Question

Two straight parallel wire are separated by 6 cm. 2.0-A current flowing in the first wire. If magnetic field is found to be zero when both wire at 2.2 cm. what is the magnitude and direction of the current of second wire?
Thanks for helping :)

Answer 1

Can you clarify your question? What do you mean when you say "when both wire at 2.2 cm"?

Answer 2

@matt101 , when the both wire separated by 2.2 cm. sorry if my words confuse you

Answer 3

If I've understood your question correctly, the current in the second wire would also be 2.0 A, just in the opposite direction.

This is just a matter of adding the magnetic fields so that they cancel, and figuring out the corresponding current required to do so. First, you need the equation for calculating a magnetic field around a current-carrying wire:

\[B={\mu_0I \over 2 \pi r}\]
Where B is the magnetic field, µ(0) is the permissivity of free space (a constant), I is the current, 2π is a constant, and r is the distance from the wire.

Since we know B=0, the B of one wire must equal the negative B of the other (remember, these fields are going in opposite directions. That means:

\[{\mu_0I_1 \over 2 \pi r_1}=-{\mu_0I_2 \over 2 \pi r_2}\]
The constants appear on can be reduced out. Based on your explanation of the question, the r can be reduced out as well because the point we're considering is the same distance from both wires, so r(1)=r(2). THIS IS THE PART OF YOUR QUESTION THAT SEEMS A BIT ODD TO ME - ARE YOU SURE IT'S COPIED DOWN CORRECTLY? DO YOU HAVE A PICTURE?

That means all you're left with is:
\[I_1=-I_2\]
Which is the answer I wrote above.

Answer 4

No i dont have any picture as it only in the question. I understand what you talking about and thanks! :)

Answer 5

The force PER UNIT LENGTH between two long straight current carrying wires separated by a distance r is given by
\[\frac{F}{l}=\frac{\mu_{o}}{4\pi}.\frac{2I_{1}I_{2}}{r}\]
This follows directly from the derivation of the force experienced by a current carrying conductor placed in a magnetic field
\[|F^{\rightarrow}|=B.I.l.\sin \theta\]