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(C) The magnetic flux density at the centre of a circular coil of radius $R$ carrying current $i$ is given by $B_0 = \frac{\mu_0 i}{2R}$.
At a distan

Vedclass · Accepted Answer

$\frac{B_0}{(1 + p^2)^{3/2}}$

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(C) The magnetic flux density at the centre of a circular coil of radius $R$ carrying current $i$ is given by $B_0 = \frac{\mu_0 i}{2R}$.
At a distance $x = pR$ from the centre on the axis,the magnetic flux density $B$ is given by $B = \frac{\mu_0 i R^2}{2(R^2 + x^2)^{3/2}}$.
Substituting $x = pR$ into the formula,we get $B = \frac{\mu_0 i R^2}{2(R^2 + (pR)^2)^{3/2}}$.
$B = \frac{\mu_0 i R^2}{2(R^2(1 + p^2))^{3/2}} = \frac{\mu_0 i R^2}{2R^3(1 + p^2)^{3/2}}$.
$B = \frac{\mu_0 i}{2R} \cdot \frac{1}{(1 + p^2)^{3/2}}$.
Since $B_0 = \frac{\mu_0 i}{2R}$,we have $B = \frac{B_0}{(1 + p^2)^{3/2}}$.

The flux density obtained at the centre of a circular coil of radius $R$ which carries a current $i$ is $B_0$. At a distance $pR$ from the centre on the axis,the flux density will be

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