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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

$R \sqrt{3}$

c

$2 R \sqrt{3}$

d

$(2 \sqrt{3}) R$

answer is B.

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Detailed Solution

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$B_{axis} = \frac{μ_{0}}{4 π} \times \frac{2 π I R^{2}}{{(R^{2} + x^{2})}^{3 / 2}} At centre, B_{centre} = \frac{μ_{0}}{4 π} \times \frac{2 π I}{R} I n t h e g i v e n p r o b l e m, \frac{μ_{0}}{4 π} \times \frac{2 π I R^{2}}{{(R^{2} + x^{2})}^{3 / 2}} = \frac{1}{8} [\frac{μ_{0}}{4 π} \times \frac{2 π I}{R}] or {(R^{2} + x^{2})}^{3 / 2} = 8 R^{3} Solving, we get x = R \sqrt{3}$

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