Force between parallel conductors (1 Viewer)

[untouchablecuz]

Two long parallel conductors carry equal currents in opposite directions. The force between them is 3F.

The current in one of the conductors is doubled, but the current in the other is reduced to a third of its original value. The distance between the conductors is halved.

What is the new force between the conductors?

my working:

initially,

[3F]=(kl)(a)(b)/d, where a and b represent the currents

the force, F', after applying changes

F'=(kl)(2a)(b/3)/(d/2)=(4/3)(kl)(a)(b)/d=(4/3)(3F)=4F

hence the new force, F', is 4F

the answer is 8F...

where did i go wrong?

 

[darkchild69]

I also get 4F ((3x2x2)/3)

 

[_GOD_]

So do i, maybe the answers wrong :S

 

[Top Secret]

i also get 4f ((3x2x2)/3)

+1

 

[untouchablecuz]

alrighty, thanks guys