IPhO 1999 · Theory problem 2

Magnetic field with a V-shaped wire

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Among the first successes of the interpretation by Ampère of magnetic phenomena, we have the computation of the magnetic field \(\mathbf B\) generated by wires carrying an electric current, as compared to early assumptions originally made by Biot and Savart.

A particularly interesting case is that of a very long thin wire, carrying a constant current \(i\), made out of two rectilinear sections and bent in the form of a "V", with angular half-span¹ \(\alpha\) (see figure). According to Ampère's computations, the magnitude \(B\) of the magnetic field in a given point P lying on the axis of the "V", outside of it and at a distance \(d\) from its vertex, is proportional to \(\tan\left(\frac{\alpha}{2}\right)\). Ampère's work was later embodied in Maxwell's electromagnetic theory, and is universally accepted.

Figure 1

Using our contemporary knowledge of electromagnetism,

  1. Find the direction of the field \(\mathbf B\) in P. [1 point]

  2. Knowing that the field is proportional to \(\tan\left(\frac{\alpha}{2}\right)\), find the proportionality factor \(k\) in \[ |\mathbf B(\mathrm P)|=k\tan\left(\frac{\alpha}{2}\right). \] [1.5 points]

  3. Compute the field \(\mathbf B\) in a point \(\mathrm P^*\) symmetric to P with respect to the vertex, i.e. along the axis and at the same distance \(d\), but inside the "V" (see figure). [2 points]

¹ Throughout this problem \(\alpha\) is expressed in radians

Figure 2
  1. In order to measure the magnetic field, we place in P a small magnetic needle with moment of inertia \(I\) and magnetic dipole moment \(\boldsymbol\mu\); it oscillates around a fixed point in a plane containing the direction of \(\mathbf B\). Compute the period of small oscillations of this needle as a function of \(B\). [2.5 points]

In the same conditions Biot and Savart had instead assumed that the magnetic field in P might have been (we use here the modern notation) \(B(\mathrm P)=\frac{i\mu_0\alpha}{\pi^2d}\), where \(\mu_0\) is the magnetic permeability of vacuum. In fact they attempted to decide with an experiment between the two interpretations (Ampère's and Biot and Savart's) by measuring the oscillation period of the magnetic needle as a function of the "V" span. For some \(\alpha\) values, however, the differences are too small to be easily measurable.

  1. If, in order to distinguish experimentally between the two predictions for the magnetic needle oscillation period \(T\) in P, we need a difference by at least 10%, namely \(T_1>1.10\,T_2\) (\(T_1\) being the Ampère prediction and \(T_2\) the Biot-Savart prediction) state in which range, approximately, we must choose the "V" half-span \(\alpha\) for being able to decide between the two interpretations. [3 points]

Hint

Depending on which path you follow in your solution, the following trigonometric equation might be useful: \[ \tan\left(\frac{\alpha}{2}\right)=\frac{\sin\alpha}{1+\cos\alpha} \]

Physical constants and general data

In addition to the numerical data given within the text of the individual problems, the knowledge of some general data and physical constants may be useful, and you may find them among the following ones. These are nearly the most accurate data currently available, and they have thus a large number of digits; you are expected, however, to write your results with a number of digits that must be appropriate for each problem.

Speed of light in vacuum: \(c=299792458\ \mathrm{m\cdot s^{-1}}\)
Magnetic permeability of vacuum: \(\mu_0=4\pi\cdot10^{-7}\ \mathrm{H\cdot m^{-1}}\)
Dielectric constant of vacuum: \(\varepsilon_0=8.8541878\ \mathrm{pF\cdot m^{-1}}\)
Gravitational constant: \(G=6.67259\cdot10^{-11}\ \mathrm{m^3/(kg\cdot s^2)}\)
Gas constant: \(R=8.314510\ \mathrm{J/(mol\cdot K)}\)
Boltzmann's constant: \(k=1.380658\cdot10^{-23}\ \mathrm{J\cdot K^{-1}}\)
Stefan's constant: \(\sigma=56.703\ \mathrm{nW/(m^2\cdot K^4)}\)
Proton charge: \(e=1.60217733\cdot10^{-19}\ \mathrm C\)
Electron mass: \(m_e=9.1093897\cdot10^{-31}\ \mathrm{kg}\)
Planck's constant: \(h=6.6260755\cdot10^{-34}\ \mathrm{J\cdot s}\)
Base of centigrade scale: \(T_K=273.15\ \mathrm K\)
Sun mass: \(M_S=1.991\cdot10^{30}\ \mathrm{kg}\)
Earth mass: \(M_E=5.979\cdot10^{24}\ \mathrm{kg}\)
Mean radius of Earth: \(r_E=6.373\ \mathrm{Mm}\)
Major semiaxis of Earth orbit: \(R_E=1.4957\cdot10^{11}\ \mathrm m\)
Sidereal day: \(d_S=86.16406\ \mathrm{ks}\)
Year: \(y=31.558150\ \mathrm{Ms}\)
Standard value of the gravitational field at the Earth surface: \(g=9.80665\ \mathrm{m\cdot s^{-2}}\)
Standard value of the atmospheric pressure at sea level: \(p_0=101325\ \mathrm{Pa}\)
Refractive index of air for visibile light, at standard pressure and \(15\,^\circ\mathrm C\): \(n_{\mathrm{air}}=1.000277\)
Solar constant: \(S=1355\ \mathrm{W\cdot m^{-2}}\)
Jupiter mass: \(M=1.901\cdot10^{27}\ \mathrm{kg}\)
Equatorial Jupiter radius: \(R_B=69.8\ \mathrm{Mm}\)
Mean radius of Jupiter’s orbit: \(R=7.783\cdot10^{11}\ \mathrm m\)
Jovian day: \(d_J=35.6\ \mathrm{ks}\)
Jovian year: \(y_J=374.32\ \mathrm{Ms}\)
\(\pi\): \(3.14159265\)

