Kvant Physics Problem 172

A rigid conducting rod of length $l = 0.

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Problem

A conducting rod is suspended horizontally by two light wires in a vertical magnetic field with induction $E = 1~\text{Тл}$ (see Fig. 4). The length of the rod is $l = 0{,}2~\text{м}$, its mass is $m = 10~\text{г}$, and the length of the wires is $l_1 = 0{,}1~\text{м}$. A capacitor of capacitance $C = 100~\text{мкФ}$, charged to a voltage $U = 100~\text{В}$, is connected to the attachment points of the wires.

  1. Determine the maximum angle of deviation of the system from the equilibrium position after the capacitor is discharged, assuming that the discharge occurs in a very short time.
  2. Determine the capacitance $C_1$ of a capacitor for which the system will deflect by an angle $\alpha = 3^\circ$, if, when discharging a capacitor charged to the same voltage with capacitance $C_0 = 10~\text{мкФ}$, the deflection angle is $\alpha_0 = 2^\circ$.

V. V. Svetozarov

Setup and Assumptions

A rigid conducting rod of length $l = 0.2~\text{m}$ and mass $m = 0.01~\text{kg}$ is suspended horizontally by two identical light wires of length $l_1 = 0.1~\text{m}$. The suspension is symmetric, so after a horizontal impulse the system moves as a physical pendulum whose effective length is $l_1$ and whose oscillation occurs in a vertical plane.

A capacitor of capacitance $C$ charged to voltage $U$ is discharged through the rod in a time much shorter than the mechanical response time of the system. The magnetic field is uniform and vertical with induction $B = 1~\text{T}$. During discharge, a current flows through the rod, producing a Lorentz force perpendicular to both the rod and the magnetic field.

Air resistance, electrical resistance details, and deformation of the wires are neglected. The discharge is assumed complete so that the total transferred charge is $Q = CU$. The impulse approximation is valid, so the mechanical motion begins with a well-defined initial velocity and zero initial angular displacement.

Physical Principles

The mechanical effect of the magnetic field on a current-carrying conductor is given by the Lorentz force integrated over the conductor length, so the force magnitude on the rod is $F = B I l$.

The total mechanical impulse delivered during discharge is equal to the time integral of force,

$$J = \int F,dt = B l \int I,dt = B l Q,$$

where $Q = CU$ is the total charge passing through the rod.

The impulse changes the linear momentum of the rod,

$$J = m v_0.$$

After the impulse, the rod and wires form a conservative gravitational system. The kinetic energy converts into gravitational potential energy,

$$\frac{1}{2} m v_0^2 = m g h,$$

where the vertical rise of the center of mass for angular displacement $\alpha$ is

$$h = l_1(1 - \cos \alpha).$$

For small angles in part two, the exact energy relation is retained so that scaling between configurations remains valid.

Derivation

The total impulse delivered by the magnetic force during discharge is

$$J = B l C U.$$

Substituting numerical values,

$$J = (1~\text{T})(0.2~\text{m})(100 \times 10^{-6}\text{F})(100\text{V}).$$

The product of capacitance and voltage gives charge,

$$CU = 100 \times 10^{-6} \cdot 100 = 10^{-2}~\text{C}.$$

Hence,

$$J = 0.2 \cdot 10^{-2} = 2 \times 10^{-3}~\text{N}\cdot\text{s}.$$

The initial velocity follows from momentum conservation,

$$v_0 = \frac{J}{m} = \frac{2 \times 10^{-3}}{0.01} = 0.2~\text{m/s}.$$

The initial kinetic energy is

$$\frac{1}{2} m v_0^2 = \frac{1}{2} \cdot 0.01 \cdot (0.2)^2 = 2 \times 10^{-4}~\text{J}.$$

At maximum deviation, all kinetic energy is converted into gravitational potential energy,

$$\frac{1}{2} m v_0^2 = m g l_1 (1 - \cos \alpha).$$

After canceling mass,

$$\frac{1}{2} v_0^2 = g l_1 (1 - \cos \alpha).$$

Substitution yields

$$\frac{1}{2} (0.2)^2 = 9.8 \cdot 0.1 \cdot (1 - \cos \alpha).$$

This gives

$$0.02 = 0.98 (1 - \cos \alpha),$$

so

$$1 - \cos \alpha = 2.0408 \times 10^{-2}.$$

Thus,

$$\cos \alpha = 0.979592.$$

The angle is

$$\alpha = \arccos(0.979592) \approx 0.202~\text{rad}.$$

Conversion to degrees gives

$$\alpha \approx 11.6^\circ.$$

For the second part, the impulse scales linearly with capacitance,

$$J \propto C,$$

so the initial velocity also scales as

$$v_0 \propto C,$$

and the kinetic energy scales as

$$\frac{1}{2} v_0^2 \propto C^2.$$

From energy conservation,

$$1 - \cos \alpha \propto C^2.$$

Therefore for two cases,

$$\frac{1 - \cos \alpha}{1 - \cos \alpha_0} = \left(\frac{C_1}{C_0}\right)^2.$$

Solving for $C_1$ gives

$$C_1 = C_0 \sqrt{\frac{1 - \cos \alpha}{1 - \cos \alpha_0}}.$$

For small angles, $1 - \cos \alpha \approx \alpha^2/2$, so

$$C_1 = C_0 \frac{\alpha}{\alpha_0}.$$

Using $\alpha_0 = 2^\circ = 0.0349066~\text{rad}$ and $\alpha = 3^\circ = 0.0523599~\text{rad}$,

$$\frac{\alpha}{\alpha_0} = 1.5.$$

Thus,

$$C_1 = 10~\mu\text{F} \cdot 1.5 = 15~\mu\text{F}.$$

Result

For the first part,

$$\alpha_{\max} = \arccos!\left(1 - \frac{(B l C U)^2}{2 m^2 g l_1}\right),$$

which numerically gives

$$\boxed{\alpha_{\max} \approx 11.6^\circ}.$$

For the second part,

$$C_1 = C_0 \sqrt{\frac{1 - \cos 3^\circ}{1 - \cos 2^\circ}},$$

which evaluates to

$$\boxed{C_1 \approx 15~\mu\text{F}}.$$

Sanity Checks

The impulse has dimensions $[B l C U] = \text{T}\cdot\text{m}\cdot\text{C} = \text{N}\cdot\text{s}$, consistent with momentum transfer. The derived velocity $0.2~\text{m/s}$ is small compared with typical mechanical speeds in gravitational pendula of length $0.1~\text{m}$, which supports the nonrelativistic treatment.

The computed angle $11.6^\circ$ corresponds to a moderate deflection where the exact cosine relation remains more accurate than a linear small-angle approximation, since $1 - \cos \alpha \approx 0.02$ is not negligibly small.

The second result depends on the quadratic scaling of energy with capacitance, so the proportionality $C_1/C_0 = \alpha/\alpha_0$ is consistent with the small-angle regime where $\alpha^2$ dominates the potential energy change. The ratio $1.5$ is physically reasonable because increasing deflection from $2^\circ$ to $3^\circ$ requires about $2.25$ times more energy, hence about $\sqrt{2.25} = 1.5$ times larger capacitance.