The needle of a deflection galvanometer shows | vedclass

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(C) For a short bar magnet in $	an A$ position,the magnetic field $B$ is given by $B = \frac{\mu_0}{4\pi} \frac{2M}{d^3}$.In a deflection galvanomete

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$	an^{-1}\left(\frac{\sqrt{3}}{8}ight)$

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(C) For a short bar magnet in $	an A$ position,the magnetic field $B$ is given by $B = \frac{\mu_0}{4\pi} \frac{2M}{d^3}$.In a deflection galvanometer,$B = H 	an 	heta$,where $H$ is the horizontal component of Earth's magnetic field.Thus,$\frac{\mu_0}{4\pi} \frac{2M}{d^3} = H 	an 	heta$ ..... $(i)$When the distance is doubled $(d' = 2d)$,the new deflection $	heta'$ is given by:$\frac{\mu_0}{4\pi} \frac{2M}{(2d)^3} = H 	an 	heta'$ ..... $(ii)$Dividing equation $(ii)$ by $(i)$:$\frac{	an 	heta'}{	an 	heta} = \frac{1}{2^3} = \frac{1}{8}$$	an 	heta' = \frac{	an 60^\circ}{8} = \frac{\sqrt{3}}{8}$$	heta' = 	an^{-1}\left(\frac{\sqrt{3}}{8}ight)$

The needle of a deflection galvanometer shows a deflection of $60^\circ$ due to a short bar magnet at a certain distance in $\tan A$ position. If the distance is doubled,the deflection is

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