A bar magnet of length `8 cm` and having a pole strengh of `1.0 A m` is placed vertically on a horizontal table with its south pole on the table. A neutral point is found on the table at a distance of 6.0 cm north of the magnet. Calculate the earth's horizontal magnetic field.

Options

A. `22xx10^(-6)T`
B. `44xx10^(-6)T`
C. `66xx10^(-6)T`
D. `88xx10^(-6)T`

Answer

A

Explanation

The magnetic field at P due to south pole
`B_(S)=mu_(0)/(4pi)m/r^(2)`
and de to north pole
`B_(N)=mu_(0)/(4pi)m/((l^(2)+r^(2)))` along NP

Its component along north is `B_(N)costheta`. On substituting the value of `theta`, it
= `mu_(0)/(4pi)m/((l^(2)+r^(2)))xxr/(sqrt(l^(2)+r^(2)))`
= `mu_(0)/(4pi)m/((l^(2)+r^(2))^(3//2))`
For neutral point at P, we have
`B_(H)=B_(S)-B_(N)costheta`
= `mu_(0)/(4pi)[m/r^(2)-(mr)/((l^(2)+r^(2))^(3/2))]`
= `10^(-7)xx1[1/0.06^(2)-0.06/((0.08^(2)+0.06^(2))^(3/2))]`
= `22xx10^(-6)T`