Two short magnets each of moment 10 are placed in end on position so that their centres are 0.1m apart. The force between them is

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Arun · Answer

Dear student F = Mu/4pi 6 M1 M2 /d^4 hence F = 0.6 N Hope it helps Regards Arun (askIITians forum expert)

Khimraj · Answer

formula for force BT parallel placed magnet f=Mo/4π(6m1m2/r4) No=4π×10^-7 therefore f=10^-7(6×10×10/(0.1^4)=0.6N

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Two short magnets each of moment 10 are placed in end on position so that their centres are 0.1m apart. The force between them is

Two short magnets each of moment 10 are placed in end on position so that their centres are 0.1m apart. The force between them is

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Two short magnets each of moment 10 are placed in end on position so that their centres are 0.1m apart. The force between them is

Two short magnets each of moment 10 are placed in end on position so that their centres are 0.1m apart. The force between them is

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