Two short bar magnets 'A and 'B' (having magnetic moments ' $\mathrm{M}_{1}$ ' and ' $\mathrm{M}_{2}$ '…

Two short bar magnets 'A and 'B' (having magnetic moments ' $\mathrm{M}_{1}$ ' and ' $\mathrm{M}_{2}$ ' respectively) are kept one above the other with their magnetic axis perpendicular to each other. If their resultant at a point on the axis of magnet ' $A$ ' is inclined at $45^{\circ}$ with the axis of magnet A then the ratio of magnetic moments $\frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}$ is $\left[\tan 45^{\circ}=1\right]$
  1. $2: 1$
  2. $2: 3$
  3. $1: 2$
  4. $3: 2$

Solution

$B_{1}=\frac{2 \mu_{0} M_{1}}{4 \pi d^{3}} \quad B_{2}=\frac{-\mu_{0} M_{2}}{4 \pi d^{3}}$ $\tan 45=\left|\frac{B_{2}}{B_{1}}\right|$ $1=\frac{M_{2}}{2 M_{1}}$ $\Rightarrow M_{2}: M_{1}=2: 1$

Asked in: MHT CET 2020 (12 Oct Shift 1)

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