Kvant Physics Problem 58
The system consists of a thin metallic ring of mass $m$ and resistance $R$, with diameter $d$, falling vertically in a magnetic field that points along the vertical direction.
Verified: yes
Verdicts: PASS + PASS
Solve time: 10m19s
Source on kvant.digital
Problem
F58. A ring with diameter $d$ and resistance $R$ falls from a great height in a magnetic field. The plane of the ring remains horizontal at all times, and the mass of the ring is $m$. Find the steady-state falling speed of the ring if the magnitude of the magnetic field induction vector varies with height $H$ according to
$$B = B_{0}(1+\alpha H),$$
where $B_{0}$ and $\alpha$ are constants.
The problem considers the motion of a thin metallic ring in a vertical magnetic field with height-dependent induction; the plane of the ring remains horizontal throughout the fall.
Setup and Assumptions
The system consists of a thin metallic ring of mass $m$ and resistance $R$, with diameter $d$, falling vertically in a magnetic field that points along the vertical direction. The magnetic field induction varies with height $H$ according to
$$B(H)=B_0(1+\alpha H),$$
where $B_0$ is the field at $H=0$ and $\alpha$ is a constant with dimensions of inverse length. The ring maintains a horizontal orientation at all times, so the plane of the ring is perpendicular to the magnetic field. The unknown quantity is the steady-state velocity $v$ of the ring as it falls.
The following assumptions are made: the ring is rigid and perfectly conducting except for the specified resistance $R$, air resistance is negligible, self-inductance of the ring is ignored because we consider the steady state, and gravitational acceleration $g$ is constant. The system is analyzed in an inertial frame with the $H$-axis pointing upward; the downward direction is taken as positive for velocity.
Since the field is specified by the linear law $B(H)=B_0(1+\alpha H)$, the field gradient is constant:
$$\frac{dB}{dH}=B_0\alpha.$$
For a thin ring whose diameter is small compared with the characteristic length $1/|\alpha|$, the variation of $B$ across the ring is negligible. The magnetic field may then be treated as uniform over the ring area at each instant, while its value changes as the ring moves through the gradient. Under this approximation the magnetic flux is simply $BA$, with $A=\pi d^2/4$.
Physical Principles
Faraday’s law of electromagnetic induction states that an electromotive force is induced in a closed conducting loop when the magnetic flux through the loop changes:
$$\mathcal E=-\frac{d\Phi}{dt},$$
where $\Phi$ is the magnetic flux through the ring.
Ohm’s law gives the induced current:
$$I=\frac{\mathcal E}{R}.$$
The expression
$$F_{\rm mag}=IBL$$
is not applicable to the ring as a whole, because the magnetic forces on different elements of the ring are directed radially and their vector sum is not obtained from such a formula. The magnetic braking force is obtained instead from energy conservation.
If the ring moves with speed $v$, the mechanical power extracted from the motion by the electromagnetic interaction is
$$P_{\rm mech}=F_{\rm mag}v.$$
Neglecting self-inductance, the induced current is quasistationary, and all electromagnetic work is converted into Joule heat. The electrical power dissipated in the ring is
$$P_{\rm el}=I^2R.$$
Hence
$$F_{\rm mag}v=I^2R.$$
In the steady state the acceleration vanishes, so the upward magnetic force balances the weight:
$$mg=F_{\rm mag}.$$
Derivation
The magnetic flux through the ring is
$$\Phi(H)=B(H)A = B_0(1+\alpha H)\frac{\pi d^2}{4},$$
where
$$A=\frac{\pi d^2}{4}.$$
As the ring falls with speed $v$,
$$\frac{d\Phi}{dt} = \frac{d\Phi}{dH}\frac{dH}{dt}.$$
Since
$$\frac{d\Phi}{dH} = A\frac{dB}{dH} = \frac{\pi d^2}{4},B_0\alpha,$$
and the magnitude of $dH/dt$ equals $v$, the magnitude of the induced emf is
$$|\mathcal E| = \frac{\pi d^2}{4},B_0\alpha,v.$$
Thus
$$I = \frac{|\mathcal E|}{R} = \frac{B_0\alpha\pi d^2}{4R},v.$$
The electrical power dissipated in the ring is
$$I^2R = \left(\frac{B_0\alpha\pi d^2}{4R}v\right)^2R = \frac{(B_0\alpha\pi d^2)^2}{16R}v^2.$$
Using the power balance relation,
$$F_{\rm mag}v=I^2R,$$
gives
$$F_{\rm mag} = \frac{I^2R}{v} = \frac{(B_0\alpha\pi d^2)^2}{16R}v.$$
The appearance of $B_0$ rather than $B(H)$ is not an additional approximation. The induced emf depends on the flux derivative
$$\frac{d\Phi}{dH} = A\frac{dB}{dH},$$
and for the given field
$$\frac{dB}{dH}=B_0\alpha$$
is a constant independent of height. Consequently the induced current and the magnetic braking force are independent of the instantaneous value of $H$.
At terminal velocity,
$$mg=F_{\rm mag},$$
hence
$$mg = \frac{(B_0\alpha\pi d^2)^2}{16R}v.$$
Solving for $v$,
$$v = \frac{16mgR}{\pi^2B_0^2d^4\alpha^2}.$$
Result
The steady-state falling speed of the ring is
$$\boxed{ v= \frac{16mgR} {\pi^2B_0^2d^4\alpha^2} }.$$
If, for example,
$$m=0.1\ {\rm kg}, \qquad R=0.5\ \Omega, \qquad d=0.1\ {\rm m}, \qquad B_0=0.2\ {\rm T}, \qquad \alpha=1\ {\rm m^{-1}},$$
then
$$v= \frac{16\cdot0.1\cdot9.8\cdot0.5} {\pi^2\cdot(0.2)^2\cdot(0.1)^4}.$$
The numerator equals
$$16\cdot0.1\cdot9.8\cdot0.5=7.84.$$
The denominator equals
$$\pi^2\cdot0.04\cdot10^{-4} = 9.8696\cdot4\times10^{-6} = 3.94784\times10^{-5}.$$
Hence
$$v = \frac{7.84}{3.94784\times10^{-5}} \approx1.99\times10^5\ {\rm m/s}.$$
Thus
$$\boxed{ v\approx2.0\times10^5\ {\rm m/s} }.$$
Sanity Checks
The dimensions are correct. Using
$$1,{\rm T} = \frac{{\rm kg}} {{\rm A,s^2}}, \qquad 1,\Omega = \frac{{\rm kg,m^2}} {{\rm A^2,s^3}},$$
the dimensions of
$$\frac{mgR}{B^2d^4\alpha^2}$$
are
$$\frac{ ( {\rm kg,m,s^{-2}}) ({\rm kg,m^2,A^{-2},s^{-3}}) }{ ({\rm kg^2,A^{-2},s^{-4}}) ({\rm m^4}) ({\rm m^{-2}}) } = {\rm m,s^{-1}}.$$
In the limit $\alpha\to0$, the field becomes uniform and
$$\frac{dB}{dH}\to0,$$
so no flux change is produced by the motion. The induced current tends to zero and the terminal speed diverges, which is consistent with the absence of electromagnetic braking.
Increasing $R$ decreases the induced current for a given speed, reducing the dissipated power and the magnetic braking force. The terminal speed is therefore proportional to $R$, in agreement with the formula.
The derivation is valid when the ring is sufficiently small that the magnetic field can be regarded as uniform over its area at each instant. Under that standard approximation, the result follows rigorously from Faraday’s law, Ohm’s law, Joule dissipation, and mechanical energy balance.