MHT CET · Physics · Magnetic Effects of Current
A circular coil of radius ' \(r\) ' and number of turns ' \(n\) ' carries a current ' \(I\) '. The magnetic fields at a small distance ' \(h\) ' along the axis of the coil \(\left(B_a\right)\) and at the centre of the coil \(\left(\mathrm{B}_{\mathrm{c}}\right)\) are measured. The ralation between \(B_c\) and \(B_a\) is
- A \(\mathrm{B}_{\mathrm{c}}=\mathrm{B}_{\mathrm{a}}\left(1+\frac{\mathrm{h}^2}{\mathrm{r}^2}\right)\)
- B \(\mathrm{B}_{\mathrm{c}}=\mathrm{B}_{\mathrm{a}}\left(1+\frac{\mathrm{h}^2}{\mathrm{r}^2}\right)^{\frac{1}{2}}\)
- C \(\mathrm{B}_{\mathrm{c}}=\mathrm{B}_{\mathrm{a}}\left(1+\frac{\mathrm{h}^2}{\mathrm{r}^2}\right)^{\frac{3}{2}}\)
- D \(\mathrm{B}_{\mathrm{c}}=\mathrm{B}_{\mathrm{a}}\left(1+\frac{\mathrm{h}^2}{\mathrm{r}^2}\right)^{-\frac{3}{2}}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{B}_{\mathrm{c}}=\mathrm{B}_{\mathrm{a}}\left(1+\frac{\mathrm{h}^2}{\mathrm{r}^2}\right)^{\frac{3}{2}}\)
Step-by-step Solution
Detailed explanation
Magnetic field along the axis of the coil is:
\(
\mathrm{B}_{\mathrm{a}}=\frac{\mu_0}{4 \pi}\left[\frac{2 \pi \mathrm{nIr} \mathrm{r}^2}{\left(\mathrm{r}^2+\mathrm{h}^2\right)^{\frac{3}{2}}}\right]
\)
Magnetic field at the centre of the coil is:
\(
\begin{aligned}
\mathrm{B}_{\mathrm{C}} & =\frac{\mu_0}{4 \pi}\left(\frac{2 \pi \mathrm{nI}}{\mathrm{r}}\right) \\
\frac{\mathrm{B}_{\mathrm{C}}}{\mathrm{B}_{\mathrm{a}}} & =\frac{\left(\mathrm{r}^2+\mathrm{h}^2\right)^{\frac{3}{2}}}{\mathrm{r}^3} \\
\therefore \quad \mathrm{B}_{\mathrm{C}} & =\mathrm{B}_{\mathrm{a}}\left[1+\frac{\mathrm{h}^2}{\mathrm{r}^2}\right]^{\frac{3}{2}}
\end{aligned}
\)
\(
\mathrm{B}_{\mathrm{a}}=\frac{\mu_0}{4 \pi}\left[\frac{2 \pi \mathrm{nIr} \mathrm{r}^2}{\left(\mathrm{r}^2+\mathrm{h}^2\right)^{\frac{3}{2}}}\right]
\)
Magnetic field at the centre of the coil is:
\(
\begin{aligned}
\mathrm{B}_{\mathrm{C}} & =\frac{\mu_0}{4 \pi}\left(\frac{2 \pi \mathrm{nI}}{\mathrm{r}}\right) \\
\frac{\mathrm{B}_{\mathrm{C}}}{\mathrm{B}_{\mathrm{a}}} & =\frac{\left(\mathrm{r}^2+\mathrm{h}^2\right)^{\frac{3}{2}}}{\mathrm{r}^3} \\
\therefore \quad \mathrm{B}_{\mathrm{C}} & =\mathrm{B}_{\mathrm{a}}\left[1+\frac{\mathrm{h}^2}{\mathrm{r}^2}\right]^{\frac{3}{2}}
\end{aligned}
\)
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