MHT CET · Physics · Electromagnetic Induction
When the number of turns in a coil is doubled without any change in the length of the coil, its self-inductance
- A becomes 4 times.
- B becomes 2 times.
- C gets halved.
- D remains unchanged.
Answer & Solution
Correct Answer
(A) becomes 4 times.
Step-by-step Solution
Detailed explanation
Self-inductance of solenoid is,
\(\frac{\mathrm{L}}{l}=\mu_0 \mathrm{~N}^2 \mathrm{~A}\)
where N is the total number of turns.
As \(\mathrm{L} \propto \mathrm{N}^2\)
\(\begin{array}{ll}
\therefore & \frac{\mathrm{L}_2}{\mathrm{~L}_1}=\left(\frac{\mathrm{N}_2}{\mathrm{~N}_1}\right)^2=(2)^2 \\
\therefore & \mathrm{~L}_2=4 \mathrm{~L}_1
\end{array}\)
\(\frac{\mathrm{L}}{l}=\mu_0 \mathrm{~N}^2 \mathrm{~A}\)
where N is the total number of turns.
As \(\mathrm{L} \propto \mathrm{N}^2\)
\(\begin{array}{ll}
\therefore & \frac{\mathrm{L}_2}{\mathrm{~L}_1}=\left(\frac{\mathrm{N}_2}{\mathrm{~N}_1}\right)^2=(2)^2 \\
\therefore & \mathrm{~L}_2=4 \mathrm{~L}_1
\end{array}\)
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