Question

Two short magnets apply a force F on each other. If the separation between them is halved then new forcebetween them is approximately..​

Answer 1

Hello Dear,

◆ Answer -

F' = 4 F

◆ Explaination -

According to Coulomb's law -

F = (μ/4π) m1.m2 / r^2

When separation between magnets is halved, i.e. r' = r/2

F' = (μ/4π) m1.m2 / r'^2

F' = (μ/4π) m1.m2 / (r/2)^2

F' = 4 × (μ/4π) m1.m2 / r^2

F' = 4 F

Therefore, force of magnetism between two magnets will be quadrupled on halving separation.

Thanks dear...

Answer 2

Answer:

16F

Explanation:

We have a special expression for this particular case..

F=3μm1m2/4π*r^4  

where r is the distance of separation

so now if we substitute r=r/2 (as the distance is halved)

r^4=r^4/16

as you can see the distance is deciding factor.

So, by inspection we can easily claim that F'(new force)=16F(old force)