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
$\mathrm{B}=\frac{{\mu }_0}{2}\cdot \frac{{\mathrm{nir}}^2}{{\left({\mathrm{d}}^2+{\mathrm{r}}^2\right)}^{3/2}}\cong \frac{{\mu }_0}{2}\cdot \frac{{\mathrm{nir}}^2}{{\mathrm{d}}^3}(\mathrm{d}>>\mathrm{r})$
$=\frac{{\mu }_0}{2\pi }\cdot \frac{\mathrm{n}\left(\pi {\mathrm{r}}^2\right)\mathrm{i}}{{\mathrm{d}}^3}=\frac{{\mu }_0}{2\pi }\cdot \frac{\mu }{{\mathrm{d}}^3}$