MHT CET · Physics · Magnetic Effects of Current
A long wire carrying a steady current is bent into a circle of single turn. The magnetic field at the centre of the coil is ' B '. If it is bent into a circular loop of radius ' \(\mathrm{r}_1{ }^{\prime}\) having ' n ' turns, the magnetic field at the centre of the coil for same current is
- A \(\frac{B}{n^2}\)
- B \(\frac{B}{n}\)
- C \(\mathrm{n}^2 \mathrm{~B}\)
- D nB
Answer & Solution
Correct Answer
(C) \(\mathrm{n}^2 \mathrm{~B}\)
Step-by-step Solution
Detailed explanation
\(R_1 = \frac{L}{2\pi}\) \(B = \frac{\mu_0 I}{2R_1} = \frac{\mu_0 I}{2(L/2\pi)} = \frac{\mu_0 I \pi}{L}\)
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