefore \quad\) Total magnetic flux of the solenoid is,
\(\begin{aligned}
& \phi_{\mathrm{net}}=\mathrm{N} \phi \\
& \phi_{\mathrm{net}}=1500 \times 2.8 \times 10^{-3}=4.2 \mathrm{~Wb}
\end{aligned}\)
Flux can also be given by, \(\phi=\mathrm{LI}\)
\(\therefore \quad\) The self-inductance of the solenoid is,
\(\begin{aligned}
& \mathrm{L}=\frac{\phi}{\mathrm{I}}=\frac{4.2}{3.5} \\
& \mathrm{~L}=1.2 \mathrm{H}
\end{aligned}\)
\(\therefore \quad\) Total magnetic flux of the solenoid is,
\(\begin{aligned}
& \phi_{\mathrm{net}}=\mathrm{N} \phi \\
& \phi_{\mathrm{net}}=1500 \times 2.8 \times 10^{-3}=4.2 \mathrm{~Wb}
\end{aligned}\)
Flux can also be given by, \(\phi=\mathrm{LI}\)
\(\therefore \quad\) The self-inductance of the solenoid is,
\(\begin{aligned}
& \mathrm{L}=\frac{\phi}{\mathrm{I}}=\frac{4.2}{3.5} \\
& \mathrm{~L}=1.2 \mathrm{H}
\end{aligned}\)
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