Q. A bar magnet is $10\ cm$ long, and is kept with its north (N)- pole pointing north. A neutral point is formed at a distance of $15\ cm$ from each pole. Given the horizontal component of earth’s field to be 0.4 Gauss, the pole strength of the magnet is
A) 9 $A$-$m$
B) 6.75 $A$-$m$
C) 27 $A$-$m$
D) 1.35 $A$-$m$
Explanation
(d) Length of magnet
$= 10\ cm = 10 \times 10^{- 2}m$
$r = {15 \times 10}^{- 2}m$
$OP = \sqrt{225 - 25} = \sqrt{200}\ cm$
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Since, at the neutral point, magnetic field due to the magnet equal to $B_{H}$
$B_{H} = \frac{\mu_{0}}{4\pi}.\frac{M}{\left( OP^{2} + AO^{2} \right)^{3/2}}$
$0.4 \times 10^{- 4} = 10^{- 7} \times \frac{M}{\left( 200 \times 10^{- 4} + 25 \times 10^{- 4} \right)^{3/2}}$
$\frac{0.4 \times 10^{- 4}}{10^{- 7}} \times \left( 225 \times 10^{- 4} \right)^{3/2} = M$
$0.4 \times 10^{3} \times 10^{- 6}(225)^{3/2} = M$
$M = 1.35\ A - m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.