Force on a Current‑Carrying Conductor – A‑Level Physics 9702
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.
Identify the relevant formula: \$F = B I L \sin\theta\$.
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.
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
Suggested diagram: Horizontal wire carrying current to the right, magnetic field vectors into the page, and the resulting upward force vector shown using the right‑hand rule.
Practice Questions
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.
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.
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.