How to Calculate Force on a Perpendicular Finite Wire Near an Infinite Wire?

AI Thread Summary
To calculate the force on a finite conducting wire perpendicular to an infinite wire, the formula F = I_f ∫(μ₀ I_i / 2πr) dr can be used, where I_f and I_i are the currents in the finite and infinite wires, respectively. The integration region needs to be clearly defined, with the infinite wire assumed to lie along the y-axis and the finite wire along the x-axis. The magnetic field B created by the infinite wire at a point on the x-axis is given by B_i = k*I_i/x, where k = μ₀/2π. The force on an elementary length dx of the finite wire is dF = I_f*B_i*dx, leading to the expression dF = k*I_f*I_i*dx/x. Integrating this from x1 to x2 will yield the total force's magnitude and direction.
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Homework Statement



I want the force on a finite conducting wire that is perpendicular to an infinite wire.

Can I do it with this:

F = I_f \int{\frac{\mu_0 I_i}{2\pi r}}{dr} where I_f, I_i are the currents in the finite and infinite wires.
 
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Yes, that’s the way to do it. But what is the region of integration? It would be easier to answer if you had described the exact picture you had in mind. Why is Ii inside the integral? I presume you are dealing with steady currents.

Let’s set it up properly. Suppose the infinite wire lies along the y-axis and Ii is toward +ve y-axis. The finite wire lies on the x-axis from x1 to x2 and If is toward the +ve x-axis.

B due to Ii at a pt x on the x-axis =Bi = k*Ii/x, where I’ve written k for mu_0/2pi. Bi points in the –z dircn.

If we consider an elementary length dx at x, then the force on this is dF = If*Bi*(sin 90)dx = k*If*Ii*dx/x

Now you can integrate from x1 to x2 and tell us the magnitude and direction of the force?
 
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