A metal bar of mass _m_ slides frictionlessly on two parallel conducting rails a distance _l_ apart. A resistor R is connected across the rails, and a uniform magnetic field __B__ pointing into the page fills the region. (a) If the bar moves to the right at speed _v_, what is the current in the resistor? (b) What is the magnetic force on the bar? In what direction? (c) If the bar starts out with speed \\( v_0 \\) at time \\( t = 0 \\), and is left to slide, what is its speed at a later time t? (d) The initial kinetic energy of the bar was, of course, \\( 1/2 m v_0 ^2 \\). Check that the energy delivered to the resistor is exactly \\( 1/2 m v_0 ^2 \\)

(a) To get the current through the resistor, calculate the flux through the loop: \[\Phi_B = B l x \\ emf = - \pdv{\Phi}{t} = - Blv\] We can always use good old Ohm's law \[I = V / R = -Blv / R\] The induced magnetic flux opposes the change in flux, so the current flows down through the resistor. (b) Force on the bar? Just use Lorentz force law \[F = I \int \dd \vec{l} \cross \vec{B} = - \frac{B^2 l^2 v}{R}\] The direction opposes _x_ (force is to the left). (c) \[v(t = 0) = v_0 \\ F = m \dv{v}{t} = - \frac{B^2 l^2}{R} v \\ \frac{\dd v}{v} = - \frac{B^2 l^2 }{m R } \dd t \\ \int_{v_0} ^v \frac{\dd v}{v} = - \frac{B^2 l^2 }{m R } \int_0 ^t \dd t \\ \ln \frac{v}{v_0} = - \frac{B^2 l^2 }{m R } t \\ v = v_0 e^{-\frac{B^2 l^2 }{m R } t}\] (d) Power dissipated in a resistor is \[P = I^2 R \] So energy delivered is \[\int_0 ^\infty I^2 R \dd t = \frac{B^2 l^2 v^2}{R^2} R \dd t \\ = \int_0 ^\infty \frac{B^2 l^2}{R} v_0 ^2 e^{-2\frac{B^2 l^2 }{m R } t} \dd t \\ = - \frac{B^2 l^2}{R} v_0 ^2 \frac{m R}{2 B^2 l^2} [ 0 - 1 ] \\ = \frac{1}{2} m v_0 ^2\] Hooray!

A fat wire, radius _a_, carries constant current _I_, uniformly distributed over its cross section. A narrow gap in the wire of width \\( w << a \\) forms a parallel-plate capacitor, as shown. Find the magnetic field in the gap, at a distance \\( s < a \\) from the axis.

Within the wire, you can draw an Amperian loop to find B within the wire with \[\int B \cdot \dd l = \mu_0 I_{enc}\] Within the gap, we need the Ampere's correction term \[\curl \vec{B} = \mu_0 \vec{J} + \mu_0 \epsilon_0 \pdv{\vec{E}}{t} \\ \oint B \cdot \dd l = \mu_0 \epsilon_0 \int \pdv{\vec{E}}{t} \cdot \dd l\] Current is flowing to the end of the wire with nowhere to go, so it must build up there. We've got a parallel plate capacitor within the gap \[B(s) \cdot 2 \pi s = \mu_0 \epsilon_0 \dv{}{t} \int \frac{\sigma (t) }{\epsilon_0} \dd a\] \[B(s) = \frac{\mu_0}{2} s \pdv{\sigma}{t} = \frac{\mu_0 s}{2} \frac{I}{\pi a^2}\] where \\( \pdv{\sigma}{t} = \dv{A}{t} \frac{1}{\pi a^2} \\) \[B(s) = \frac{\mu_0 I}{2 \pi a^2} s \hat{\phi}\]