= Solution
The differential form of <Faraday's law of induction> is
$$
\nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t}.
$$
By <Stokes theorem>, for a stationary oriented circuit it becomes $\mathcal E=\oint\mathbf E\cdot d\boldsymbol\ell=-d\Phi_B/dt$, 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 $\boldsymbol\omega$ 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 gives
$$
V_b=\int_0^rB\omega s\,ds=\frac12B\omega r^2,
\qquad Q=CV_b=\frac12BC\omega r^2.
$$
The <Lorentz force> on this positive <electric charge> is radially outward. Its ratio to the required centripetal term is $QB/(m\omega)\lesssim CB^2L^2/(2m)$, 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 $T_w\sin\theta=m\omega^2r$, giving $\cos\theta=g/(\omega^2L)$. Hence the <magnetically charged conical pendulum> has
$$
\boxed{Q_0=\frac12BC\left(\omega L^2-\frac{g^2}{\omega^3}\right).}
$$
A nonzero-radius orbit requires $\omega^2L>g$. 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> $Q_0E$, and neglect all magnetic-force feedback. Then
$$
\cos\theta(E)=\frac{mg-Q_0E}{m\omega^2L},\qquad
\frac{\delta r}{r}=-\frac{\cos\theta\,\delta(\cos\theta)}{\sin^2\theta}
=\frac{Q_0E g}{m\omega^4r^2}
=\boxed{\frac{CBgE}{2m\omega^3}.}
$$
This predicts an increased radius for $Q_0E>0$. 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 become
$$
Q=C\!\left[\frac12B\omega L^2(1-s^2)-ELs\right],\qquad
\omega^2Ls+\frac Qm(B\omega Ls+E)=g.
$$
The 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 $CB^2L^2/m$, with $s_0=g/(\omega^2L)$, gives
$$
\delta s=-\frac{CBLE}{2m\omega}(1-3s_0^2),\qquad
\boxed{\frac{\delta r}{r}=\frac{CBgE}{2m\omega^3}\,
\frac{1-3s_0^2}{1-s_0^2}.}
$$
The extra factor arises from the added field changing $Q$ by $-CELs_0$ and therefore changing the radial magnetic <force>. That term is linear in $E$ too; the heavy-bob assumption alone does not justify retaining $Q_0E$ 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.
Back to article page
