How Do Magnetic Forces Affect Parallel Current-Carrying Wires?

[eku_girl83]

Three parallel wires each carry current I in the direction shown in the figure (file attached). If the separation between adjacent wires is d, calculate the magnitude and direction of the net magnetic force per unit length on each wire.
a) What is the magnetic force on the top wire?
I think the direction is up, but I don't know how to find the magnitude.
b) What is the magnetic force on the middle wire?
I know this is zero
c) What is the magnetic force on the bottom wire?

I also know that net magnetic force per unit length is equal to (mu0*I*I')/(2*pi*r).
mu0=4*pi*10^-7

Any help would be appreciated!

 

[member 553083]

a) The magnetic force on the top wire is downwards, with a magnitude of (mu0*I^2)/(2*pi*d). b) The middle wire experiences no net magnetic force. c) The force on the bottom wire is upwards, with a magnitude of (mu0*I^2)/(2*pi*d).

 

[M98Ranger]

a) The magnetic force on the top wire can be calculated using the formula for the net magnetic force per unit length. The direction of the force will be upwards, as the current in the top wire is going into the page, and the magnetic field created by the other two wires is into the page as well. The magnitude of the force can be found by plugging in the values into the formula. So, the magnetic force on the top wire would be (4*pi*10^-7 * I * I)/(2*pi*d) = (2*10^-7 * I^2)/d, where I is the current in each wire and d is the separation between adjacent wires.

b) The magnetic force on the middle wire will indeed be zero, as the magnetic fields created by the other two wires will cancel each other out. This can be seen by drawing the magnetic field lines and noticing that they are in opposite directions and cancel each other out.

c) The magnetic force on the bottom wire can also be calculated using the same formula as in part a. However, in this case, the direction of the force will be downwards, as the current in the bottom wire is going out of the page, and the magnetic field created by the other two wires is into the page. So, the magnitude of the force would be the same as in part a, but the direction will be downwards.