A bar magnet 8 cm long is placed in the magnetic meridian with the N pole pointing toward geographical north.Two neutral points separated by a distance of 6 cm are obtained on the equatorial axis of the magnet.If horizontal component of earth's field is `3.2xx10^(-5) T`,then pole strength of magnet is

Options

A. 5 ab amp cm
B. 3 ab amp cm
C. 7 ab amp cm
D. 2 ab amp cm

Answer

B

Explanation

2l=8 cm `therefore`l=4cm
`r=3cm,B_(H)=3.2xx10^(-5)T,m=?`
In case of N to N position neutral point is obtained at equator of dipole.At neutral point
`B_(H)=B_(eq)`
`B_(H)=(mu_(0))/(4pi)(M)/((r^(2)+l^(2))^(3//2))`
`B_(H)=(M)/((r^(2)+l^(2))^(3//2)`(in C.G.S. system)
`therefore m.2l=(B_(H)(r^(2)+l^(2))^(3//2))/(2l)=(3.2xx10^(-5)xx10^(4)(9+16)^(3//2))/(8)`
`m=(3.2xx10^(-1)xx(5^(2))^(3//2))/(8)=5ab A cm`