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Q.

The magnetic field normal to the plane of a wire of n turns and radius r which carries a current l is measured on the axis of the coil at a small distance h from the centre of the coil. This is smaller than the magnetic field at the centre by the fraction

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a

$\frac{3 r^{2}}{2 h^{2}}$

b

$\frac{2 r^{2}}{3 h^{2}}$

c

$\frac{3 h^{2}}{2 r^{2}}$

d

$\frac{2 h^{2}}{3 r^{2}}$

answer is D.

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Detailed Solution

As $B_{c} = \frac{μ_{0}}{4 π} \frac{2 π n I}{r}$ and $B_{h} = \frac{μ_{0}}{4 π} \frac{2 π n I r^{2}}{{(r^{2} + h^{2})}^{3 / 2}}$ so $\frac{B_{h}}{B_{c}} = {(1 + \frac{h^{2}}{r^{2}})}^{- 3 / 2}$

Fractional decrease in the magnetic field will be

$= \frac{B_{c} - B_{h}}{B_{c}} = (1 - \frac{B_{h}}{B_{c}}) = [1 - {(1 + \frac{h^{2}}{r^{2}})}^{- 3 / 2}] = 1 - (1 - \frac{3 h^{2}}{2 r^{2}}) = \frac{3 h^{2}}{2 r^{2}}$

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