recall and use F = BQv sin θ

Force on a Current‑Carrying Conductor – Cambridge AS & A Level Physics (9702)

Learning Objective

Recall and apply the magnetic‑force relationships for a moving charge and for a straight conductor carrying a steady current, use the appropriate right‑hand rules to determine force direction, and relate these ideas to magnetic torque, motors and other common applications.

20.1 – Concept of a Magnetic Field

• Magnetic flux density, B – a vector quantity that describes the strength and direction of a magnetic field at a point.

  • SI unit: tesla (T) = N A⁻¹ m⁻¹.
  • Field‑line visualisation:

    • Lines point in the direction of B.
    • Line density ∝ magnitude of B.
    • In a uniform field the lines are parallel and equally spaced (e.g. into the page shown by crosses ✗).

• Mathematical definition (Cambridge Syllabus 20.1):

  \[

  \mathbf{F}=q\,\mathbf{v}\times\mathbf{B}

  \]

  where q is the charge, v its velocity and “×” denotes the vector cross‑product.

20.3 – Force on a Moving Charge

• Magnitude:

  \[

  F = B\,q\,v\sin\theta

  \]

  where θ is the angle between the velocity vector v and the field B.

• Direction – right‑hand rule for a positive charge:

  1. Point fingers in the direction of v.
  2. Rotate wrist so that fingers can be swept toward B (the smallest angle).
  3. Thumb points in the direction of the force F.

  (For electrons the force is opposite to the thumb direction.)

20.2 – Force on a Straight Conductor

Key Equations

• Scalar form:

  \[

  F = B\,I\,L\sin\theta

  \]

• Vector (cross‑product) form:

  \[

  \boxed{\mathbf{F}=I\,\mathbf{L}\times\mathbf{B}}

  \]

  where 𝐿 is a vector of magnitude L in the direction of conventional current.

Derivation from the Charge‑Based Formula

1. Charge passing a point in time t is \(Q = I t\).
2. Drift speed v gives a travelled length \(L = v t\).
3. Substituting into \(F = B Q v \sin\theta\):

   \[

   F = B\,(I t)\,\left(\frac{L}{t}\right)\sin\theta = B I L \sin\theta.

   \]

4. Writing the relationship with vectors gives \(\mathbf{F}=I\mathbf{L}\times\mathbf{B}\).

Sign Convention

• Use conventional current (flow of positive charge) when applying the formula and the right‑hand rule.
• If the problem explicitly refers to electron flow, reverse the direction of the resulting force.

Magnetic Torque

When the conductor is part of a rotating coil the force produces a torque:

\[

\boxed{\tau = I L B \sin\theta}

\]

For a rectangular coil of N turns, area A and field B (common in DC motors):

\[

\boxed{\tau = N I A B \sin\theta}

\]

Direction – Right‑Hand Rule for a Current‑Carrying Wire

1. Stretch the right hand so that the thumb points in the direction of conventional current (or the vector 𝐿).
2. Point the fingers in the direction of the magnetic field B.
3. The palm (or the direction your palm pushes) gives the direction of the force F on the wire.

Variables Summary

20.4 – Magnetic Fields Produced by Currents

• Biot–Savart Law (point source):

  \[

  d\mathbf{B}= \frac{\mu_0}{4\pi}\,\frac{I\,d\mathbf{l}\times\hat{\mathbf{r}}}{r^{2}}

  \]

  where μ₀ = 4π × 10⁻⁷ T m A⁻¹.

• Long straight conductor (Cambridge Syllabus 20.4):

  \[

  B = \frac{\mu_0 I}{2\pi r}

  \]

  – field circles the wire (right‑hand grip rule).

• Solenoid (many turns):

  \[

  B = \mu_0 n I \qquad (n = \text{turns per unit length})

  \]

  – field is uniform and parallel to the axis inside the solenoid.

20.5 – Electromagnetic Induction (Faraday’s & Lenz’s Laws)

• Faraday’s law (Cambridge Syllabus 20.5):

  \[

  \mathcal{E} = -\frac{d\Phi}{dt}

  \]

  where \(\Phi = B A \cos\theta\) is the magnetic flux.

• Lenz’s law – the induced emf produces a current whose magnetic field opposes the change in flux.
• Relevance to the force law:

  • In a DC motor the rotating coil experiences a torque \(\tau = N I A B \sin\theta\) while the changing flux induces a back‑emf \(\mathcal{E}= -N\frac{d\Phi}{dt}\).
  • In a galvanometer the deflection is proportional to the magnetic force on the moving coil, which is directly linked to the induced current.

