phys qu (1 Viewer)

[sikeveo]

Any takers?

 

[speed2]

any chance of a pic dat we can read?

 

[sikeveo]

[img=http://img190.imageshack.us/img190/6888/physqu9ui.th.jpg]

oops sorry!

 

[serge]

0.0047N repulsive

but since they're asking for each wire
it would be half of that force

 

[sikeveo]

thats wrong =/

 

[helper]

1.87e-5N, sound correct?
Straight application of Amperes Law

Not half the force as they are a Newton's third law pair.

 

[sikeveo]

correct thanks

 

[serge]

yep, you're right... (first time i did i messed up cancelling out the 2x2 amps with the 2E7)

F/L=(I1xI2x2E-7)/d

F= (2E-7x4x.35)/0.015

= 1.87x10^-5

 

[Haku]

never came across these types of questions. is it hsc questions? would it be featured.

i am kinda slow, so could someone please elaborate on the way to do it using both words and formulas?

 

[Jago]

It's faraday's experiment i think, i don't know how you would do it in words but the formula way is described above (serge).

 

[sunjet]

F = k l I I / d
Direction = Up

 

[Haku]

by the part saying direction is up? wouldn;t it be to the right for the right wire and left for the left wire since there is repulsion as the current is going in opposite directions for the two straight part of the wire, and the force experienced by each wire is the same.

oh and by refering to 2E-7, does that mean "k"? like 2 x 10 to the -7?

 

[serge]

yep, thats what i meant, just typing in calculator talk

 

[Haku]

lol i c, thanks for clarifying