IPhO 1985 · Theory problem 2

Hall effect in an InSb bar

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In a long bar having the shape of a rectangular parallelepiped with sides \(a\), \(b\), and \(c\) (\(a\gg b\gg c\)), made from the semiconductor InSb flows a current \(I\) parallel to the edge \(a\). The bar is in an external magnetic field \(B\) which is parallel to the edge \(c\). The magnetic field produced by the current \(I\) can be neglected. The current carriers are electrons. The average velocity of electrons in a semiconductor in the presence of an electric field only is \(v=\mu E\), where \(\mu\) is called mobility. If the magnetic field is also present, the electric field is no longer parallel to the current. This phenomenon is known as the Hall effect.

a) Determine what the magnitude and the direction of the electric field in the bar is, to yield the current described above.

b) Calculate the difference of the electric potential between the opposite points on the surfaces of the bar in the direction of the edge \(b\).

c) Find the analytic expression for the DC component of the electric potential difference in case b) if the current and the magnetic field are alternating (AC); \(I=I_0\sin\omega t\) and \(B=B_0\sin(\omega t+\delta)\).

d) Design and explain an electric circuit which would make possible, by exploiting the result c), to measure the power consumption of an electric apparatus connected with the AC network.

Data: The electron mobility in InSb is \(7.8\ \mathrm{m^2T/Vs}\), the electron concentration in InSb is \(2.5\cdot10^{22}\ \mathrm{m^{-3}}\), \(I=1.0\ \mathrm A\), \(B=0.10\ \mathrm T\), \(b=1.0\ \mathrm{cm}\), \(c=1.0\ \mathrm{mm}\), \(e_0=-1.6\cdot10^{-19}\ \mathrm{As}\).

Official English text of the problem.

Answers

Choose right-handed axes with \(+\hat{\mathbf x}\) along the conventional current and \(+\hat{\mathbf z}\) along the magnetic field. Let \(e=-e_0>0\), let \(n\) be the electron concentration, and define the Hall voltage as \[ U_H=V(y=+b/2)-V(y=-b/2). \]

  • a) The electric field is \[ \boxed{\mathbf E=\frac{I}{nebc\mu}\hat{\mathbf x} -\frac{IB}{nebc}\hat{\mathbf y} =(3.2\hat{\mathbf x}-2.5\hat{\mathbf y})\ \mathrm{V\,m^{-1}}.} \] Its magnitude is \(\boxed{4.1\ \mathrm{V\,m^{-1}}}\). It is perpendicular to \(\mathbf B\) and tilted through \[ \boxed{\theta=\tan^{-1}(\mu B)=38^\circ} \] from the current direction toward \(-\hat{\mathbf y}\), the direction of \(\mathbf j\times\mathbf B\).

  • b) \[ \boxed{U_H=\frac{IB}{nec}=2.5\times10^{-2}\ \mathrm V=25\ \mathrm{mV}.} \] The \(+y\) face is at higher potential.

  • c) \[ \boxed{U_{H,\mathrm{DC}}=\frac{I_0B_0}{2nec}\cos\delta.} \]

  • d) Use a series current coil to produce \(B(t)=k_I i_L(t)\), and drive the Hall bar through a resistive voltage-sensing branch so that \(I_H(t)=k_Uu(t)\). An averaging voltmeter across the \(b\)-faces then reads \[ \boxed{\overline{U_H}=\frac{k_Uk_I}{nec}P,\qquad P=U_{\mathrm{rms}}I_{L,\mathrm{rms}}\cos\varphi.} \] Calibrate its scale directly in watts.

Solution

We consider the uniform central region of the long bar, neglecting contact effects and the magnetic field generated by its current. The transverse faces are open-circuit, so there is no steady transverse current.

For numerical calculations, the mobility is interpreted as \(7.8\ \mathrm{m^2/(V\,s)}\), as required by \(v=\mu E\); the unit \(\mathrm{m^2T/Vs}\) given in the data is dimensionally inconsistent with this relation.

a) Magnitude and direction of the electric field

The current density and electron drift velocity are related by \[ \mathbf j=-ne\mathbf v_e. \] Since the cross-sectional area is \(bc\), \[ \mathbf j=\frac{I}{bc}\hat{\mathbf x}, \qquad \mathbf v_e=-v_d\hat{\mathbf x}, \qquad v_d=\frac{I}{nebc}. \]

The given mobility law relates speed to field magnitude. In vector form, electrons drift opposite to the electric field when no magnetic field is present.

In the usual isotropic mobility model, collisions exert a mean resisting force \(-e\mathbf v_e/\mu\). The steady force balance in both fields is therefore \[ 0=-e\left(\mathbf E+\mathbf v_e\times\mathbf B\right) -\frac{e}{\mu}\mathbf v_e. \] Consequently, \[ \mathbf E=-\frac{\mathbf v_e}{\mu}-\mathbf v_e\times\mathbf B. \]

With \(\mathbf B=B\hat{\mathbf z}\), \[ \mathbf v_e\times\mathbf B =(-v_d\hat{\mathbf x})\times(B\hat{\mathbf z}) =v_dB\hat{\mathbf y}, \] so \[ \boxed{\mathbf E=\frac{v_d}{\mu}\hat{\mathbf x} -v_dB\hat{\mathbf y}.} \]

The longitudinal component maintains the drift against collisions. The transverse component balances the magnetic force. In particular, the magnetic force initially pushes electrons toward the \(-y\) face; their accumulation produces the required field in the \(-y\) direction.

