Past exam of the physics 2 paper of the University of Oxford / 2019 resit / 8

Faraday's law of induction states that the oriented electromotive force around a closed material circuit is minus the rate of change of its linked magnetic flux:The surface orientation and circuit direction obey the right-hand rule. The minus sign expresses Lenz's law: the induced electric current, when allowed to flow, produces a magnetic field opposing the change of linked magnetic flux. For a stationary circuit this reduces to $\oint \mathbf{E}\cdot d\mathbf{\ell }=-d{\Phi }_B/dt$.

The source's “Lorenz force” is the Lorentz force, $q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. A wire element of cross-section $A_c$ contains $n{A}_{c}d\ell$ carriers of charge $q$ with drift velocity ${\mathbf{v}}_d$. Adding their magnetic forces and using $I=nq{A}_c{v}_d$ givesfor a straight wire in a uniform magnetic field. The length vector follows conventional electric current, irrespective of the carrier sign.

For the unlabelled upward-motion request, define $\hat{\mathbf{t}}=(\cos \varphi ,0,-\sin \varphi )$ down the slope, $\hat{\mathbf{b}}=(0,1,0)$ across the rails and $\hat{\mathbf{n}}=(\sin \varphi ,0,\cos \varphi )$ out of their plane. Choose the vertical magnetic field as $\mathbf{B}=B\hat{\mathbf{z}}$. An electric current $I$ across the bar in direction $-\hat{\mathbf{b}}$ gives a horizontal magnetic force $-IBs\hat{\mathbf{x}}$. The other forces are weight $-mg\hat{\mathbf{z}}$ and the rail normal force $N\hat{\mathbf{n}}$. Uniform upward velocity requiresConsequentlyEach symmetric rail supplies half of $N$. The electric current must make $I\mathbf{L}\times \mathbf{B}$ point horizontally toward the uphill side; for the coordinates just defined it flows along $-\hat{\mathbf{b}}$. Reversing the vertical magnetic field reverses this current direction. The external supply must maintain that electric current against the motional electromotive force during ascent. For $\cos \varphi =0$, no finite horizontal magnetic force can support uniform motion along the vertical rails.

Use the coordinates in 8. At downhill speed $v$, the moving bar has $\mathbf{v}\times \mathbf{B}=-Bv\cos \varphi \hat{\mathbf{b}}$. The electromotive force and induced electric current are thereforeHere $I$ is positive along $-\hat{\mathbf{b}}$ across the sliding bar, then returns through the rails and resistor. This direction makes the magnetic force oppose downhill motion, as required by Lenz's law. The factor $\cos \varphi$ is essential: only the component of magnetic field normal to the rail plane threads the circuit. Rail and bar electrical resistances and circuit self-inductance are neglected relative to the stated large $R$ model.

The component of magnetic force along the slope has magnitude $IBs\cos \varphi$. The rate at which the sliding bar does work against this magnetic braking isBy Ohm's law, $\mathcal{E}=IR$, soThis exactly equals the Joule heating in the resistor in the negligible-self-inductance approximation. The normal force does no work because it is perpendicular to the velocity.

The gravitational power is instead $mgv\sin \varphi$. By the work-energy theorem,It equals the Joule heating only at constant speed. For a concrete counterexample, during the initial acceleration the kinetic energy increases, so some gravitational power remains mechanical.

Project Newton's second law down the slope. The normal reaction has no component along the rails, while the magnetic term opposes motion:This first-order linear differential equation has the release-from-rest solutionThe electrical self-inductance has been neglected; retaining it introduces a coupled electric current equation rather than this instantaneous resistive response.

At the terminal velocity, the acceleration in iii vanishes. ThusFor $0<\varphi <\pi /2$ and nonzero $B$, the magnetic drag increases linearly with speed and this terminal state is stable. Horizontal rails have no gravitational drive. Vertical rails have no linked vertical magnetic flux change, so the model gives free fall rather than a finite terminal velocity.