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Q.

A circular coil of wire of radius ‘r’ has ‘n’ turns and carries a current ‘I’. The magnetic induction (B) at a point on the axis of the coil at a distance $\sqrt{3 r}$ from its centre is

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a

$\frac{μ_{0} In}{8 r}$

b

$\frac{μ_{0} In}{32 r}$

c

$\frac{μ_{0} nI}{16 r}$

d

$\frac{μ_{0} In}{4 r}$

answer is C.

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Detailed Solution

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$\begin{array}{l} B = \frac{μ_{0} {nir}^{2}}{2 {(r^{2} + x^{2})}^{\frac{3}{2}}} \\ B = \frac{μ_{0} {nir}^{2}}{2 {(r^{2} + 3 r^{2})}^{\frac{3}{2}}} = \frac{μ_{0} \times n \times i \times r^{2}}{2 \times 8 r^{3}} = \frac{μ_{0} ni}{16 r} \end{array}$

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