MHT CET · Physics · Magnetic Effects of Current
Two circular coils \(P\) and \(Q\) are made from similar wire, but radius of \(Q\) is twice that of \(P\). What should be the value of potential difference across them so that the magnetic induction at their centre may be same?
- A \(V_{q}=2 V_{p}\)
- B \(V_{q}=3 V_{p}\)
- C \(V_{q}=4 V_{p}\)
- D \(V_{q}=\frac{1}{4} V_{p}\)
Answer & Solution
Correct Answer
(C) \(V_{q}=4 V_{p}\)
Step-by-step Solution
Detailed explanation
\(B_{1}=\frac{\mu_{0}}{4 \pi}\left[\frac{2 \pi I_{1}}{T_{1}}\right]\) and \(B_{2}=\frac{\mu_{0}}{4 \pi}\left[\frac{2 \pi I_{2}}{r_{2}}\right]\)
Since, \(\quad B_{1}=B_{2}\) and \(r_{2}=2 r_{1}, I_{2}=2 l_{1}\)
\(
\frac{V_{2}}{V_{1}}=\frac{I_{2} R_{2}}{I_{1} R_{1}}=\frac{2 I_{1}}{I_{1}} \times \frac{2 R_{1}}{R_{1}}=\frac{4}{1}
\)
Since, \(\quad B_{1}=B_{2}\) and \(r_{2}=2 r_{1}, I_{2}=2 l_{1}\)
\(
\frac{V_{2}}{V_{1}}=\frac{I_{2} R_{2}}{I_{1} R_{1}}=\frac{2 I_{1}}{I_{1}} \times \frac{2 R_{1}}{R_{1}}=\frac{4}{1}
\)
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