Two short magnets AB and CD are in the xy-plane and are parallel to X-axis and coordinates of their centres respectively are (0,2) and (2,0). Line joining the north-south poles of CD is opposite to that of AB and lies along the positive X-axis. The resultant field induction due to AB and CD at a point P(2,2) is `100xx10^(-7)` T. When the poles of the magnet CD are reversed, the resultant field induction is ` 50xx10^(-7) T`. The value of magnetic moments of AB and CD (in `Am^(2)`) are

Options

A. 300,200
B. 600,400
C. 200,100
D. 300,150

Answer

A

Explanation

`B=B_(1)+B_(2)=(mu_(0))/(4pi)(2M_(1))/(d^(3))+(mu_(0))/(4pi)(2M_(2))/(d^(3))=(mu_(0))/(4pi)(1)/(d^(3))[2M_(1)+M_(2)]`
`implies 100xx10^(-7)=10^(-7)[(2M_(1))/(8)+(M_(2))/(8)]`
`2M_(1)+M_(2)=800`
When the poles of the magnet CD are reversed then
`(mu_(0))/(4pid^(3))[2M_(1)-M_(2)]=50xx10^(-7)`
`2M_(1)-M_(2)=400`
On solving equations (1) and (2), we get,
`M_(1)=300Am^(2),M_(2)=200 Am^(2)`