Numerically, \[ bc=(1.0\times10^{-2})(1.0\times10^{-3}) =1.0\times10^{-5}\ \mathrm{m^2}, \] \[ v_d= \frac{1.0} {(2.5\times10^{22})(1.6\times10^{-19})(1.0\times10^{-5})} =25\ \mathrm{m\,s^{-1}}. \] Thus \[ E_x=\frac{25}{7.8}=3.2\ \mathrm{V\,m^{-1}}, \qquad E_y=-25(0.10)=-2.5\ \mathrm{V\,m^{-1}}, \qquad E_z=0. \]

The magnitude is \[ E=\sqrt{E_x^2+E_y^2} =\frac{I}{nebc\mu}\sqrt{1+(\mu B)^2} =\boxed{4.1\ \mathrm{V\,m^{-1}}}. \] The angle from \(+\hat{\mathbf x}\) toward \(-\hat{\mathbf y}\) satisfies \[ \tan\theta=\frac{|E_y|}{E_x}=\mu B=0.78, \] giving \[ \boxed{\theta=38^\circ}. \]

b) Transverse potential difference

Choose opposite points on the two \(b\)-faces at the same longitudinal coordinate. Since \(\mathbf E=-\boldsymbol{\nabla}V\), \[ U_H=-\int_{-b/2}^{b/2}E_y\,dy =-E_yb=v_dBb. \] Substituting the drift speed gives \[ \boxed{U_H=\frac{IB}{nec}.} \]

Numerically, \[ U_H= \frac{(1.0)(0.10)} {(2.5\times10^{22})(1.6\times10^{-19})(1.0\times10^{-3})} =\boxed{0.025\ \mathrm V=25\ \mathrm{mV}}. \]

The \(-y\) face accumulates electrons and is at lower potential; the \(+y\) face is at higher potential.

c) DC component for alternating current and magnetic field

Assume quasistatic operation: the electron drift and Hall charge distribution follow the applied signals sufficiently rapidly, and inductive pickup in the voltage leads is negligible. The instantaneous Hall voltage is then \[ U_H(t)=\frac{I(t)B(t)}{nec} =\frac{I_0B_0}{nec}\sin\omega t\,\sin(\omega t+\delta). \]

Using \[ \sin A\sin B=\frac12\left[\cos(A-B)-\cos(A+B)\right], \] we obtain \[ U_H(t)=\frac{I_0B_0}{2nec} \left[\cos\delta-\cos(2\omega t+\delta)\right]. \]

The second term averages to zero over a cycle. Hence \[ \boxed{U_{H,\mathrm{DC}} =\overline{U_H} =\frac{I_0B_0}{2nec}\cos\delta.} \] Equivalently, \[ \overline{U_H} =\frac{I_{\mathrm{rms}}B_{\mathrm{rms}}}{nec}\cos\delta. \]

The output contains a DC component and a ripple at twice the applied frequency. Its DC component vanishes when the two inputs are in quadrature.

d) Circuit for measuring AC power

The result in part c) makes the Hall element a multiplier: its transverse voltage is proportional to the product of its longitudinal current and the magnetic field.

A suitable circuit has these connections:

  1. Connect a field coil in series with the apparatus. Place the Hall bar inside the coil, with its \(c\)-direction parallel to the field. Use a linear, preferably air-cored coil, so that \[ B(t)=k_I i_L(t), \] where \(i_L(t)\) is the apparatus current and \(k_I\) is the coil’s field constant.

  2. Connect a separate branch across the AC network, consisting of a large noninductive resistor \(R\) in series with the Hall bar’s two longitudinal end contacts. The bar’s longitudinal resistance follows from part a): \[ r_H=\frac{E_xa}{I_H}=\frac{a}{ne\mu bc}. \] Thus, with negligible branch reactance, \[ I_H(t)=\frac{u(t)}{R+r_H}=k_Uu(t). \] Choosing \(R\gg r_H\) gives \(k_U\simeq1/R\) and limits the sensing current.

  3. Connect opposite Hall contacts on the \(b\)-faces, at the same longitudinal position, to a high-input-impedance averaging voltmeter. A buffered low-pass filter followed by a DC meter is also suitable. Average over many AC cycles.

Neglect the field coil’s voltage drop and the loading of the voltage-sensing branch. Then \(u(t)\) is also the apparatus voltage, and \[ U_H(t)=\frac{I_H(t)B(t)}{nec} =\frac{k_Uk_I}{nec}\,u(t)i_L(t). \] Its average is therefore \[ \boxed{\overline{U_H}=\frac{k_Uk_I}{nec}\, \overline{u(t)i_L(t)} =\frac{k_Uk_I}{nec}P.} \]

For sinusoidal operation, write \[ u(t)=U_0\sin\omega t,\qquad i_L(t)=I_{L0}\sin(\omega t-\varphi). \] Then \[ P=\frac{U_0I_{L0}}2\cos\varphi =U_{\mathrm{rms}}I_{L,\mathrm{rms}}\cos\varphi. \] The meter consequently measures the average power consumed, including the power-factor dependence. Its watt calibration is \[ \boxed{P=\frac{nec}{k_Uk_I}\,\overline{U_H}.} \]

The solution was checked against the official solution.