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

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Correct answer: C. ^ - 1 [ 3 8 ]

For a short bar magnet in tan A& position,

\dfrac{\mu0}{4\pi}\dfrac{2M}{d3} = Htan \theta

...(i)

When distance is doubled, then new deflection

\theta′

is given by,

\dfrac{\mu0}{4\pi}\dfrac{2M}{\left(2d\right)3} = H{tanθ}^{'}

....(ii)

∴ \dfrac{tan {θ}^{′}}{tan \theta} = \dfrac{1}{8}

⇒ {tan\theta}^{′} = \dfrac{tan\theta}{8} = \dfrac{\tan 60 °}{8} = \dfrac{\sqrt{3}}{8}

⇒

{\theta}^{′} = \tan^{- 1}$ \left[\dfrac{\sqrt{3}}{8}\right]

A deflection magnetometer is placed with its arm along the...

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