Question

A bar magnet is suspended by a thin wire in a uniform magnetic field. On twisting the upper end of wire by ${160}^o$, the magnet is displaced from its initial position by ${30}^o$. How much should the upper end of the wire be twisted so that the magnet is displaced by ${90}^o$ from its initial position?

Answer 1

in this question since the Upper end is displaced by ${160}^0$
and magnet is displaced by ${30}^0$
Therefore avg twisting is ${160}^0-{30}^0$=${130}^0$
Let C is constant for every $1^0$ turn then for every ${130}^0$
T=Cx${130}^0$
And for Magnet we have
F=MxBsin(30)
for equilibrium we have
F=T

MxB sin(30) = C x 130...................(1)

And upper end is displaced by $\theta$ and magnet by ${90}^0$
therefore MBsin90 = C($\theta$-90) ............(2)

dividing 2\3 we get
$\theta$-${90}^0$=${260}^0$
therefore $\theta$=${350}^0$