A short bar magnet is arranged with its North pole pointing geographical North. It is found that the horizontal component of Earths magnetic induction (BH ) is balanced by the magnetic induction of the magnet at a point which is at a distance of 20 cm from its centre. The magnetic moment of the magnet is (if BH=4× 10-5Wb text -m-2 )

Question

A short bar magnet is arranged with its North pole pointing geographical North. It is found that the horizontal component of Earths magnetic induction (BH ) is balanced by the magnetic induction of the magnet at a point which is at a distance of 20 cm from its centre. The magnetic moment of the magnet is (if BH=4× 10-5Wb text -m-2 ) (A) 3.2 A- m2 (B) 1.6 A- m2 (C) 6.4 A- m2 (D) 0.8 A- m2

Tardigrade Admin · Accepted Answer

BH=(μ 0/4π ).(M/r3) 4× 10-5=10-7× (M/(20)310-6) ((20)3× 4× 10-5× 10-6/10-7)=M M=3.2 Am2

3.2 A- m2

1.6 A- m2

6.4 A- m2

0.8 A- m2

BH​=4πμ0​​.r3M​ 4×10−5=10−7×(20)310−6M​ 10−7(20)3×4×10−5×10−6​=M M=3.2Am2

$B_{H} = \frac{μ _{0}}{4 π} . \frac{M}{r ^{3}}$ $4 \times 10^{- 5} = 10^{- 7} \times \frac{M}{( 20 ) ^{3} 10 ^{- 6}}$ $\frac{( 20 ) ^{3} \times 4 \times 10 ^{- 5} \times 10 ^{- 6}}{10 ^{- 7}} = M$ $M = 3.2 A m^{2}$