Question

A charge of $1$ coulomb is placed at one end of a non-conducting rod of length $0.6m$. The rod is rotated in a vertical plane about a horizontal axis passing through the other end of the rod with an angular velocity ${10}^4\pi$ rad/sec. Find the magnetic field at a point on the axis of rotation at a distance of $0.8m$ from the center of the path is $N\pi \times {10}^{-4}T$, then find the value of $N$.

Answer 1

Magnetic field on the axis of a circular coil of radius R and magnetic moment M is given by

$B=\frac{{\mu }_0}{4\pi }\ast \frac{2M}{(R^2+x^2{)}^{\frac{3}{2}}}$

In our case

$I=\frac{q}{t}=\frac{q}{\left(\frac{2\pi }{w}\right)}$

$\therefore I=\frac{qw}{2\pi }=\frac{1\ast {10}^4\pi }{2\pi }$

$=0.5\ast {10}^{4}A$

$R=0.6m$

$x=0.8m$

and $M=I\pi R^2=0.5\ast {10}^4\ast \pi \ast (0.6{)}^2$

$=0.18\pi \ast {10}^4$

$\therefore B={10}^{-7}\ast \frac{2\ast 0.18\pi \ast {10}^4}{\left[\right(0.6{)}^2+(0.8{)}^2{]}^{\frac{3}{2}}}$

$=\frac{1.1304\ast {10}^{-3}}{(1{)}^{\frac{3}{2}}}=1.1304\ast {10}^{-3}T$