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 points where the wires are attached.

  1. Determine the maximum angle of deviation of the system from its 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, upon discharge, the system will deflect by an angle $\alpha = 3^\circ$, if when a capacitor charged to the same voltage with capacitance $C_0 = 10~\text{мкФ}$ is discharged, the angle of deflection is $\alpha_0 = 2^\circ$.

V. V. Svetozarov

Setup and Assumptions

A rigid conducting rod of length $l = 0.2,\text{m}$ is suspended horizontally by two identical light wires of length $l_1 = 0.1,\text{m}$. The system is placed in a uniform vertical magnetic field with induction $B = 1,\text{T}$. A capacitor of capacitance $C = 100,\mu\text{F}$, initially charged to voltage $U = 100,\text{V}$, is connected to the suspension points of the wires, so that during discharge a current pulse flows through the rod.

The rod mass is $m = 10,\text{g} = 0.01,\text{kg}$. The discharge time is assumed negligible compared with the mechanical oscillation period, so the magnetic force acts as an impulse.

The motion after discharge is treated as planar motion of a rigid body in a gravitational field, with the wires behaving as massless constraints that define a pendulum-like motion with effective length $l_1$. Air resistance and electrical resistance effects are neglected in the mechanical evolution after the impulse.

Physical Principles

The impulse of the magnetic force acting on a current-carrying conductor in a uniform magnetic field is determined by

$\mathbf{F} = I,\mathbf{l} \times \mathbf{B}.$

The time-integrated impulse is

$\mathbf{J} = \int \mathbf{F},dt = B l \int I,dt,$

where $\int I,dt$ equals the total charge $Q$ passing through the rod during discharge.

For a capacitor,

$Q = CU.$

The mechanical energy immediately after the impulse is converted into gravitational potential energy at the turning point,

$\frac{1}{2}mv^2 = mg h,$

where the vertical rise for a pendulum of length $l_1$ is

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

Derivation

The total charge transferred during discharge is

$Q = CU.$

The magnetic impulse delivered to the rod is

$J = B l Q = B l C U.$

This impulse equals the momentum of the rod,

$mv = B l C U,$

so the initial speed is

$v = \frac{B l C U}{m}.$

The kinetic energy immediately after discharge is

$\frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{B l C U}{m}\right)^2 = \frac{B^2 l^2 C^2 U^2}{2m}.$

At the maximum deflection angle $\alpha$, energy conservation gives

$\frac{B^2 l^2 C^2 U^2}{2m} = mg l_1(1 - \cos\alpha).$

Solving for the angle dependence,

$1 - \cos\alpha = \frac{B^2 l^2 C^2 U^2}{2m^2 g l_1}.$

First part

Substituting numerical values:

$Q = CU = 100 \cdot 10^{-6},\text{F} \cdot 100,\text{V} = 1.0 \times 10^{-2},\text{C}.$

Impulse:

$J = B l Q = 1,\text{T} \cdot 0.2,\text{m} \cdot 1.0 \times 10^{-2},\text{C} = 2.0 \times 10^{-3},\text{N·s}.$

Velocity:

$v = \frac{J}{m} = \frac{2.0 \times 10^{-3},\text{N·s}}{1.0 \times 10^{-2},\text{kg}} = 0.2,\text{m/s}.$

Energy:

$\frac{1}{2}mv^2 = \frac{1}{2}\cdot 1.0\times 10^{-2},\text{kg}\cdot (0.2,\text{m/s})^2 = 2.0 \times 10^{-4},\text{J}.$

Potential energy scale:

$mgl_1 = 1.0\times 10^{-2},\text{kg}\cdot 9.8,\text{m/s}^2 \cdot 0.1,\text{m} = 9.8\times 10^{-3},\text{J}.$

Thus

$1 - \cos\alpha = \frac{2.0\times 10^{-4}}{9.8\times 10^{-3}} = 2.04\times 10^{-2}.$

So

$\cos\alpha = 0.9796,$

which gives

$\alpha \approx 11.5^\circ.$

Second part

The relation derived above shows

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

Hence for two cases,

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

Using $\alpha_0 = 2^\circ$, $\alpha = 3^\circ$, $C_0 = 10,\mu\text{F}$:

$\cos 2^\circ = 0.9993908,\quad 1 - \cos 2^\circ = 6.092\times 10^{-4},$

$\cos 3^\circ = 0.9986295,\quad 1 - \cos 3^\circ = 1.3705\times 10^{-3}.$

Ratio:

$\frac{1 - \cos 3^\circ}{1 - \cos 2^\circ} = \frac{1.3705\times 10^{-3}}{6.092\times 10^{-4}} = 2.25.$

Therefore,

$\frac{C_1}{C_0} = \sqrt{2.25} = 1.5,$

so

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

Result

First part:

$\alpha = \arccos!\left(1 - \frac{B^2 l^2 C^2 U^2}{2m^2 g l_1}\right) \approx 11.5^\circ.$

Second part:

$C_1 = C_0 \sqrt{\frac{1 - \cos 3^\circ}{1 - \cos 2^\circ}} = 15,\mu\text{F}.$

Final answers:

$\boxed{\alpha \approx 11.5^\circ}, \qquad \boxed{C_1 \approx 15,\mu\text{F}}.$

Sanity Checks

The impulse scales as $B l C U$, giving units $\text{T}\cdot\text{m}\cdot\text{C} = \text{N·s}$, consistent with momentum transfer. The derived angle depends on the dimensionless ratio $v^2/(g l_1)$, ensuring dimensional consistency.

The first result gives a moderate angle, consistent with a small but non-negligible conversion of electrical energy $CU^2/2 = 0.5,\text{J}$ into mechanical energy $2\times 10^{-4},\text{J}$, indicating low mechanical efficiency of conversion through the impulse mechanism.

In the second part, the proportionality $1 - \cos\alpha \propto C^2$ predicts that a modest increase in angle from $2^\circ$ to $3^\circ$ requires a capacitance increase by a factor close to $1.5$, consistent with the computed value.