Force on a Current‑Carrying Conductor (Cambridge AS & A Level 9702)
Learning Objective (20.1 – 20.4)
Students will be able to:
- Define a magnetic field B and represent it graphically.
- Calculate the magnetic field produced by common current configurations.
- Apply the Lorentz‑force law to a single moving charge and to a conductor carrying a current.
- Predict the magnitude and direction of the force using F = I L B \sin\theta and Fleming’s left‑hand rule.
- Explain the Hall‑effect probe and evaluate the Hall voltage.
- Use the torque expression \(\tau = N I A B \sin\phi\) for a current‑carrying coil.
20.1 Magnetic Field – Definition & Representation
20.1.1 Definition
The magnetic field \(\mathbf{B}\) at a point is defined as the force per unit positive charge moving with velocity \(\mathbf{v}\) at that point:
\[
\mathbf{B}= \frac{\mathbf{F}}{q\,v\sin\theta}\qquad\text{or}\qquad
\mathbf{F}=q\,\mathbf{v}\times\mathbf{B}.
\]
- Units: 1 T = 1 N A\(^{-1}\) m\(^{-1}\) (tesla).
- Direction: given by the tangent to magnetic‑field lines; the sense is the direction a north‑pole of a compass needle would point.
20.1.2 Field‑line diagrams
- Lines emerge from the north pole and enter the south pole of a magnet.
- Density of lines ∝ field strength.
20.1.3 Right‑hand rule for field direction
For a straight conductor carrying current \(I\), point the thumb of the right hand in the direction of the conventional current; the curled fingers give the direction of the magnetic field lines encircling the wire.
20.1.4 Field of a straight conductor (link to 20.4)
Using Ampère’s law (or the Biot–Savart law) the magnitude of the field at a distance \(r\) from a long, straight wire is
\[
B=\frac{\mu_{0} I}{2\pi r},
\qquad\mu_{0}=4\pi\times10^{-7}\ \text{T·m·A}^{-1}.
\]
20.2 Force on a Current‑Carrying Conductor
20.2.1 Force on a single moving charge
The Lorentz force on a charge \(q\) moving with velocity \(\mathbf{v}\) in a magnetic field \(\mathbf{B}\) is
\[
\mathbf{F}=q\,\mathbf{v}\times\mathbf{B},
\qquad
F = qvB\sin\theta .
\]
20.2.2 Derivation of the force on a straight conductor
- Current definition (20.2.2.1):
\[
I = n q A v_{d},
\]
where \(n\) = charge‑carrier number density (m\(^{-3}\)), \(q\) = charge of each carrier (C), \(A\) = cross‑sectional area (m\(^2\)), \(v_{d}\) = drift speed (m s\(^{-1}\)).
- Number of carriers in a segment of length \(L\) (20.2.2.2):
\[
N = n A L .
\]
- Force on one carrier (20.2.2.3):
\[
\mathbf{f}=q\,\mathbf{v}_{d}\times\mathbf{B}.
\]
- Total force (20.2.2.4):
\[
\mathbf{F}=N\mathbf{f}
=(nAL)q\,\mathbf{v}_{d}\times\mathbf{B}
=(nqAv_{d})\,L\,\hat{\mathbf{I}}\times\mathbf{B}
=I\,L\,\hat{\mathbf{I}}\times\mathbf{B}.
\]
\(\hat{\mathbf{I}}\) is a unit vector in the direction of the conventional current.
20.2.3 Resulting formula (20.2.3)
Magnitude:
\[
F = I L B \sin\theta .
\]
Special case required by the syllabus (wire ⟂ field): \(\theta = 90^{\circ}\) ⇒ \(\sin\theta = 1\), so
\[
F_{\perp}= I L B .
\]
20.2.4 Direction – Fleming’s Left‑Hand Rule (20.2.4)
- First finger (index) → direction of \(\mathbf{B}\) (north → south).
- Second finger (middle) → direction of conventional current \(\mathbf{I}\) (positive → negative).
- Thumb → direction of the force \(\mathbf{F}\) on the conductor.
20.2.5 Hall‑effect probe (20.2.5)
When a current‑carrying conductor is placed in a magnetic field, the magnetic force on the charge carriers builds up a transverse electric field \(E_{H}\). At equilibrium
\[
qE_{H}=qv_{d}B \;\;\Longrightarrow\;\; E_{H}=v_{d}B .
\]
The measurable Hall voltage across a plate of width \(w\) is
\[
V_{H}=E_{H}w = v_{d} B w .
\]
Since \(I = n q A v_{d}\), the Hall voltage can also be written as
\[
V_{H}= \frac{I B w}{n q A}.
\]
20.3 Force on a Moving Charge & Hall Effect (Worked Example)
Example 3 – Hall‑voltage calculation
Problem: A rectangular copper strip (width \(w = 2.0\ \text{mm}\), thickness \(t = 0.5\ \text{mm}\)) carries a current \(I = 3.0\ \text{A}\). It is placed in a uniform magnetic field \(B = 0.40\ \text{T}\) directed into the page. The free‑electron density in copper is \(n = 8.5\times10^{28}\ \text{m}^{-3}\). Find the Hall voltage across the width of the strip.
- Cross‑sectional area \(A = w t = (2.0\times10^{-3})(5.0\times10^{-4}) = 1.0\times10^{-6}\ \text{m}^2\).
