Two short magnets of magnetic moment `M_(1)` and `M_(2)` are fixed on a table as shown.What will be the direction and magnitude of magnetic induction produced by these magnets at the point 'P' `(M_(1)2.7 Am^(2)=3.2Am^(2),d_(1)=0.3m,d_(2)=0.4 m)`

Options

A. `10.04 xx10^(-7) T,tan^(1)(2)`
B. `11.04xx10^(-7) T,tan^(1)(2)`
C. `111.8xx10^(-7) T,tan^(-1) (1//2)`
D. `13.04xx10^(-7)T,tan^(1)(2)`

Answer

C

Explanation

`B_(1)=(mu_(0))/(4pi)(M_(1))/(r_(1)^(3))`
`=(10^(-7)xx2.7)/(27xx10^(-3))=1xx10^(-5)T`
`B_(2)=(mu_(0))/(4pi)(M_(2))/(r_(2)^(3))`
`(10^(-7)xx3.2)/(64xx10^(-3))=0.5xx10^(-5)T`
`=0.5xx10^(-5)T`
Now B=`sqrt(B_(1)^(2)+B_(2)^(2))`
`sqrt(1xx10^(-10)+0.25xx10^(-10)`
`=sqrt(1.25xx10^(-10))`
`=1.118xx10^(-5)`
`=111.8xx10^(-7)T`
Now `tan beta=(B_(2))/(B_(1))=(0.5xx10^(-5))/(1xx10^(-5))`
`therefore beta=tan^(-1)(0.5)`
`=tan^(-1)((1)/(2))`