Two bar magnets of same size with magnetic moments M $_{1}$ and M $_{2}$ (M $_{1}$ > M $_{2}$ ) are simultaneously used at the tan A position in a DMM. When the magnets are placed with unlike poles in contact the deflection is 30 $^{0}$ and when like poles are in contact the deflection is 60 $^{0}$ . Then $\frac{M_{1}}{M_{2}} :$

\frac{3}{1}

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\frac{3}{4}

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\frac{6}{1}

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\frac{2}{1}

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Solution

The correct option is D $\frac{2}{1}$
The deflection off the magnetic needle in tan A position by a short magnet is given by
$\frac{μ_{o}}{4 π} \frac{2 M}{d^{3}} = B_{H} t a n θ$

$\frac{t a n θ_{A}}{t a n θ_{B}} = \frac{M_{A}}{M_{B}}$

$\frac{t a n θ_{A}}{t a n θ_{B}} = \frac{M_{1} - M_{2}}{M_{1} + M_{2}}$

$\frac{M_{1} - M_{2}}{M_{1} + M_{2}} = \frac{1}{3}$

$\frac{M_{1}}{M_{2}} = \frac{2}{1}$

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Two bar magnets of same size with magnetic moments M1 and M2 (M1 > M2 ) are simultaneously used at the tan A position in a DMM. When the magnets are placed with unlike poles in contact the deflection is 300 and when like poles are in contact the deflection is 600 . Then M1M2:

The correct option is D 21The deflection off the magnetic needle in tan A position by a short magnet is given by μo4π2Md3=BHtanθtanθAtanθB=MAMBtanθAtanθB=M1−M2M1+M2M1−M2M1+M2=13M1M2=21