A bar magnet of magnetic moment $M_{1}$ is axially cut into two equal parts. If these two pieces are arranged perpendicular to each other, the resultant magnetic moment is $M_{2}$ , then the value of $\frac{M_{1}}{M_{2}}$ is

\frac{1}{2 \sqrt{2}}

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\frac{1}{\sqrt{2}}

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\sqrt{2}

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Solution

The correct option is D $\sqrt{2}$
$M = \sqrt{M_{1}^{2} + M_{2}^{2} + M_{1} M_{2} cos θ}$
$M_{2} = \sqrt{M_{1}^{2} / 4 + M_{1}^{2} / 4 + M_{1}^{2} / 4 cos 90^{0}}$
$M_{2} = \sqrt{M_{1}^{2} / 2}$
$∴ M_{2} = M_{1} / \sqrt{2}$

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A bar magnet of magnetic moment M1 is axially cut into two equal parts. If these two pieces are arranged perpendicular to each other, the resultant magnetic moment is M2, then the value of M1M2 is

The correct option is D √2M=√M21+M22+M1M2cosθM2=√M21/4+M21/4+M21/4cos900M2=√M21/2∴M2=M1/√2

A bar magnet of magnetic moment $M_{1}$ is axially cut into two equal parts. If these two pieces are arranged perpendicular to each other, the resultant magnetic moment is $M_{2}$ , then the value of $\frac{M_{1}}{M_{2}}$ is

The correct option is D $\sqrt{2}$
$M = \sqrt{M_{1}^{2} + M_{2}^{2} + M_{1} M_{2} cos θ}$
$M_{2} = \sqrt{M_{1}^{2} / 4 + M_{1}^{2} / 4 + M_{1}^{2} / 4 cos 90^{0}}$
$M_{2} = \sqrt{M_{1}^{2} / 2}$
$∴ M_{2} = M_{1} / \sqrt{2}$