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
$B=\frac{\left(\mu \right)_{0}}{2}\cdot \frac{\left(nir\right)^{2}}{\left(d^{2} + r^{2}\right)^{3 / 2}}\cong\frac{\left(\mu \right)_{0}}{2}\cdot \frac{\left(nir\right)^{2}}{d^{3}}\left(\right.d>>r\left.\right)$
$=\frac{\left(\mu \right)_{0}}{2 \pi }\cdot \frac{n \left(\left(\pi r\right)^{2}\right) i}{d^{3}}=\frac{\left(\mu \right)_{0}}{2 \pi }\cdot \frac{\mu }{d^{3}}$