Official English text of the problem.

Answers

Let \(\hat{\mathbf z}\) point out of the page, towards the reader, and let \(\mu=|\boldsymbol\mu|\).

  1. The field at P is perpendicular to the wire's plane, out of the page.

  2. \[ \boxed{k=\frac{\mu_0i}{2\pi d}}, \qquad \boxed{\mathbf B(\mathrm P)= \frac{\mu_0i}{2\pi d}\tan\left(\frac{\alpha}{2}\right)\hat{\mathbf z}}. \]

  3. \[ \boxed{\mathbf B(\mathrm P^*)= -\frac{\mu_0i}{2\pi d}\cot\left(\frac{\alpha}{2}\right)\hat{\mathbf z}}, \] directed into the page.

  4. \[ \boxed{T=2\pi\sqrt{\frac{I}{\mu B}}}. \]

  5. For the stated strict inequality \(T_1>1.10T_2\), \[ \boxed{0<\alpha<\alpha_c,\qquad \alpha_c\simeq0.768\ \mathrm{rad}\simeq44.0^\circ}. \] At \(\alpha=\alpha_c\), the difference is exactly 10%.

Solution

We treat the two long, thin arms as semi-infinite straight wires in vacuum. The current enters the vertex along the lower arm and leaves along the upper arm, as shown. Take the vertex as the origin, with the positive \(x\)-axis pointing into the V and \(\hat{\mathbf z}\) pointing out of the page.

1. Direction of the field at P

The modern Biot–Savart law is \[ d\mathbf B=\frac{\mu_0i}{4\pi} \frac{d\boldsymbol\ell\times\mathbf R}{R^3}, \] where \(d\boldsymbol\ell\) follows the current and \(\mathbf R\) points from the current element to the observation point.

Both vectors lie in the plane of the figure, so every contribution is perpendicular to that plane. At P, the right-hand rule gives an outward contribution from each arm: the lower arm carries current towards the vertex, while the upper arm carries current away from it. Thus their contributions add: \[ \boxed{\mathbf B(\mathrm P)\text{ points out of the page}.} \]

2. Determination of \(k\)

The angular dependence is given: \[ B(\mathrm P)=k\tan\left(\frac{\alpha}{2}\right). \]

Consider the limiting configuration \(\alpha=\pi/2\). The two arms then form one infinite straight wire, and P lies at perpendicular distance \(d\) from it.

For this straight wire, cylindrical symmetry and Ampère's circuital law give \[ \oint\mathbf B\cdot d\boldsymbol\ell =B(2\pi d)=\mu_0i. \] Consequently, \[ B=\frac{\mu_0i}{2\pi d}. \] Since \(\tan(\pi/4)=1\), this determines \[ \boxed{k=\frac{\mu_0i}{2\pi d}}. \]

Using the supplied permeability, \[ k=\left(2\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\right)\frac{i}{d}. \] Therefore, \[ \boxed{\mathbf B(\mathrm P)= \frac{\mu_0i}{2\pi d}\tan\left(\frac{\alpha}{2}\right)\hat{\mathbf z}}. \]

3. Field at \(\mathrm P^*\)

At \(\mathrm P^*\), the right-hand rule gives a field into the page from each original arm.

