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Standard XII

Physics

Magnetic Moment

The small mag...

Question

The small magnets each of magnetic moment $10 A - m^{2}$ are placed end on position 0.1m apart from their centres.The force acting between them is :

0.6 \times 10^{7} N

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0.06 \times 10^{7} N

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0.6 N

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0.06 N

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Solution

The correct option is A $0.6 N$
Given,

$M_{1} = M_{2} = 10 A m^{2}$

$r = 0.1 m$

The force acting between the magnet is given by,

$F = \frac{μ_{0}}{4 π} . \frac{6 M_{1} M_{2}}{r^{4}}$

$F = 10^{- 7} \times \frac{6 \times 10 \times 10}{(0.1)^{4}}$

$F = 0.6 N$

The correct option is C.

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The small magnets each of magnetic moment 10A−m2 are placed end on position 0.1m apart from their centres.The force acting between them is :

The correct option is A 0.6NGiven,M1=M2=10Am2r=0.1mThe force acting between the magnet is given by,F=μ04π.6M1M2r4F=10−7×6×10×10(0.1)4F=0.6NThe correct option is C.

The small magnets each of magnetic moment $10 A - m^{2}$ are placed end on position 0.1m apart from their centres.The force acting between them is :

The correct option is A $0.6 N$
Given,

$M_{1} = M_{2} = 10 A m^{2}$

$r = 0.1 m$

The force acting between the magnet is given by,

$F = \frac{μ_{0}}{4 π} . \frac{6 M_{1} M_{2}}{r^{4}}$

$F = 10^{- 7} \times \frac{6 \times 10 \times 10}{(0.1)^{4}}$

$F = 0.6 N$

The correct option is C.