CIE AS/A Level Physics 20.2 Force on a current-carrying conductor Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 20.2 Force on a current-carrying conductor Study Notes – New Syllabus
CIE AS/A Level Physics 20.2 Force on a current-carrying conductor Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics latest syllabus with Candidates should be able to:
- understand that a force might act on a current-carrying conductor placed in a magnetic field
- recall and use the equation \( F = B I L \sin \theta \), with directions as interpreted by Fleming’s left-hand rule
- define magnetic flux density as the force acting per unit current per unit length on a wire placed at right-angles to the magnetic field
Force on a Current-Carrying Conductor in a Magnetic Field
When a conductor carrying an electric current is placed in a magnetic field, it can experience a magnetic force. This is a fundamental phenomenon of electromagnetism and forms the basis of electric motors, loudspeakers, galvanometers, and many other devices.
Why the Force Occurs:
- An electric current is a flow of moving charges (electrons).
- Moving charges produce their own magnetic field around the conductor.
- This magnetic field interacts with the external magnetic field in the region.
- The interaction produces a force on the conductor — a magnetic force.
This force is sometimes called the motor effect.
Key Idea:
A current-carrying conductor placed in a magnetic field will experience a force if the current and field are not parallel.
Conditions for a Force:
- There must be a magnetic field.
- The conductor must carry a current.
- The conductor must not be aligned parallel to the magnetic field (θ ≠ 0°, 180°). Maximum force occurs when θ = 90° (current and field perpendicular).
Direction of Force:
Given by the left-hand rule (Fleming’s left-hand rule): Thumb → Force First finger → Field Second finger → Current (All at right angles)
Real-World Examples:
- Electric motors (coil experiences a force causing rotation)
- Loudspeakers (current in coil interacts with magnetic field to move the cone)
- Rail guns (current between rails produces a large magnetic force)
Example
A wire carrying current is placed perpendicular to a magnetic field. The wire jumps sideways. Explain why the wire moves.
▶️ Answer / Explanation
The current in the wire creates a magnetic field that interacts with the external magnetic field. The interaction produces a force on the wire, causing it to move sideways.
Example
A current-carrying conductor is placed parallel to a magnetic field. Will the conductor experience a force? Explain your reasoning.
▶️ Answer / Explanation
No force acts because the conductor’s current is parallel to the magnetic field. The force depends on the component of current perpendicular to the field, and in this case that component is zero.
Example
A straight wire carrying current is placed at 30° to a magnetic field. Explain qualitatively why the force is smaller than the force when the wire is perpendicular to the field.
▶️ Answer / Explanation
The force on a current-carrying conductor depends on the perpendicular component of the current relative to the magnetic field.
The perpendicular component is:
\( \mathrm{I_{\perp} = I \sin(30^\circ)} \)
Since \( \mathrm{\sin(30^\circ) = 0.5} \), only half the current contributes to the force compared to the perpendicular (90°) case.
Therefore, the force is smaller because the conductor is not fully perpendicular to the magnetic field.
Using the Equation \( \mathrm{F = B I L \sin\theta} \) and Fleming’s Left-Hand Rule
The force on a current-carrying conductor in a magnetic field is given by the equation:
\( \mathrm{F = B I L \sin\theta} \)
- \( \mathrm{F} \) = magnetic force (N)
- \( \mathrm{B} \) = magnetic flux density (T)
- \( \mathrm{I} \) = current (A)
- \( \mathrm{L} \) = length of conductor in the field (m)
- \( \mathrm{\theta} \) = angle between the current direction and the magnetic field
Meaning of the Equation:
- If the conductor is perpendicular to the field (θ = 90°): \( \mathrm{\sin 90^\circ = 1} \) → maximum force
- If the conductor is parallel to the field (θ = 0°): \( \mathrm{\sin 0^\circ = 0} \) → no force
- The force depends on the perpendicular component of current in the field.
Direction of Force → Fleming’s Left-Hand Rule
Thumb → Force
First finger → Magnetic field (N → S)
Second finger → Current (positive → negative)
The three directions must be at right angles to each other.
Example
A conductor of length \( \mathrm{0.50\ m} \) carries a current of \( \mathrm{4\ A} \) in a field of \( \mathrm{0.20\ T} \). The conductor is perpendicular to the magnetic field. Calculate the force.