- Using \(V_{H}= \dfrac{I B w}{n q A}\) with \(q = 1.60\times10^{-19}\ \text{C}\):
\[
V_{H}= \frac{(3.0)(0.40)(2.0\times10^{-3})}
{(8.5\times10^{28})(1.60\times10^{-19})(1.0\times10^{-6})}
\approx 1.8\times10^{-6}\ \text{V}=1.8\ \mu\text{V}.
\]
- The Hall voltage is 1.8 µV, with the positive side on the side toward which the magnetic force on the electrons pushes them.
20.4 Magnetic Fields Due to Currents
20.4.1 Field of a long straight conductor
\[
B = \frac{\mu_{0} I}{2\pi r},
\qquad\text{direction given by the right‑hand rule (thumb = current).}
\]
20.4.2 Field inside a solenoid
For a tightly wound solenoid of \(n\) turns per unit length carrying current \(I\):
\[
B = \mu_{0} n I,
\]
field lines are parallel to the axis; the direction follows the right‑hand rule (curl fingers in the sense of the winding, thumb points along \(\mathbf{B}\)).
20.4.3 Field of a toroid (optional extension)
\[
B = \frac{\mu_{0} N I}{2\pi r},
\]
where \(N\) is the total number of turns and \(r\) the mean radius.
20.4.4 Summary of field‑generation formulas
| Configuration | Formula for \(|\mathbf{B}|\) | Direction rule |
|---|
| Long straight wire | \(\displaystyle B=\frac{\mu_{0}I}{2\pi r}\) | Right‑hand thumb = current |
| Solenoid (length ≫ diameter) | \(\displaystyle B=\mu_{0} n I\) | Right‑hand curl = current loops |
| Toroid (circular coil) | \(\displaystyle B=\frac{\mu_{0} N I}{2\pi r}\) | Right‑hand thumb = axis of toroid |
Worked Example 1 – Force on a Straight Wire
Problem: A horizontal wire 0.30 m long carries a current of 5.0 A to the east. It is placed in a uniform magnetic field of magnitude 0.80 T directed vertically upward. Find the magnitude and direction of the force.
- Current \(\mathbf{I}\): east (+x).
Magnetic field \(\mathbf{B}\): upward (+z).
Angle \(\theta = 90^{\circ}\) ⇒ \(\sin\theta = 1\).
- Magnitude: \(F = I L B = (5.0)(0.30)(0.80) = 1.2\ \text{N}\).
- Using Fleming’s left‑hand rule: thumb points toward the north (+y).
Force = 1.2 N northward.
Worked Example 2 – Torque on a Rectangular Motor Coil
Problem: A rectangular coil of 4 turns has dimensions \(l = 0.10\ \text{m}\) and \(w = 0.05\ \text{m}\). It carries a current of 2.0 A and is placed in a uniform magnetic field of 0.60 T. The field makes an angle \(\phi = 30^{\circ}\) with the normal to the coil. Find the torque.
- Area of one turn: \(A = l w = 5.0\times10^{-3}\ \text{m}^2\).
- Torque on a coil: \(\tau = N I A B \sin\phi\).
\[
\tau = (4)(2.0)(5.0\times10^{-3})(0.60)\sin30^{\circ}
= 1.2\times10^{-2}\ \text{N·m}.
\]
- The coil tends to rotate so that its plane becomes parallel to \(\mathbf{B}\) (principle of an electric motor).
Applications (Illustrative Only)
- Electric motors: Rotating armature coils experience the torque \(\tau = N I A B \sin\phi\).
- Railguns: Parallel rails carry very large currents; the magnetic force \(\mathbf{F}=I L B\) accelerates a projectile.
- Galvanometers & Hall probes: Deflection of a moving coil or Hall voltage provides a direct measurement of current.
Summary Table
| Quantity | Symbol | Expression | Units |
| Current | I | I = n q A v_{d} | A (ampere) |
| Magnetic field (definition) | \(\mathbf{B}\) | \(\mathbf{F}=q\,\mathbf{v}\times\mathbf{B}\) | T (tesla) |
| Field of a straight wire | \(B\) | \(B=\dfrac{\mu_{0} I}{2\pi r}\) | T |
| Field inside a solenoid | \(B\) | \(B=\mu_{0} n I\) | T |
| Force on a straight conductor | \(\mathbf{F}\) | \(\mathbf{F}= I L\,\hat{\mathbf{I}}\times\mathbf{B}\) | N |
| Magnitude of force | F | \(F = I L B \sin\theta\) | N |
| Torque on a coil (N turns) | \(\tau\) | \(\tau = N I A B \sin\phi\) | N·m |
| Hall voltage | V_{H} | \(V_{H}= \dfrac{I B w}{n q A}\) | V |
Key Take‑aways
- The magnetic field is defined by the force on a moving charge; its direction is given by field lines.
- Current is the collective drift of many charges; the total magnetic force on a conductor is the vector sum of the individual Lorentz forces.
- For a straight conductor: \(F = I L B \sin\theta\); the force is maximal when the wire is perpendicular to the field.
- Direction of the force is given by the cross product \(\hat{\mathbf{I}}\times\mathbf{B}\) or conveniently by Fleming’s left‑hand rule.
- The Hall effect provides a practical way to measure magnetic field strength or current using the transverse voltage that develops across a current‑carrying plate.
- Torque on a current‑carrying coil, \(\tau = N I A B \sin\phi\), explains the operation of electric motors and many other electromagnetic devices.