To find its magnitude, continue both arms through the vertex, preserving their current directions, so that each becomes an infinite straight wire. The perpendicular distance from \(\mathrm P^*\) to either completed wire is \[ \rho=d\sin\alpha. \] Each completed wire produces a field into the page of magnitude \[ \frac{\mu_0i}{2\pi\rho}. \] Their total field therefore has magnitude \[ B_{\mathrm{complete}}=\frac{\mu_0i}{\pi d\sin\alpha}. \]

The two added half-lines form a left-opening V, which is the mirror image of the original wire. Relative to this added V, \(\mathrm P^*\) is an exterior axial point at distance \(d\). Its field consequently has magnitude \(B(\mathrm P)\), and the right-hand rule shows that it also points into the page.

Superposition thus gives \[ B(\mathrm P^*)+B(\mathrm P) =\frac{\mu_0i}{\pi d\sin\alpha}. \] Substituting the result of part 2, \[ B(\mathrm P^*)= \frac{\mu_0i}{2\pi d} \left[\frac{2}{\sin\alpha} -\tan\left(\frac{\alpha}{2}\right)\right]. \] Using \[ \tan\left(\frac{\alpha}{2}\right) =\frac{1-\cos\alpha}{\sin\alpha}, \] the bracket becomes \[ \frac{2-(1-\cos\alpha)}{\sin\alpha} =\frac{1+\cos\alpha}{\sin\alpha} =\cot\left(\frac{\alpha}{2}\right). \] Hence \[ \boxed{\mathbf B(\mathrm P^*)= -\frac{\mu_0i}{2\pi d} \cot\left(\frac{\alpha}{2}\right)\hat{\mathbf z}}. \]

4. Period of small oscillations

Treat the small needle as a magnetic dipole in a field uniform over its length, neglecting damping and other torques. Let \(\theta\) be the angle between its magnetic moment and the field.

The magnetic torque is \[ \boldsymbol\tau=\boldsymbol\mu\times\mathbf B. \] The equilibrium with \(\boldsymbol\mu\) parallel to \(\mathbf B\) is stable. Taking \(\theta\) as displacement from this equilibrium, the rotational equation of motion is \[ I\ddot\theta=-\mu B\sin\theta. \]

For small oscillations, \(|\theta|\ll1\) rad and \(\sin\theta\simeq\theta\), so \[ \ddot\theta+\frac{\mu B}{I}\theta=0. \] This is simple harmonic motion with angular frequency \[ \omega_0=\sqrt{\frac{\mu B}{I}}, \] and therefore \[ \boxed{T=2\pi\sqrt{\frac{I}{\mu B}}}. \] With SI units for \(I\), \(\mu\), and \(B\), the period is in seconds.

5. Angular range giving a period difference greater than 10%

The two field predictions at P are \[ B_1=\frac{\mu_0i}{2\pi d}\tan\left(\frac{\alpha}{2}\right), \qquad B_2=\frac{\mu_0i\alpha}{\pi^2d}, \] where the second expression is the historical assumption quoted in the problem.

Since \(T\propto B^{-1/2}\), \[ \frac{T_1}{T_2} =\sqrt{\frac{B_2}{B_1}} =\sqrt{\frac{2\alpha}{\pi\tan(\alpha/2)}}. \] The experimental requirement becomes \[ \frac{2\alpha}{\pi\tan(\alpha/2)}>1.10^2=1.21. \]

The period ratio decreases as \(\alpha\) increases. To verify this, put \(x=\alpha/2\); its square is proportional to \(x\cot x\), and \[ \frac{d}{dx}(x\cot x) =\frac{\sin x\cos x-x}{\sin^2x}<0, \] because \(\sin(2x)<2x\) for the positive angles considered.

Its limiting values are \[ \lim_{\alpha\to0^+}\frac{T_1}{T_2} =\frac{2}{\sqrt\pi}\simeq1.1284, \qquad \left.\frac{T_1}{T_2}\right|_{\alpha=\pi/2}=1. \] Thus a single threshold separates the distinguishable and indistinguishable ranges.

Solving \[ \frac{2\alpha_c}{\pi\tan(\alpha_c/2)}=1.21 \] numerically gives \[ \alpha_c\simeq0.7682\ \mathrm{rad} \simeq44.01^\circ. \] Therefore the required half-span is approximately \[ \boxed{0<\alpha<0.77\ \mathrm{rad}\quad \text{(approximately below }44^\circ\text{)}.} \]

The exact threshold is included if “at least 10%” is interpreted inclusively; it is excluded by the explicitly printed strict inequality. The degenerate value \(\alpha=0\) is excluded because both fields vanish and there is no finite magnetic oscillation period.

The solution was checked against the official solution.