▶️ Answer / Explanation
Since θ = 90°, \( \mathrm{\sin\theta = 1} \)
\( \mathrm{F = B I L = 0.20 \times 4 \times 0.50 = 0.40\ N} \)
Force = 0.40 N
Example
A wire of length \( \mathrm{0.30\ m} \) carries a current of \( \mathrm{6\ A} \) and sits in a \( \mathrm{0.50\ T} \) magnetic field at an angle of 45°. Find the magnetic force.
▶️ Answer / Explanation
Use:
\( \mathrm{F = B I L \sin\theta} \)
\( \mathrm{F = 0.50 \times 6 \times 0.30 \times \sin 45^\circ} \)
Since \( \mathrm{\sin 45^\circ = 0.707} \):
\( \mathrm{F = 0.50 \times 6 \times 0.30 \times 0.707 = 0.636\ N} \)
Force ≈ 0.64 N
Example
A conductor of length \( \mathrm{0.80\ m} \) carries an unknown current. It experiences a force of \( \mathrm{1.2\ N} \) when placed in a magnetic field of \( \mathrm{0.40\ T} \). The angle between the current and the field is 30°. Calculate the current \( \mathrm{I} \).
▶️ Answer / Explanation
Use:
\( \mathrm{F = B I L \sin\theta} \)
Rearrange:
\( \mathrm{I = \dfrac{F}{B L \sin\theta}} \)
Substitute values:
\( \mathrm{I = \dfrac{1.2}{0.40 \times 0.80 \times \sin 30^\circ}} \)
Since \( \mathrm{\sin 30^\circ = 0.5} \):
\( \mathrm{I = \dfrac{1.2}{0.40 \times 0.80 \times 0.5}} \)
\( \mathrm{I = \dfrac{1.2}{0.16} = 7.5\ A} \)
Current = 7.5 A
Magnetic Flux Density
Magnetic flux density is a measure of the strength of a magnetic field. It tells us how much force the field produces on a current-carrying conductor.
Definition:
Magnetic flux density \( \mathrm{B} \) is defined as the force per unit current per unit length on a straight conductor placed at right-angles to the magnetic field.
\( \mathrm{B = \dfrac{F}{IL}} \quad (\text{when } \theta = 90^\circ) \)
- \( \mathrm{B} \) = magnetic flux density (tesla, T)
- \( \mathrm{F} \) = force on conductor (N)
- \( \mathrm{I} \) = current in conductor (A)
- \( \mathrm{L} \) = length of conductor in field (m)
Key Points:
- If \( \mathrm{I} \), \( \mathrm{L} \), or \( \mathrm{F} \) increases, \( \mathrm{B} \) increases proportionally.
- Definition only applies when the conductor is perpendicular to the magnetic field.
- 1 tesla (1 T) is a very strong magnetic field.
Physical Meaning:
A magnetic field of 1 T exerts a force of 1 N on a 1 m wire carrying 1 A of current at right-angles to the field.
Connection to Motor Effect:
The definition comes directly from the motor-effect force equation \( \mathrm{F = BIL\sin\theta} \). For \( \mathrm{\theta = 90^\circ} \), \( \mathrm{F = BIL} \Rightarrow B = F/(IL) \).
Example
A conductor of length 1 m carries 1 A of current and experiences a force of 0.6 N when placed at right-angles to a magnetic field. Calculate the magnetic flux density.
▶️ Answer / Explanation
\( \mathrm{B = \dfrac{F}{IL} = \dfrac{0.6}{1 \times 1} = 0.6\ T} \)
Magnetic flux density = 0.6 T
Example
A 0.40 m wire carries a current of 5 A and is placed perpendicular to a magnetic field. The magnetic flux density is 0.25 T. Find the force acting on the wire.
▶️ Answer / Explanation
Use \( \mathrm{F = BIL} \):
\( \mathrm{F = 0.25 \times 5 \times 0.40 = 0.50\ N} \)
Force = 0.50 N
Example
A magnetic field exerts a force of \( \mathrm{2.4\ N} \) on a \( \mathrm{0.60\ m} \) conductor carrying \( \mathrm{8\ A} \). The conductor is at right-angles to the field. Calculate the magnetic flux density.
▶️ Answer / Explanation
\( \mathrm{B = \dfrac{F}{IL} = \dfrac{2.4}{8 \times 0.60}} \)
\( \mathrm{B = \dfrac{2.4}{4.8} = 0.50\ T} \)
Magnetic flux density = 0.50 T