A circular current carrying coil produced a magnetic field of induction $1 μ T$ and $27 μ T$ at a point P on axis and at the centre of a coil respectively. If the radius of the coil is 20cm then distance of ‘P’ from center of the coil is

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$40 (\sqrt{2}) c m$

$20 (\sqrt{2}) c m$

$20 c m$

$10 \sqrt{2} c m$

answer is B.

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Detailed Solution

$B_{A} = \frac{B_{0}}{{(1 + \frac{x^{2}}{r^{2}})}^{3 / 2}}$

$1 μ T = \frac{27 μ T}{{(1 + \frac{x^{2}}{r^{2}})}^{3 / 2}}$

$(1 + \frac{x^{2}}{r^{2}}) = 3^{2} \Rightarrow x = 40 \sqrt{2} c m$

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