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Questions  >>  CBSE XII  >>  Physics  >>  Magnetism and matter

A short bar magnet placed in a horizontal plane has its axis aligned along the magnet north - south direction . Null points are found on the axis of the of the magnet at $14\;cm$ from the center of the magnet . The earth magnetic field at the place is $0.36\;G$ and the cycle of dip is zero. If the bar magnet is turned around by $180^{\circ}$Where will the new null points be located ?

$\begin{array}{1 1} 1.11 \;cm \\ 111\;cm \\ 11.1 \;cm \\ none\;of\;the\;above \end{array} $

1 Answer

Solution :
When the bar magnet is turned by $ 180^{\circ}$ then the null points are obtained on the equatorial line .
$\therefore B^1= \large\frac{\mu_0}{4 \pi} \frac{m}{d^3}$
This is equal to the horizontal component of earth's magnetic field .
$\therefore B^1= H$
=> $ \large\frac{\mu_0}{4 \pi } \frac{ m}{d^3}$$=H$-------(1)
Initially the null points are on the axis of the magnet .
Hence $B_1 = \large\frac{ \mu_0}{4 \pi} .\frac{2 m}{d^3}$$=H$----------(2)
equating (1) and (2)
$\large\frac{ \mu_0}{4 \pi } .\frac{m}{(d)^3} =\frac{\mu_0}{4 \pi} .\frac{2m}{d^3}$
=> $ \large\frac{1}{(d)^3}=\frac{2}{d^3}$
$\therefore (d^1)^3 =\large\frac{d^3}{2}=\frac{(14)^3}{(2)}$
$\therefore d^1 = \large\frac{14}{(2)^{1/3}}$$=11.1\;cm$
Answer :$11.1 \;cm $
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