IPhO 1991 · Theory problem 2

Charged balls moving along a square loop in an electric field (relativity)

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In a square loop with a side length \(L\), a large number of balls of negligible radius and each with a charge \(q\) are moving at a speed \(u\) with a constant separation \(a\) between them, as seen from a frame of reference that is fixed with respect to the loop. The balls are arranged on the loop like the beads on a necklace, \(L\) being much greater than \(a\), as indicated in the figure 2.1. The no conducting wire forming the loop has a homogeneous charge density per unit length in the in the frame of the loop. Its total charge is equal and opposite to the total charge of the balls in that frame.

Consider the situation in which the loop moves with velocity \(v\) parallel to its side \(AB\) (fig. 2.1) through a homogeneous electric field of strength \(E\) which is perpendicular to the loop velocity and makes an angle \(\theta\) with the plane of the loop.

Figure 1

Figure 2.1

Taking into account relativistic effects, calculate the following magnitudes in the frame of reference of an observer who sees the loop moving with velocity \(v\):

a) The spacing between the balls on each of the side of the loop, \(a_{AB}\), \(a_{BC}\), \(a_{CD}\), y \(a_{DA}\).

b) The value of the net charge of the loop plus balls on each of the side of the loop: \(Q_{AB}\), \(Q_{BC}\), \(Q_{CD}\) y, \(Q_{DA}\)

c) The modulus \(M\) of the electrically produced torque tending to rotate the system of the loop and the balls.

d) The energy \(W\) due to the interaction of the system, consisting of the loop and the balls with the electric field.

All the answers should be given in terms of quantities specified in the problem.

Note. The electric charge of an isolated object is independent of the frame of reference in which the measurements takes place. Any electromagnetic radiation effects should be ignored.

Some formulae of special relativity

Consider a reference frame \(S'\) moving with velocity \(V\) with reference to another reference frame \(S\). The axes of the frames are parallel, and their origins coincide a \(t=0\). \(V\) is directed along the positive direction of the \(x\) axis.

Relativistic sum of velocities

If a particle is moving with velocity \(u'\) in the \(x'\) direction , as measured in \(S'\), the velocity of the particle measured in \(S\) is given by:

\[ u=\frac{u'+V}{1+\dfrac{u'V}{c^2}} \]

Relativistic Contraction

If an object at rest in frame \(S\) has length \(L_0\) in the \(x\)-direction, an observer in frame \(S'\) (moving at velocity \(V\) in the \(x\)-direction} will measure its length to be:

\[ L=L_0\sqrt{1-\frac{v^2}{c^2}} \]

Official English text of the problem.

Answers

The results use the continuous charge approximation justified by \(L\gg a\), with the directions shown in Fig. 2.1. Here \(c\) is the speed of light.

  • a) \[ \boxed{ a_{AB}=\frac{a\sqrt{1-v^2/c^2}}{1+uv/c^2}, \qquad a_{BC}=a, \qquad a_{CD}=\frac{a\sqrt{1-v^2/c^2}}{1-uv/c^2}, \qquad a_{DA}=a.} \]

  • b) \[ \boxed{ Q_{AB}=\frac{qLuv}{ac^2}, \qquad Q_{BC}=0, \qquad Q_{CD}=-\frac{qLuv}{ac^2}, \qquad Q_{DA}=0.} \]

  • c) \[ \boxed{M=\frac{|q|L^2uvE}{ac^2}\sin\theta.} \]

  • d) For the field direction shown, whose in-plane component points from \(AB\) toward \(CD\), \[ \boxed{W=\frac{qL^2uvE}{ac^2}\cos\theta.} \]

Solution

Let \(S'\) be the rest frame of the loop and \(S\) the observer’s frame. Take the \(x\)-axis along \(A\to B\), so that the loop moves with velocity \(v\hat{\mathbf x}\). Take \(\hat{\mathbf y}\) along \(A\to D\), perpendicular to \(AB\) within the loop’s plane.

Define \[ \gamma=\frac{1}{\sqrt{1-v^2/c^2}}, \qquad \kappa=\frac{uv}{c^2}. \]

Since \(L\gg a\), we replace the discrete balls by a continuous line charge on each straight side, neglecting fluctuations associated with individual balls entering or leaving a corner. In \(S'\), the ball and wire charge densities are respectively \[ \lambda'_{\mathrm b}=\frac qa, \qquad \lambda'_{\mathrm w}=-\frac qa. \]

a) Spacing between the balls

The spacing must be measured between positions recorded simultaneously in \(S\). Such events are generally not simultaneous in \(S'\).

