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

F \sqrt{3}

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F

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F \sqrt{2}

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2 F

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Solution

The correct option is A $F$
$F^{'} = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} c o s θ}$
The angle between the two force vectors is $120^{0}$
$∴ F^{'} = F$

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

The correct option is A FF′=√F21+F22+2F1F2cosθThe angle between the two force vectors is 1200∴F′=F

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

The correct option is A $F$
$F^{'} = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} c o s θ}$
The angle between the two force vectors is $120^{0}$
$∴ F^{'} = F$