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

Options

A. `9 A-m`
B. `6.75 A-m`
C. `27 A-m`
D. `1.35 A-m`

Answer

D

Explanation

`d=sqrt((15)^(2)-(5)^(2))=sqrt200=10sqrt2cm`
from tanget law,
`B_(H)=(mu_(0)" "2lm)/(4pi(d^(2)+t^(2))^(3/2))`
`B_(H)=0.4xx10^(-4),l=10cm` Given
`B_(H)=(mu" "(2l)(m))/(4pi(d^(2)+t^(2))^(3/2))`
`implies0.4xx10^(-4)=(4pixx10^(-7))/(4pi)xx(0.1xxm)/((200xx10^(-4)+25xx10^(-4))^(3/2)`
`implies m=(0.4xx10^(-4)xx(225xx10^(-4))^(3/2))/(10^(-7))`
`thereforem=13.5 A-m`