Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / Section E

Gauss's law statesThe surface is closed and oriented outward, and $Q_{\mathrm{enc}}$ includes all enclosed charge. The integral law follows from the local form through the divergence theorem. It is universally valid in classical electromagnetism, but calculating the field directly from the flux requires enough symmetry.

For spherical charge symmetry, $E(r)=Q(<r)/(4\pi {ϵ}_0{r}^2)$. A uniform infinite line gives $E=\lambda /(2\pi {ϵ}_{0}r)$ from a coaxial cylinder. An infinite sheet gives $E=\sigma /(2{ϵ}_0)$ on either side from a pillbox, with the direction set by charge sign. The normal field jump is $\sigma /{ϵ}_0$; the electrostatic field inside a conductor is zero. Zero enclosed charge implies zero net flux, not generally zero field. ^[ref]

For a steady thin-wire current in vacuum, the Biot-Savart law isThe line element follows conventional current direction. More generally replace $Id{\mathbf{\ell }}^{\prime }$ with $\mathbf{J}({\mathbf{r}}^{\prime })d^3{r}^{\prime }$. The cross product gives the circulating field direction and superposition allows contributions from separate wire segments to be added.

Integration gives $B={\mu }_{0}I/(2\pi r)$ around a long straight wire. On the axis of a circular loop of radius a, $B_z={\mu }_{0}I{a}^2/[2(a^2+z^2{)}^{3/2}]$, including the centre value ${\mu }_{0}I/(2a)$. Coil turns multiply the field when their geometries coincide; distributed coils and solenoids follow by integration. This is a magnetostatic or quasistatic calculation: rapidly changing currents require retarded electrodynamics and appropriate displacement-current effects, not an instantaneous wire formula. ^[ref]

For a stationary closed circuit, Faraday's law of induction isThe loop orientation and magnetic-flux normal obey the right-hand convention. For a moving material circuit, the emf is instead $\oint (\mathbf{E}+\mathbf{v}\times \mathbf{B})\cdot d\mathbf{\ell }$, with the same negative derivative of the flux through the moving circuit. The additional magnetic force term accounts for motional emf.

Lenz's law expresses the minus sign physically: an induced current opposes the change in flux that produces it. It does not always oppose the field itself. Increasing positive flux drives a current producing negative flux; decreasing it reverses that response. A straight conducting rod of length l moving across a perpendicular uniform field has emf magnitude Blv. A resistive return circuit then develops a magnetic braking force, and the supplied mechanical power equals Joule heating in the ideal steady limit. This connects induction with conservation of energy. ^[ref]

A conservative force has work depending only on the endpoints, so it can be writtenThe arbitrary additive constant in potential energy has no effect on force. A smooth curl-free force is conservative on a simply connected domain; a hole or excluded singularity requires checking global circulation as well.

Uniform gravity has $U=mgz$, central gravity has $U=-GMm/r$ with zero at infinity, and a linear spring has $U=k{x}^2/2$ relative to equilibrium. If the potential is time independent and no other force does work, $K+U$ is conserved by the work-energy theorem. A time-dependent potential or nonconservative work can change mechanical energy; having an instantaneous gradient force alone is not sufficient for energy conservation. ^[ref]

Treat Earth as a stationary spherical source, with $\mu =G{M}_E$, and neglect drag and other bodies. The conserved quantities are total mechanical energy and the angular momentum vector, implying a fixed orbital plane and constant areal velocity. The Kepler problem additionally conserves the Laplace-Runge-Lenz vector. The orbiting object's linear momentum is not constant because gravity exerts a force.

For a circular orbit, $m{v}_c^2/r_0=\mu m/r_0^2$, so $K=\mu m/(2r_0)=-U/2$. HenceAt the closest approach of a noncircular orbit, the velocity is tangential and $v_0=k\sqrt{\mu /r_0}$. Its energy and specific angular momentum areThe pericentre-speed qualification of Kepler orbit classes requires $k\ge 1$ when $r_0$ is genuinely the closest approach. For an ellipse, conservation of h gives $v_m=v_0{r}_0/r_m$. Equating energies at the two turning points and writing $x=r_m/r_0$ givesThe distinct apocentre root is thereforeFor an unbound orbit, $v^2(r)=2E/m+2\mu /r$ decreases toward infinity. The asymptotic minimum speed of an unbound Kepler orbit isIt is the limiting infimum, reached only asymptotically: positive for a hyperbola and zero for a parabola. The turning-point interpretation of $r_0$ is essential; if $0<k<1$, the same instantaneous energy describes a bound orbit whose $r_0$ is the apocentre instead. ^[ref]

The orbit is circular for $\boxed{k=1}$. The speed equals $\sqrt{G{M}_E/r_0}$ and radial acceleration balances the changing direction of the velocity.

A noncircular ellipse with $r_0$ at pericentre has $\boxed{1<k<\sqrt{2}}$. Its energy is negative. Merely specifying $k<1$ would make the stated closest-approach point an apocentre, so it is excluded by the geometry; the circular case is listed separately.

The orbit is unbound for $\boxed{k\ge \sqrt{2}}$: parabolic at equality and hyperbolic above it. The pericentre speed has reached or exceeded the local escape velocity.