A deflection magnetometer is placed with its arm along the east-west direction ( tan A position) and a short bar magnet is placed symmetrically along its axis at some distance with its north pole poin

Question

A deflection magnetometer is placed with its arm along the east-west direction ( tan A position) and a short bar magnet is placed symmetrically along its axis at some distance with its north pole pointing towards east. In this position the needle of the magnetometer shows a deflection of 60 ° . If we double the distance of the bar magnet, then the deflection will be

TestHub · Accepted Answer

For a short bar magnet in tan A position,

$\frac{μ_{0}}{4 π} \frac{2 M}{d^{3}} = H tanθ$ ...(i)

When distance is doubled, then new deflection $θ'$ is given by,

$\frac{μ_{0}}{4 π} \frac{2 M}{{(2 d)}^{3}} = H {tanθ}^{'}$ ....(ii)

$∴ \frac{\tan θ^{'}}{\tan θ} = \frac{1}{8}$

$\Rightarrow {t a n θ}^{'} = \frac{t a n θ}{8} = \frac{\tan 60 °}{8} = \frac{\sqrt{3}}{8}$

$\Rightarrow θ^{'} = \tan^{- 1} [\frac{\sqrt{3}}{8}]$

Physics - Magnetism Question with Solution | TestHub

Options:

Doubts & Discussion