Four magnets of equal length but magnetic moments $M, 2 M, 3 M$ and $4 M$ respectively are arranged in the form of a square such that their unlike poles are in contact. Then the resultant magnetic moment of the system is-

2 M

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2 \sqrt{2} M

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\sqrt{2} M

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\sqrt{3} M

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Solution

The correct option is B $2 \sqrt{2} M$

$M_{r e s} = \sqrt{(2 M)^{2} + (2 M)^{2}} = \sqrt{8 M^{2}}$

$= 2 \sqrt{2} M$

 Hence, $(B)$ is the correct answer.

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Four magnets of equal length but magnetic moments M, 2M, 3M and 4M respectively are arranged in the form of a square such that their unlike poles are in contact. Then the resultant magnetic moment of the system is-

The correct option is B 2√2M Mres=√(2M)2+(2M)2=√8M2 =2√2M  Hence, (B) is the correct answer.

Four magnets of equal length but magnetic moments $M, 2 M, 3 M$ and $4 M$ respectively are arranged in the form of a square such that their unlike poles are in contact. Then the resultant magnetic moment of the system is-

The correct option is B $2 \sqrt{2} M$

$M_{r e s} = \sqrt{(2 M)^{2} + (2 M)^{2}} = \sqrt{8 M^{2}}$

$= 2 \sqrt{2} M$

 Hence, $(B)$ is the correct answer.