When a wire carrying an electric current I lies in a magnetic field B, every moving charge inside the wire experiences a magnetic force.
The net force on the whole wire is given by the cross‑product of the current direction and the magnetic field:
\$F = I\,L\,\mathbf{\hat{L}}\times\mathbf{B}\$
Here L is the length of the wire inside the field and \$\mathbf{\hat{L}}\$ is a unit vector along the wire.
If the field is perpendicular to the wire, the magnitude simplifies to \$F = I\,L\,B\$.
Analogy: Imagine a row of tiny cars (the electrons) driving along a road (the wire). If a strong wind (the magnetic field) blows across the road, the cars are pushed sideways. The stronger the wind or the more cars driving, the greater the sideways push on the whole road.
When the wire is placed in a magnetic field, the moving charges are deflected to one side of the conductor. This creates a transverse electric field that balances the magnetic force. The resulting potential difference across the conductor is called the Hall voltage (\$V_H\$).
The Hall effect is used to measure magnetic fields, carrier density, and carrier type (electrons or holes). It is also the principle behind many magnetic sensors.
• Wire of thickness \$t\$ (distance between the two faces).
• Current \$I\$ flows along the \$x\$-axis.
• Magnetic field \$B\$ points along the \$z\$-axis, perpendicular to the wire.
For a charge \$q\$ moving with drift velocity \$v_d\$, the magnetic force is
\$\mathbf{F}m = q\,\mathbf{v}d \times \mathbf{B}\$
Since \$\mathbf{v}_d\$ is along \$x\$ and \$B\$ along \$z\$, the force points along \$y\$ (towards one side of the wire).
Charges accumulate until an electric force balances the magnetic force:
\$q\,Ey = q\,vd\,B \;\;\Rightarrow\;\; Ey = vd\,B\$
The Hall voltage is the potential difference across the thickness:
\$VH = Ey\,t = v_d\,B\,t\$
Current density is \$J = n\,q\,v_d\$, where \$n\$ is the number density of charge carriers.
The total current is \$I = J\,A = n\,q\,v_d\,A\$, with \$A\$ the cross‑sectional area (\$A = w\,t\$, width \$w\$).
Solving for \$v_d\$:
\$v_d = \frac{I}{n\,q\,A} = \frac{I}{n\,q\,w\,t}\$
\$V_H = \left(\frac{I}{n\,q\,w\,t}\right) B\,t = \frac{B\,I}{n\,q\,w}\$
If we consider the Hall voltage across the full thickness (i.e., \$w = 1\$ m for a unit width), we obtain the standard expression:
\$\boxed{V_H = \frac{B\,I}{n\,t\,q}}\$
Here \$t\$ is the thickness of the conductor.
Key take‑away: The Hall voltage is directly proportional to the magnetic field and the current, and inversely proportional to the product of carrier density, thickness, and charge magnitude. This relationship lets us measure any one of these quantities if the others are known.
| Symbol | Description | Units |
|---|---|---|
| \$V_H\$ | Hall voltage (potential difference) | Volts (V) |
| \$B\$ | Magnetic flux density | Tesla (T) |
| \$I\$ | Electric current | Amperes (A) |
| \$n\$ | Carrier density (number of charge carriers per cubic metre) | m⁻³ |
| \$t\$ | Thickness of the conductor | metres (m) |
| \$q\$ | Charge of a single carrier (e.g., \$-e\$ for electrons) | Coulombs (C) |
Practical tip: In a lab, you can measure \$V_H\$ with a multimeter while applying a known current and magnetic field. Rearranging the formula lets you calculate the carrier density \$n\$—a key property of the material.
Answers will reinforce the concepts and give you confidence in applying the Hall effect in real‑world scenarios. Happy learning! 🚀