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?

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((1/2)) (B)  tan -1((1/3)) (C)  tan -1(1) (D) 0°

Tardigrade Admin · Accepted Answer

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} \times B_{H} = M_{2} \times B_{H} .

If

θ

is the angle between

M_{1}

and

B_{H}

will be

(9 0^{\circ} - θ);

so

M_{1} B_{H} sin θ = M_{2} B_{H} sin

(9 0^{\circ} - θ)

\Rightarrow tan θ = \frac{M _{2}}{M _{1}} = \frac{M}{3 M} = \frac{1}{3}

\Rightarrow θ = tan^{- 1} (\frac{1}{3})

$tan^{- 1} (\frac{1}{2})$

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

$tan^{- 1} (1)$

$0^{\circ}$

tan−1(21​)

tan−1(31​)

tan−1(1)

0∘

$tan^{- 1} (\frac{1}{2})$

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

$tan^{- 1} (1)$

$0^{\circ}$