MHT CET · Physics · Magnetic Effects of Current
A long wire carries a steady current. It is bent into a coil of one turn such that magnetic induction at centre is \(B\), then same wire is bent to from a coil of smaller radius of \(n\) turns such that magnetic induction at centre is \(B\), then
- A \(B^{\prime}=B / n^2\)
- B \(B^{\prime}=n B\)
- C \(B^{\prime}=B\)
- D \(B^{\prime}=n^2 B\)
Answer & Solution
Correct Answer
(D) \(B^{\prime}=n^2 B\)
Step-by-step Solution
Detailed explanation
Initially, \(B=\frac{\mu_0 I}{2 r}\)
For coil with smaller radius \(r^{\prime}=\frac{r}{n}\), such that \(2 \pi r=n\left(2 \pi r^{\prime}\right)\) :
\(\begin{aligned}
& \Rightarrow B^{\prime}=n\left(\frac{\mu_0 I}{2 r^{\prime}}\right)=\frac{\mu_0 n I}{2(r / n)} \\
& \Rightarrow B^{\prime}=n^2 B
\end{aligned}\)
For coil with smaller radius \(r^{\prime}=\frac{r}{n}\), such that \(2 \pi r=n\left(2 \pi r^{\prime}\right)\) :
\(\begin{aligned}
& \Rightarrow B^{\prime}=n\left(\frac{\mu_0 I}{2 r^{\prime}}\right)=\frac{\mu_0 n I}{2(r / n)} \\
& \Rightarrow B^{\prime}=n^2 B
\end{aligned}\)
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