Two short bar magnets A and B (having magneti | vedclass

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Question

(A) Let the point be at a distance $d$ from the center of the magnets along the axis of magnet $A$.The magnetic field due to magnet $A$ (axial positio

Vedclass · Accepted Answer

$2: 1$

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(A) Let the point be at a distance $d$ from the center of the magnets along the axis of magnet $A$.The magnetic field due to magnet $A$ (axial position) is $B_{1} = \frac{\mu_{0}}{4\pi} \frac{2M_{1}}{d^{3}}$.The magnetic field due to magnet $B$ (equatorial position) at the same point is $B_{2} = \frac{\mu_{0}}{4\pi} \frac{M_{2}}{d^{3}}$.Since the axes are perpendicular,the resultant field makes an angle $	heta$ with the axis of magnet $A$ such that $	an 	heta = \frac{B_{2}}{B_{1}}$.Given $	heta = 45^{\circ}$,so $	an 45^{\circ} = 1$.Therefore,$1 = \frac{(\mu_{0} M_{2}) / (4\pi d^{3})}{(2\mu_{0} M_{1}) / (4\pi d^{3})} = \frac{M_{2}}{2M_{1}}$.This gives $\frac{M_{2}}{M_{1}} = 2$,or $M_{2} : M_{1} = 2 : 1$.

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