On either side parallel to the motion, let the ball velocity in \(S'\) be \(w'\), where \[ w'=+u \quad\text{on }AB, \qquad w'=-u \quad\text{on }CD. \]

Choose two neighbouring balls ordered along increasing \(x'\). For events on their trajectories, \[ \Delta x'=a+w'\Delta t'. \]

The Lorentz transformation gives \[ \Delta t=\gamma\left(\Delta t'+\frac{v\Delta x'}{c^2}\right). \] Requiring \(\Delta t=0\) yields \[ \Delta t'=-\frac{v\Delta x'}{c^2}. \] Consequently, \[ \Delta x'=a-\frac{vw'}{c^2}\Delta x', \qquad \Delta x'=\frac{a}{1+vw'/c^2}. \]

The spatial transformation then gives \[ \Delta x =\gamma(\Delta x'+v\Delta t') =\gamma\Delta x'\left(1-\frac{v^2}{c^2}\right) =\frac{\Delta x'}{\gamma}. \] Thus \[ a_{\parallel}=\frac{a}{\gamma(1+vw'/c^2)}. \]

Substituting the two velocities, \[ \boxed{ a_{AB}=\frac{a\sqrt{1-v^2/c^2}}{1+uv/c^2}, \qquad a_{CD}=\frac{a\sqrt{1-v^2/c^2}}{1-uv/c^2}.} \]

On \(BC\) or \(DA\), all balls on a given side have the same \(x'\). Therefore \(\Delta x'=0\), and simultaneous observations in \(S\) also have \(\Delta t'=0\). Transverse distances are unchanged by the Lorentz transformation, so \[ \boxed{a_{BC}=a_{DA}=a.} \]

b) Net charge on each side

In \(S\), the sides parallel to the motion are contracted: \[ \ell_{AB}=\ell_{CD}=\frac L\gamma, \qquad \ell_{BC}=\ell_{DA}=L. \]

Each wire side contains a fixed charge \[ Q_{\mathrm w}=-\frac{qL}{a}. \] This is the same in both frames because charge is invariant and the wire charges do not move between sides.

The number of balls on each side is its observed length divided by the corresponding observed spacing: \[ N_{AB} =\frac{L/\gamma}{a/[\gamma(1+\kappa)]} =\frac La(1+\kappa), \] \[ N_{CD} =\frac{L/\gamma}{a/[\gamma(1-\kappa)]} =\frac La(1-\kappa), \] and \[ N_{BC}=N_{DA}=\frac La. \]

Adding the wire charge to the ball charge gives \[ Q_{AB} =q\frac La(1+\kappa)-\frac{qL}{a} =\frac{qL}{a}\kappa, \] \[ Q_{CD} =q\frac La(1-\kappa)-\frac{qL}{a} =-\frac{qL}{a}\kappa. \] The two transverse sides remain neutral. Hence \[ \boxed{ Q_{AB}=\frac{qLuv}{ac^2},\quad Q_{BC}=0,\quad Q_{CD}=-\frac{qLuv}{ac^2},\quad Q_{DA}=0.} \]

Their sum is zero, as required by invariance of the total charge. The charge of an individual side need not remain zero because the balls occupying that side at a simultaneous instant differ between frames.

c) Electrically produced torque

Write \[ Q=\frac{qLuv}{ac^2}, \] so that the net charges on \(AB\) and \(CD\) are \(Q\) and \(-Q\).

In the continuous approximation these charges are uniformly distributed along their respective sides. Their resultants act at the side midpoints. The separation between those midpoints is \(L\hat{\mathbf y}\), since this distance is perpendicular to the loop’s motion and is not contracted.

The electric dipole moment is therefore \[ \mathbf p =Q\mathbf r_{AB}-Q\mathbf r_{CD} =-QL\hat{\mathbf y}. \]

In a uniform electric field the total force is zero, while the torque is \[ \boldsymbol{\tau}=\mathbf p\times\mathbf E. \]

The field component within the loop’s plane is parallel to \(\hat{\mathbf y}\) and produces no torque. The component perpendicular to the plane is \(E\sin\theta\), giving a torque about an axis parallel to \(AB\), of magnitude \[ M=|Q|LE\sin\theta. \] Thus \[ \boxed{M=\frac{|q|L^2uvE}{ac^2}\sin\theta.} \]

This has units of \(\mathrm{N\,m}\). If \(q\) is understood as a positive charge magnitude, \(|q|\) can simply be written \(q\).

d) Interaction energy

For a uniform external electric field, \[ \phi_{\mathrm{ext}}(\mathbf r) =\phi_0-\mathbf E\cdot\mathbf r. \] The interaction energy is \[ W=\int \phi_{\mathrm{ext}}\,dq. \] Since the system’s total charge is zero, the constant \(\phi_0\) cancels: \[ W=-\mathbf E\cdot\int\mathbf r\,dq =-\mathbf p\cdot\mathbf E. \]

The angle marker in Fig. 2.1 places the in-plane component of \(\mathbf E\) along \(A\to D\), so \[ \mathbf E\cdot\hat{\mathbf y}=E\cos\theta. \] Using \(\mathbf p=-QL\hat{\mathbf y}\), \[ W=QLE\cos\theta. \] Therefore \[ \boxed{W=\frac{qL^2uvE}{ac^2}\cos\theta.} \]

This has units of joules and vanishes when the field is perpendicular to the loop. For positive \(q\), the positive sign follows because the electric dipole moment opposes the field’s in-plane component.

The solution was checked against the official solution.