TS EAMCET · Physics · Electromagnetic Induction
A coil of wire of radius \(r\) has 600 turns and a self inductance of \(108 \mathrm{mH}\). The self inductance of a coil with same radius and 500 turns is
- A \(80 \mathrm{mH}\)
- B \(75 \mathrm{mH}\)
- C \(108 \mathrm{mH}\)
- D \(90 \mathrm{mH}\)
Answer & Solution
Correct Answer
(B) \(75 \mathrm{mH}\)
Step-by-step Solution
Detailed explanation
The given, \(L_1=108 \mathrm{mH}, N_1=600\) turns, \(N_2=500\) turns and \(L_2=\) ? By self-inductance of a plane coil \[ \begin{aligned} & L_1=\frac{\mu_0 \pi N_1^2 a_1}{2} \\ & L_2=\frac{\mu_0 \pi N_2^2 a_2}{2} \end{aligned} \] From the Eqs. (i) and (ii), we get…
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