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Hall Effect

The Hall Effect: Drude Model in a Magnetic Field

When a current flows through a conductor placed in a magnetic field, the charge carriers experience a sideways push due to the Lorentz force. This builds up a charge imbalance on the edges of the material, creating a transverse voltage known as the Hall Voltage. By extending the Drude model to include magnetic forces, we can derive the Hall Coefficient and determine the sign of the charge carriers.

1. Geometry and Forces

Let us define our coordinate system and the forces acting on a single electron.

Setup

  • Drift Velocity: We assume electrons drift along the negative y-axis: \[ \mathbf{v}_d = v_0(-\hat{y}) \]
  • Magnetic Field: A constant field is applied along the z-axis: \[ \mathbf{B} = B_0 \hat{z} \]

The Lorentz Force

The magnetic force on an electron is given by \(\mathbf{F}_L = -e(\mathbf{v} \times \mathbf{B})\).
First, we compute the cross product: \[ \mathbf{v} \times \mathbf{B} = (-\hat{y}) \times (\hat{z}) = -\hat{x} \]

Substituting this back into the force equation: \[ \mathbf{F}_L = -e(-\hat{x}) v_0 B_0 \]

\[ \mathbf{F}_L = e v_0 B_0 \hat{x} \]

Physical Implication: This force pushes electrons toward the \(+x\) direction, causing them to accumulate on one side of the material.

2. The Drude Equation with Magnetic Field

In the standard Drude model, friction balances the electric force. Now, we must add the magnetic term. In the steady state, the net change in momentum is zero (\(d\mathbf{p}/dt = 0\)).

The equation of motion becomes:

\[ 0 = -\frac{\mathbf{p}}{\tau} - e\mathbf{E} - e(\mathbf{v} \times \mathbf{B}) \]

3. Expressing in Terms of Current Density

We want a relationship between Electric Field (\(\mathbf{E}\)) and Current Density (\(\mathbf{j}\)), not momentum.

Step 1: Definitions
Current density is defined as \(\mathbf{j} = -ne\mathbf{v}\).
Therefore, velocity is \(\mathbf{v} = -\frac{\mathbf{j}}{ne}\).
Momentum is \(\mathbf{p} = m\mathbf{v} = -\frac{m}{ne}\mathbf{j}\).

Step 2: Substitution
Substitute these into the Drude equation: \[ 0 = -\frac{1}{\tau}\left( -\frac{m}{ne}\mathbf{j} \right) - e\mathbf{E} - e\left( -\frac{\mathbf{j}}{ne} \times \mathbf{B} \right) \]

Simplifying terms: \[ 0 = \frac{m}{ne\tau}\mathbf{j} - e\mathbf{E} + \frac{1}{n}(\mathbf{j} \times \mathbf{B}) \]

Step 3: Solve for E
Rearranging to isolate the electric field: \[ e\mathbf{E} = \frac{m}{ne\tau}\mathbf{j} + \frac{1}{n}(\mathbf{j} \times \mathbf{B}) \] Dividing by \(e\):

\[ \mathbf{E} = \frac{m}{ne^2\tau}\mathbf{j} + \frac{1}{ne}(\mathbf{j} \times \mathbf{B}) \]

4. Physical Meaning of the Equation

This boxed equation tells us that the total electric field has two distinct components:

  • First Term (Longitudinal): \(\frac{m}{ne^2\tau}\mathbf{j}\). This is the standard Ohmic term (resistance) that drives the current forward.
  • Second Term (Transverse): \(\frac{1}{ne}(\mathbf{j} \times \mathbf{B})\). This is the Hall term. It represents an electric field perpendicular to the current flow.

5. Component Form & Hall Coefficient

If we align our current along the y-axis (\(\mathbf{j} = j_y \hat{y}\)) and magnetic field along z (\(\mathbf{B} = B_0 \hat{z}\)), then \(\mathbf{j} \times \mathbf{B} = j_y B_0 \hat{x}\).

The electric field components are:

\[ \begin{aligned} E_y &= \frac{m}{ne^2\tau} j_y \quad \text{(Longitudinal / Ohmic)} \\ E_x &= \frac{1}{ne} j_y B_0 \quad \text{(Transverse / Hall)} \end{aligned} \]

The Hall Coefficient (\(R_H\))

The Hall coefficient is defined as the ratio of the transverse Hall field (\(E_H\) or \(E_x\)) to the product of current density and magnetic field.

\[ R_H = \frac{E_x}{j_y B_0} \]

Substituting our result for \(E_x\): \[ R_H = \frac{\frac{1}{ne} j_y B_0}{j_y B_0} \]

\[ R_H = -\frac{1}{ne} \]
Significance: The negative sign confirms the charge carriers are electrons. \(R_H\) depends only on carrier density \(n\), not on relaxation time \(\tau\). This makes the Hall effect a powerful tool for measuring carrier concentration.

6. Force Balance: The Physical Mechanism

Why does a transverse field \(E_H\) exist? It is a result of the system self-organizing to reach equilibrium.

  1. Start: A current \(j\) is driven externally.
  2. Deflection: The magnetic field exerts a Lorentz force \(-e(\mathbf{v}_d \times \mathbf{B})\) sideways on the electrons.
  3. Accumulation: Electrons pile up on one edge of the wire, leaving positive ions on the other.
  4. Field Creation: This charge separation creates an internal electric field \(\mathbf{E}_H\).
  5. Equilibrium: The field grows until the electric force exactly cancels the magnetic force.

The equilibrium condition is: \[ \text{Net Transverse Force} = 0 \] \[ -e\mathbf{E}_H - e(\mathbf{v}_d \times \mathbf{B}) = 0 \]

This leads to the final vector relationship for the Hall field: \[ \mathbf{E}_H = -(\mathbf{v}_d \times \mathbf{B}) \]

Summary Diagrams

Force Balance

Visual Description: An electron in the center of a wire.
• Magnetic Force (Red Arrow): Pushes right (\(+x\)).
• Hall Electric Force (Blue Arrow): Pushes left (\(-x\)).
In steady state, Red Arrow = Blue Arrow.

Charge Accumulation

Visual Description: Cross-section of the conductor.
• Left edge: Positive charges (+).
• Right edge: Negative charges (-).
• Electric Field vector pointing from Left to Right.