Question

Two magnets 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 arrangement 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

• A. $\tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)$
• B. $\tan ^ { - 1 } \left( \frac { 1 } { 3 } \right)$
• C. $\tan ^ { - 1 } ( 1 )$
• D. 0⁰

Answer 1

Answer: (B)

$\tan ^ { - 1 } \left( \frac { 1 } { 3 } \right)$

Answer 2

For equilibrium of the system torques on $M _ { 1 }$ and $B _ { H }$ then the angle between $B _ { H }$ will be (90 – θ)

so $M _ { 1 } B _ { H } \sin \theta = M _ { 2 } B _ { H } \sin ( 90 - \theta )$

⇒ $\tan \theta = \frac { M _ { 2 } } { M _ { 1 } } = \frac { M } { 3 M } = \frac { 1 } { 3 } \Rightarrow \theta = \tan ^ { - 1 } \left( \frac { 1 } { 3 } \right)$