recall and use F = BQv sin θ

Force on a Current‑Carrying Conductor

Learning Objective

Recall and apply the magnetic force equation

$$F = B\,Q\,v\sin\theta$$

and its equivalent forms for a straight conductor carrying a steady current.

Key Concepts

• Magnetic field (B): Vector quantity measured in tesla (T).
• Charge (Q): Total charge moving through the conductor.
• Velocity (v): Speed of the charge carriers relative to the magnetic field.
• Angle (θ): Angle between the direction of motion of the charge and the magnetic field lines.
• Current (I): Rate of charge flow, $I = \dfrac{Q}{t}$.
• Length of conductor in the field (L): The portion of the wire that experiences the magnetic field.

Derivation for a Straight Conductor

For a conductor of length $L$ carrying a current $I$, the total charge passing a point in time $t$ is $Q = I t$. The charge carriers travel with drift speed $v$ so that $L = v t$. Substituting $Q$ and $v$ into the basic force law gives:

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

This is the form most often used in A‑Level problems.

Variables Summary

Symbol	Quantity	SI Unit	Typical Symbol in Exam
$F$	Magnetic force	newton (N)	F
$B$	Magnetic flux density	tesla (T)	B
$I$	Current	ampere (A)	I
$L$	Length of conductor in field	metre (m)	L
$\theta$	Angle between $I$ (or $v$) and $B$	radians (or degrees)	θ
$Q$	Charge	coulomb (C)	Q
$v$	Drift speed of charge carriers	metre per second (m s⁻¹)	v

Worked Example

Problem: A straight horizontal wire of length $0.30\;\text{m}$ carries a current of $5.0\;\text{A}$ and is placed in a uniform magnetic field of $0.80\;\text{T}$ directed into the page. The wire is oriented at $30^\circ$ to the field direction. Calculate the magnitude and direction of the magnetic force on the wire.

1. Identify the relevant formula: $F = B I L \sin\theta$.
2. Substitute the given values:

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

3. Calculate $\sin30^\circ = 0.5$ and evaluate:

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

4. Determine direction using the right‑hand rule:
   • Thumb points in the direction of conventional current (to the right).
   • Fingers point into the page (direction of $B$).
   • Palm faces upward, indicating the force is upward.
5. State the answer: $F = 0.60\;\text{N}$ upward.

Common Pitfalls

• Forgetting the $\sin\theta$ factor when the wire is not perpendicular to $B$.
• Mixing up the direction of $B$ and the direction of the force; always apply the right‑hand rule.
• Using $v$ instead of $I$ without converting correctly; remember $I = Q/t$ and $L = v t$.
• Neglecting unit consistency, especially when $B$ is given in millitesla (mT) or $L$ in centimetres.

Suggested Diagram

Practice Questions

1. A 0.15 m long segment of a wire carries a current of $2.0\;\text{A}$ in a magnetic field of $0.50\;\text{T}$ that is perpendicular to the wire. Find the magnitude of the force.
2. A rectangular loop of wire (sides $0.10\;\text{m}$ and $0.20\;\text{m}$) carries a current of $3.0\;\text{A}$ and lies in a uniform magnetic field of $0.40\;\text{T}$ directed into the page. The plane of the loop is parallel to the field. Determine the net force on the loop and explain why.
3. In a motor, a conductor of length $0.05\;\text{m}$ carries a current of $10\;\text{A}$ and rotates in a magnetic field of $0.30\;\text{T}$. At an instant when the conductor makes an angle of $45^\circ$ with the field, calculate the instantaneous torque about the axis of rotation (assume the conductor is a radius of a circular loop).

Summary

The magnetic force on a current‑carrying conductor is given by $F = B I L \sin\theta$, which is directly derived from $F = B Q v \sin\theta$. Mastery of this relationship, together with the right‑hand rule for direction, enables accurate analysis of forces in electromagnetic devices such as motors and galvanometers.