Worked Example – Force on a Wire

Problem: A horizontal wire 0.30 m long carries a current of 5.0 A to the right. It is placed in a uniform magnetic field of 0.80 T directed into the page. The wire makes an angle of \(30^{\circ}\) with the field direction. Calculate the magnitude and direction of the magnetic force.

1. Use the scalar form \(F = B I L \sin\theta\).
2. Insert the data:

   \[

   F = (0.80\;\text{T})(5.0\;\text{A})(0.30\;\text{m})\sin30^{\circ}

   \]

   \(\sin30^{\circ}=0.5\).

3. Calculate:

   \[

   F = 0.80 \times 5.0 \times 0.30 \times 0.5 = 0.60\;\text{N}

   \]

4. Direction (right‑hand rule):

   • Thumb → right (current).
   • Fingers → into the page (field).
   • Palm faces upward → force is upward.

5. Answer: \(F = 0.60\;\text{N}\) upward.

Worked Example – Torque on a Motor Coil

Problem: In a simple DC motor a single rectangular coil of 2 turns has side length 0.05 m (radius of rotation) and carries a current of 10 A. The magnetic field is uniform, \(B = 0.30\;\text{T}\). Find the instantaneous torque when the plane of the coil makes an angle of \(45^{\circ}\) with the field.

1. Torque for a coil: \(\tau = N I A B \sin\theta\) where \(A = L \times r\).

   Here \(L = 0.05\;\text{m}\) (length of each side) and the effective radius is also 0.05 m, so \(A = 0.05 \times 0.05 = 2.5\times10^{-3}\;\text{m}^2\).

2. Insert values:

   \[

   \tau = (2)(10\;\text{A})(2.5\times10^{-3}\;\text{m}^2)(0.30\;\text{T})\sin45^{\circ}

   \]

   \(\sin45^{\circ}=0.707\).

3. Calculate:

   \[

   \tau = 2 \times 10 \times 2.5\times10^{-3} \times 0.30 \times 0.707 \approx 1.06\times10^{-2}\;\text{N·m}

   \]

4. Answer: \(\tau \approx 1.1\times10^{-2}\;\text{N·m}\) (counter‑clockwise if current is as shown).

Practical Applications (AO3)

• DC motors – the force on each conductor produces a continuous torque; the back‑emf from Faraday’s law limits the current and defines the speed‑torque characteristic.
• Galvanometer – a moving coil in a magnetic field experiences a force proportional to the current; the deflection is a direct visualisation of \(F = B I L\).
• Velocity selector – uses perpendicular electric and magnetic fields; only particles with velocity \(v = \frac{E}{B}\) experience zero net force, a classic Cambridge exam scenario.
• Mass spectrometer – the magnetic force bends a charged particle’s path; the radius \(r = \frac{mv}{qB}\) follows from \(F = qvB\).

Practice Questions

1. A 0.15 m long straight wire carries 2.0 A perpendicular to a uniform magnetic field of 0.50 T. Find the magnitude of the force.
2. A rectangular loop (sides 0.10 m and 0.20 m) carries 3.0 A and lies in a uniform field of 0.40 T directed into the page. The plane of the loop is parallel to the field. Determine the net force on the loop and explain why it is zero.
3. A velocity‑selector is set up with a magnetic field of 0.20 T directed into the page and an electric field of 3.0 × 10⁴ V m⁻¹ directed upward. What speed must a positively charged particle have to pass through undeflected?
4. In a DC motor the armature consists of a single rectangular coil of 4 turns, area 2.0 × 10⁻³ m², carrying 5.0 A in a 0.25 T field. Calculate the maximum torque produced by the motor.
5. A long straight wire carries 8.0 A. Find the magnetic field 5.0 cm from the wire and the direction of the field using the right‑hand grip rule.

Summary

The magnetic force on a moving charge is \(\mathbf{F}=q\,\mathbf{v}\times\mathbf{B}\). For a straight conductor carrying a steady current this becomes

\[

\boxed{F = B I L \sin\theta \quad\text{or}\quad \mathbf{F}=I\,\mathbf{L}\times\mathbf{B}}

\]

When the conductor forms part of a rotating coil the same force gives a torque \(\tau = N I A B \sin\theta\). Understanding these relationships, the associated right‑hand rules, and the related concepts of magnetic fields produced by currents (Biot–Savart, Ampère) and electromagnetic induction (Faraday & Lenz) equips you to tackle all Cambridge AS & A Level exam items on topic 20 – from motor torque to particle deflection and velocity selection.