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Q.

A circular current carrying coil has a radius R. The distance from the center of the coil on the axis where the magnetic induction will be ${(1 / 8)}^{th}$ of its value at the center of the coil, is

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a

$R / \sqrt{3}$

b

$2 R \sqrt{3}$

c

$R \sqrt{3}$

d

$(2 \sqrt{3}) R$

answer is B.

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Detailed Solution

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$B_{a x i s} = \frac{μ_{0}}{4 π} \times \frac{2 π I R^{2}}{{(R^{2} + x^{2})}^{3 / 2}}$

At center, $B_{c e n t r e} = \frac{μ_{0}}{4 π} \times \frac{2 π I}{R}$
In the given problem,

$B_{a x i a l} = \frac{1}{8} B_{c e n t e r}$
$\Rightarrow \frac{μ_{0}}{4 π} \times \frac{2 π I R^{2}}{{(R^{2} + x^{2})}^{3 / 2}} = \frac{1}{8} [\frac{μ_{0}}{4 π} \times \frac{2 π I}{R}]$
Or ${(R^{2} + x^{2})}^{3 / 2} = 8 R^{3}$
Solving, we get $x = R \sqrt{3}$ .

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