Question
A bar magnet $$8\,cms$$ long is placed in the magnetic merdian with the $$N$$-pole pointing towards geographical north. Two neutral points separated by a distance of $$6\,cms$$ are obtained on the equatorial axis of the magnet. If horizontal component of earth’s field $$ = 3.2 \times {10^{ - 5}}T,$$ then pole strength of magnet is
A.
$$5\,ab{\text{-}}amp \times cm$$
B.
$$10\,ab{\text{-}}amp \times cm$$
C.
$$2.5\,ab{\text{-}}amp \times cm$$
D.
$$20\,ab{\text{-}}amp \times cm$$
Answer :
$$5\,ab{\text{-}}amp \times cm$$
Solution :
$${\text{Here,}}\,\,2\ell = 8\,cm,\ell = 4\,cm,\;d = \frac{6}{2} = 3\,cm.$$
At neutral point,
$$\eqalign{
& H = B = \frac{{{\mu _0}}}{{4\pi }}\frac{M}{{{{\left( {{d^2} + {\ell ^2}} \right)}^{\frac{3}{2}}}}} \cr
& = {10^{ - 7}}\frac{M}{{{{\left( {5 \times {{10}^{ - 2}}} \right)}^3}}} = \frac{M}{{1250}} \cr
& \therefore M = 1250\,H = 1250 \times 3.2 \times {10^{ - 5}}A{m^2} \cr
& m = \frac{M}{{2\ell }} = \frac{{1250 \times 3.2 \times {{10}^{ - 5}}}}{{8 \times {{10}^{ - 2}}}}Am. = 0.5\,Am = 0.5 \times \frac{1}{{10}}ab\,amp \times 100\,cm \cr
& = 5\,ab{\text{-}}amp\,cm. \cr} $$