Two poles of same strength attract each other with a force of magnitude F when placed at the two corners of an equilateral triangle. If a north pole of same strength is placed at the third vertex, it experiences a force of magnitude equal to

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$F \sqrt{2}$

$F \sqrt{3}$

answer is B.

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Detailed Solution

$F_{w} = F_{s} = \frac{μ_{0}}{4 π} \frac{m^{2}}{a^{2}} = F θ = 120^{\circ} F_{R} = \sqrt{F^{2} + F^{2} + 2 F^{2} \cos 120^{0}} (∵ \cos 120^{\circ} = - \frac{1}{2}) F_{R} = F$

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