KCET · Physics · Magnetic Effects of Current
Magnetic field at the centre of a circular coil of radius \(R\) due to \(i\) flowing through it is \(B\). The magnetic field at a point along the axis at distance \(R\) from the centre is
- A \(\frac{B}{2}\)
- B \(\frac{B}{4}\)
- C \(\frac{\mathrm{B}}{\sqrt{8}}\)
- D \(\sqrt{8} \mathrm{~B}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{B}}{\sqrt{8}}\)
Step-by-step Solution
Detailed explanation
The magnetic field at a point along the axis at distance \(R\) from the centre of a circular coil of radius \(R\) carrying current \(i\) is
\(\begin{aligned}\mathrm{B}_{\mathrm{A}} &=\frac{\mu_{0} 2 \pi \mathrm{R}^{2}}{4 \pi\left(\mathrm{R}^{2}+\mathrm{R}^{2}\right)^{3 / 2}} \\&=\frac{\mu_{0} \mathrm{i}}{2 \sqrt{8} \mathrm{R}} \\&=\frac{\mathrm{B}}{\sqrt{8}} \quad\left[\mathrm{~B}_{\text {centre }}=\mathrm{B}=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{R}}\right]\end{aligned}\)
\(\begin{aligned}\mathrm{B}_{\mathrm{A}} &=\frac{\mu_{0} 2 \pi \mathrm{R}^{2}}{4 \pi\left(\mathrm{R}^{2}+\mathrm{R}^{2}\right)^{3 / 2}} \\&=\frac{\mu_{0} \mathrm{i}}{2 \sqrt{8} \mathrm{R}} \\&=\frac{\mathrm{B}}{\sqrt{8}} \quad\left[\mathrm{~B}_{\text {centre }}=\mathrm{B}=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{R}}\right]\end{aligned}\)
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