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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

Ritheesh , 6 Years ago
Grade 12
anser 2 Answers
Arun
Dear student
 
F = Mu/4pi 6 M1 M2 /d^4
 
hence F = 0.6 N
 
Hope it helps
 
Regards
Arun (askIITians forum expert)
 
Last Activity: 6 Years ago
Khimraj
 
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
Last Activity: 6 Years ago
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