Question

A bar magnet is $10 \,cm$ long 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 is $0.4$ 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

Answer 1

Answer: (D)

1.35 A-m

Answer 2

The correct option is (D): 1.35 A-m
Length of magnet $= 10 \, cm = 10 \times 10^{-2} m$,
$r = 15 \times 10^{-2} \, m$
$OP = \overline{225-25} = \overline{200} \,cm$
Since, at the neutral point, magnetic field due to the magnet is equal to $B_H$
$B_{H} = \frac{\mu_{0} }{4\pi} . \frac{M}{OP^{2} + AO^{2 32}}$
$0.4 \times10^{3} \times 10^{-7} \times \frac{M}{200\times 10^{-4} + 25 \times 10^{4 32}}$
$\frac{0.4 \times 10^{-4}}{10^{-7}} \times 225 \times 10^{-4 32} = M$
$0.4 \times 10^{3} \times 10^{-6} 225^{32} = M$
$M = 1.35 \,A - m$