Question

A point $P$ lies on the axis of a flat coil carrying a current. The magnetic moment of the coil is $\mu .$ What will be the magnetic field at point $P$ ? It is given that the distance of $ \, P$ from the centre of coil is $d$ , which is large compared to the radius of the coil.

• A. $\frac{\left(\mu \right)_{0}}{2 \pi }\left(\frac{\mu }{d^{3}}\right)$
• B. $\frac{\left(\mu \right)_{0}}{4 \pi }\left(\frac{\mu }{d^{3}}\right)$
• C. $\frac{\left(\mu \right)_{0}}{6 \pi }\left(\frac{\mu }{d^{2}}\right)$
• D. $\frac{\left(\mu \right)_{0}}{8 \pi }\left(\frac{\mu }{d^{2}}\right)$

Answer 1

Answer: (A)

The magnetic field at $P$ due to the flat coil of $n$ turns, radius $r$ , carrying current i is
$\mathrm{B}=\frac{\mu_0}{2} \cdot \frac{\operatorname{nir}^2}{\left(\mathrm{~d}^2+\mathrm{r}^2\right)^{3 / 2}} \cong \frac{\mu_0}{2} \cdot \frac{\operatorname{nir}^2}{\mathrm{~d}^3}(\mathrm{~d}>>\mathrm{r})$
$=\frac{\mu_0}{2 \pi} \cdot \frac{\mathrm{n}\left(\pi \mathrm{r}^2\right) \mathrm{i}}{\mathrm{d}^3}=\frac{\mu_0}{2 \pi} \cdot \frac{\mu}{\mathrm{d}^3}$