A circular coil of radius 'r' and number of t | vedclass

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(C) The magnetic field at a distance '$h$' along the axis of a circular coil is given by:$B_{a} = \frac{\mu_0 n I r^2}{2(r^2 + h^2)^{3/2}}$The magneti

Vedclass · Accepted Answer

$B_{c}=B_{a}\left(1+\frac{h^2}{r^2}\right)^{\frac{3}{2}}$

Vedclass · Answer

(C) The magnetic field at a distance '$h$' along the axis of a circular coil is given by:$B_{a} = \frac{\mu_0 n I r^2}{2(r^2 + h^2)^{3/2}}$The magnetic field at the centre of the coil $(h=0)$ is:$B_{c} = \frac{\mu_0 n I}{2r}$Taking the ratio of $B_c$ to $B_a$:$\frac{B_c}{B_a} = \frac{\mu_0 n I / 2r}{\mu_0 n I r^2 / 2(r^2 + h^2)^{3/2}}$$\frac{B_c}{B_a} = \frac{(r^2 + h^2)^{3/2}}{r^3} = \left(\frac{r^2 + h^2}{r^2}\right)^{3/2} = \left(1 + \frac{h^2}{r^2}\right)^{3/2}$Therefore,$B_c = B_a \left(1 + \frac{h^2}{r^2}\right)^{3/2}$.

$A$ circular coil of radius '$r$' and number of turns '$n$' carries a current '$I$'. The magnetic fields at a small distance '$h$' along the axis of the coil $(B_a)$ and at the centre of the coil $(B_c)$ are measured. The relation between $B_c$ and $B_a$ is

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