Question

A bar magnet of length $10$ cm and pole strength $20$ A m is deflected through ${30}^0$ from the magnetic meridian. If earth's horizontal component field is $\frac{320}{4\pi }T$. Find the moment of the deflecting couple.

Answer 1

For voltage $V$ and electron mass and charge=$m$ and $e$

Electric field $E=\frac{V}{r}$

Force exerted on the electrons $F=eE$

Acceleration of electrons=$a=\frac{eV}{rm}$

From equations of motion:$=v^2=u^2+2as⟹s=\frac{v^2}{2a}$

$⟹v=\sqrt{\frac{2eV}{m}}$

Time taken by electron from circular path, $T=\frac{2\pi m}{eB}$

Velocity remains $V$, hence distance $d=v\times T$

$=\sqrt{\frac{2eV}{m}}\times \frac{2\pi m}{eB}$

$=\sqrt{\frac{8{\pi }^{2}mV}{e{B}^2}}⟹N=8$