Question

3) Consider two perpendicular wires as shown. There's an infinite wire with current Ii flowing from left to right, and a wire stub of length L with current I2 flowing towards the top of the page. There's a gap of length D between the two wires. a) Calculate the force on wire 2 (the wire with current /2) due to wire 1. If the everyday statement of Newton's third law holds, what should be the force on wire 1 due to wire 2? b) Calculate the magnetic field made by wire 2 at any arbitrary point on wire 1. For clarity, let wire 2 be at x0 and find B(x) at points on wire 1 (x being the horizontal coordinate). Take the limit as x ->0 and make sure B has the behavior it should have. c) Find the magnetic force exerted on wire 1 by wire 2. d) If you did (a) and (c) right, you'll have discovered another apparent violation of Newton's third law. In class we resolved our first such violation by noting that Newton's 3rd is really about conservation of momentum, and electromagnetic fields have momentum density that goes like E × B, and that gave us some hope that we could patch everything up. But in this situation at first glance there's only a B and no E, so that won't help. Or will it? Resolve the contradiction as best you can (qualitatively at least). Hint: You can't just punt by saying There's no such thing physically as a wire stub." There certainly can be, under the right circumstances.

Answer 1

The magnetic field due to the first wire changes along the length of the wire stub

The magnetic field of an infinite wire is given by:

$B=\frac{\mu_0 I_1}{2\pi s}$

Where "s" is the distance from the wire.

The magnetic force exerted on a current carrying conductor is given by:

$F=I\vec{l}\times\vec{B}$

Since the current is perpendicular to the magnetic field here (field is into the plane of the paper)

$F=I_2lB_1$

or in differential form:

$dF=I_2B_1dl$

$dF=I_2\frac{\mu_o I_1}{2\pi l}dl$

$F=\int_{l=D}^{l=D+L}I_2\frac{\mu_o I_1}{2\pi l}dl$

$F=\frac{\mu_0 I_1I_2}{2\pi}\int_{l=D}^{l=D+L}\frac{dl}{l}$

$F=\frac{\mu_0 I_1I_2}{2\pi}ln\frac{D+L}{D}$