Past exam of the physics B course of the University of Cambridge / 2015 / Paper 1 / Section B / B7

The differential form of Faraday's law of induction isBy Stokes theorem, for a stationary oriented circuit it becomes $\mathcal{E}=\oint \mathbf{E}\cdot d\mathbf{\ell }=-d{\Phi }_B/dt$, where ${\Phi }_B=\int \mathbf{B}\cdot d\mathbf{S}$ is magnetic flux and $\mathcal{E}$ is electromotive force. Circuit orientation and surface normal obey the right-hand rule. Lenz's law is the physical meaning of the minus sign: the induced current's magnetic field opposes the change in linked magnetic flux. For a moving circuit, the force per unit electric charge is $\mathbf{E}+\mathbf{v}\times \mathbf{B}$, and its line integral is the corresponding electromotive force. Motion through a static magnetic field can therefore induce a voltage even when ${\partial }_t\mathbf{B}=0$.

Let $\theta$ be the wire's angle from the downward vertical, $r=L\sin \theta$, and choose $\mathbf{\omega }$ and $\mathbf{B}$ upward, as in the figure. The motional term along the wire has radial component $B\omega s$ at cylindrical radius $s$. In steady electrical equilibrium, integrating from the grounded axis to the bob givesThe Lorentz force on this positive electric charge is radially outward. Its ratio to the required centripetal term is $QB/(m\omega )\lesssim C{B}^2{L}^2/(2m)$, so the stated heavy-bob condition makes its effect on the zero-field conical pendulum geometry negligible. With wire tension $T_w$, $T_w\cos \theta =mg$ and $T_w\sin \theta =m{\omega }^{2}r$, giving $\cos \theta =g/({\omega }^{2}L)$. Hence the magnetically charged conical pendulum hasA nonzero-radius orbit requires ${\omega }^{2}L>g$. Reversing the sense of rotation or the magnetic field reverses the sign of $Q_0$.

The response to the added electric field requires an electrical convention. In the usual mechanical approximation, treat the pre-existing electric charge as $Q_0$, include its upward electric force $Q_{0}E$, and neglect all magnetic-force feedback. ThenThis predicts an increased radius for $Q_{0}E>0$. It is the result when the newly applied field is used only as a probe force on the original electric charge.

There is a first-order qualification if the bob remains electrically connected to the grounded pivot and the uniform added field's potential is included consistently. Put $s=\cos \theta$ and set the external electric potential to zero at the pivot. The external potential at the bob is then $ELs$, so the charging relation and full force balance becomeThe second equation follows by eliminating the wire tension between vertical balance and radial balance, including the outward magnetic Lorentz force. Expanding both equations to first order in $E$, and to leading order in $C{B}^2{L}^2/m$, with $s_0=g/({\omega }^{2}L)$, givesThe extra factor arises from the added field changing $Q$ by $-CEL{s}_0$ and therefore changing the radial magnetic force. That term is linear in $E$ too; the heavy-bob assumption alone does not justify retaining $Q_{0}E$ while dropping it. The two expressions refer to the two electrical approximations just specified. The expansion requires a small change relative to the initial radius and is nonuniform as that radius tends to zero.

For an electrical measurement, route the grounded connection through a sensitive electrometer operated near zero input voltage, and integrate the charging electric current when $B$ or $\omega$ is changed. Starting with the field off, the transferred electric charge is $Q_0$ in this ideal capacitance model. Reversing $B$ separates the induced signal from offsets. Alternatively, isolate the charged bob briefly and measure its electric potential with a high-input-impedance electrometer, accounting for the instrument's extra capacitance and for the fact that isolation changes the electrical boundary condition.