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Q. If $B_c$ is the magnetic induction at the centre of a circular coil carrying current, then the magnetic induction at a point on the axis of the coil at a distance equal to the radius of the coil is

KEAMKEAM 2021

Solution:

At a point situated at a distance $r$ from the centre of a current carrying circular coil along the axial line, the magnetic field is given as
$B=\frac{\mu_0 i R^2}{2\left(R^2+r^2\right)^{3 / 2}}$
where, $i$ is the current in the coil and $R$ is the radius of the coil.
Given, $r=R$
$ \Rightarrow B=\frac{\mu_0 i R^2}{2\left(R^2+R^2\right)^{3 / 2}} $
$=\frac{\mu_0 i}{2 \cdot 2^{3 / 2} R} \ldots \text {(i) }$
Since, magnetic field due to a current carrying circular coil at centre is given as
$B=\frac{\mu_0 i}{2 R}=B_c \text { (given) ...(ii) }$
From Eqs. (i) and (ii), we get
$B=\frac{B_c}{2 \sqrt{2}}$