Two magnet of equal mass are joined at right angles to each other as shown the magnet 1 has a magnetic moment 3 times that of magnet 2. This arrangment is pivoted so that it is free to rotate in the horizontal plane. In equilibrium what angle will the magnet 1 subtend with the magnetic meridian?

Options

A. `tan^(-1)(1/2)`
B. `tan^(-1)(1/3)`
C. `tan^(-1)(1)`
D. `0^(@)`

Answer

B

Explanation

For equilibrium of the system torques on `M_(1)` and `M_(2)` due to `B_(H)` must counter balance each other i.e., `M_(1)xxB_(H)=M_(2)xxB_(H)`. If theta is the angle between `M_(1)` and `B_(H)` will be `(90-theta)`, so `M_(1)B_(H) sintheta=M_(2)B_(H) sin (90-theta)`
`implies tantheta=M_(2)/M_(1)=M/(3M)=1/3 implies theta=tan^(-1)(1/3)`