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Problems

Problem 7.7

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 v0 v_0 at time t=0 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, 1/2mv02 1/2 m v_0 ^2 . Check that the energy delivered to the resistor is exactly 1/2mv02 1/2 m v_0 ^2

(a) To get the current through the resistor, calculate the flux through the loop:

ΦB=Blxemf=−∂Φ∂t=−Blv \Phi_B = B l x \\ emf = - \pdv{\Phi}{t} = - Blv

We can always use good old Ohm's law

I=V/R=−Blv/R 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∫dl×B=−B2l2vR 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)=v0F=md vd t=−B2l2Rvdvv=−B2l2mRdt∫v0vdvv=−B2l2mR∫0tdtln⁡vv0=−B2l2mRtv=v0e−B2l2mRt 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=I2R P = I^2 R

So energy delivered is

∫0∞I2Rdt=B2l2v2R2Rdt=∫0∞B2l2Rv02e−2B2l2mRtdt=−B2l2Rv02mR2B2l2[0−1]=12mv02 \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!

Problem 7.34

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 w << a forms a parallel-plate capacitor, as shown. Find the magnetic field in the gap, at a distance s<a s < a from the axis.

Within the wire, you can draw an Amperian loop to find B within the wire with

∫B⋅dl=μ0Ienc \int B \cdot \dd l = \mu_0 I_{enc}

Within the gap, we need the Ampere's correction term

∇×B=μ0J+μ0ϵ0∂E∂t∮B⋅dl=μ0ϵ0∫∂E∂t⋅dl \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)⋅2πs=μ0ϵ0d d t∫σ(t)ϵ0da B(s) \cdot 2 \pi s = \mu_0 \epsilon_0 \dv{}{t} \int \frac{\sigma (t) }{\epsilon_0} \dd a

B(s)=μ02s∂σ∂t=μ0s2Iπa2 B(s) = \frac{\mu_0}{2} s \pdv{\sigma}{t} = \frac{\mu_0 s}{2} \frac{I}{\pi a^2}

where ∂σ∂t=d Ad t1πa2 \pdv{\sigma}{t} = \dv{A}{t} \frac{1}{\pi a^2}

B(s)=μ0I2πa2sϕ^ B(s) = \frac{\mu_0 I}{2 \pi a^2} s \hat{\phi}