The Force Between Current-Carrying Conductors

[vrobins1]

Homework Statement

Consider two straight conductors of length 1 m, each of them carrying a current of I = 2 A. If they are placed 1 m apart, what is the magnitude and direction of the force between these two conductors if there currents are
a) in the same direction;
b) in the opposite direction.

Homework Equations

I know that

B = $$\mu$$₀*I/2$$\pi$$r

I also know that that force on a conductor in general is F = ILBsin$$\theta$$.

The Attempt at a Solution

I calculated the magnitude of B according to the equation above and got 3.95^-5 T.

I'm not sure how to find the force of TWO conductors together though.

Would you just do F = ILB for both conductors, adding the two F values of they are going in the same direction and subtracting if they are going in opposite directions?

Is the direction determined by the RHR?

Any insight would be much appreciated. Thanks!

 

[Doc Al]

I'm not sure how to find the force of TWO conductors together though.

Would you just do F = ILB for both conductors, adding the two F values of they are going in the same direction and subtracting if they are going in opposite directions?

Not exactly. To find the force on wire #2, start by finding the field from wire #1, then find the force it exerts on wire #2. That's it: the force that wire #1 exerts on wire #2. Of course, wire #2 exerts the same force on wire #1. (Don't add/subtract the F values.)

Is the direction determined by the RHR?

Yes. The direction of the field created is determined by the RHR, and so is the direction of the force that each wire exerts on the other.

 

[nlightner]

I would like to point out that the force between two current-carrying conductors is a well-known phenomenon known as the Lorentz force. It is caused by the interaction between the magnetic fields generated by the two currents.

To answer the given problem, we can use the formula for the force between two parallel current-carrying conductors, which is given by F = μ0*I1*I2*L/2πd, where μ0 is the permeability of free space, I1 and I2 are the currents in the two conductors, L is the length of the conductors, and d is the distance between the conductors.

a) In this case, since the currents are in the same direction, the force will be attractive. Plugging in the given values, we get F = (4π*10^-7)*(2 A)*(2 A)*(1 m)/2π(1 m) = 4π*10^-7 N. The direction of the force can be determined using the right-hand rule, where the thumb points in the direction of the current in one conductor, the fingers point in the direction of the current in the other conductor, and the palm gives the direction of the force.

b) If the currents are in opposite directions, the force will be repulsive. Using the same formula, we get F = (4π*10^-7)*(2 A)*(-2 A)*(1 m)/2π(1 m) = -4π*10^-7 N. The direction of the force can be determined using the same right-hand rule, but in this case, the force will be in the opposite direction as compared to the previous case.

I would also like to mention that there is a more general formula for the force between two current-carrying conductors, which takes into account the angle between the two conductors. This formula is given by F = μ0*I1*I2*L*sinθ/2πd, where θ is the angle between the two conductors. However, in this problem, since the conductors are parallel, θ = 0 and hence the sinθ term becomes 0, resulting in the simplified formula used above.

In conclusion, the force between two current-carrying conductors can be calculated using the formula F = μ0*I1*I2*L/2πd and the direction can be determined using the