Force on a current-carrying conductor

Key Definitions

• Magnetic Flux Density ($B$): The force per unit current per unit length acting on a straight conductor placed perpendicular to the magnetic field.
• Tesla (T): The SI unit of magnetic flux density. One Tesla is the magnetic flux density that exerts a force of one newton per metre on a conductor carrying a current of one ampere at right angles to the field ($1\text{ T} = 1\text{ N A}^{-1}\text{ m}^{-1}$).
• Magnetic Field: A region of space where a magnetic pole, a current-carrying conductor, or a moving charge experiences a magnetic force.
• Conventional Current: The theoretical flow of positive charge from the positive terminal to the negative terminal. In metals, this is opposite to the actual direction of electron flow.
• Motor Effect: The phenomenon where a current-carrying conductor experiences a force when placed in an external magnetic field.

Content

3.1 The Nature of the Force (The "Catapult Field")

The force on a conductor is not a direct "attraction" or "repulsion" in the electrostatic sense, but a result of the interaction between two magnetic fields:

1. The External Field: Usually a uniform field between two magnetic poles (North to South).
2. The Induced Field: A circular magnetic field created by the current in the wire (determined by the Right-Hand Grip Rule).

When the wire is placed in the external field, these two fields superimpose. On one side of the wire, the circular field lines point in the same direction as the external field lines, creating a region of high flux density (stronger field). On the opposite side, the circular field lines oppose the external field, creating a region of low flux density (weaker field).

Magnetic field lines behave like stretched elastic bands; they try to straighten and shorten. The concentration of field lines on one side "pushes" the conductor toward the weaker region. This resultant force is often called the catapult force.

Visualizing the Catapult Field: Imagine a cross-section of a wire with current flowing into the page ($\otimes$). The circular field is clockwise. If the external field points left to right, the field lines reinforce each other above the wire and cancel each other below the wire. The resulting force $\mathbf{F}$ acts downwards.

3.2 Fleming’s Left-Hand Rule (FLHR)

To predict the direction of the force, use your left hand. Ensure your fingers are mutually perpendicular (at $90^\circ$ to each other).

• First Finger: Direction of the external Magnetic Field ($N \rightarrow S$).
• SeCond Finger: Direction of the conventional Current ($+ \rightarrow -$).
• Thumb: Direction of the resulting Motion (Force).

Important Geometry:

• If the current is parallel to the magnetic field ($\theta = 0^\circ$ or $180^\circ$), the force is zero.
• The force is always perpendicular to both the current direction and the magnetic field direction.

3.3 The Magnitude of the Force

The magnitude of the magnetic force $F$ depends on the strength of the field, the magnitude of the current, the length of the wire within the field, and the orientation of the wire.

$$F = BIL \sin \theta$$

• $F$: Magnetic force (N)
• $B$: Magnetic flux density (T)
• $I$: Current (A)
• $L$: Length of the conductor exposed to the magnetic field (m)
• $\theta$: The angle between the conductor and the magnetic field lines.

Understanding the $\sin \theta$ Component: The force is generated only by the component of the magnetic field that is perpendicular to the current. If the wire is at an angle $\theta$ to the field $B$, the perpendicular component of the field is $B \sin \theta$. Alternatively, you can view $L \sin \theta$ as the "effective length" of the wire that is perpendicular to the field.

3.4 Defining Magnetic Flux Density ($B$)

In the Cambridge 9702 syllabus, $B$ is defined by rearranging the force equation for the specific case where the wire is at right angles to the field ($\theta = 90^\circ$, so $\sin \theta = 1$):

$$B = \frac{F}{IL}$$

Exam Requirement: When asked to define $B$, do not just provide the formula. You must state: "The force per unit current per unit length acting on a straight conductor placed at right angles to the magnetic field." The phrase "at right angles" (or perpendicular) is essential for full marks.

3.5 Experimental Measurement: The Top-Pan Balance

A common A-Level experiment involves placing a U-shaped "yoke" containing two magnets on a digital top-pan balance. A stiff wire is held stationary between the poles of the magnets using a stand and clamps.

1. When current flows through the wire, the magnetic field exerts a force $F$ on the wire (e.g., upwards).
2. According to Newton’s Third Law, the wire exerts an equal and opposite force $F$ on the magnets (e.g., downwards).
3. This downward force on the magnets increases the "mass" reading on the balance.
4. The change in mass $\Delta m$ is recorded. The magnetic force is calculated as $F = \Delta m g$.
5. By plotting a graph of $F$ against $I$, the gradient ($BL$) can be used to determine the magnetic flux density $B$.

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3.6 Worked Examples

Worked Example 1 — Force on a Power Line

A horizontal overhead power line carries a current of $1200\text{ A}$ from West to East. The Earth’s magnetic field at that location has a flux density of $5.0 \times 10^{-5}\text{ T}$ and is directed North. Calculate the magnitude and direction of the force acting on a $50\text{ m}$ section of the wire.

Step 1: Identify variables $I = 1200\text{ A}$ $B = 5.0 \times 10^{-5}\text{ T}$ $L = 50\text{ m}$ $\theta = 90^\circ$ (West-East is perpendicular to North)

Step 2: Calculate magnitude $$F = BIL \sin \theta$$ $$F = (5.0 \times 10^{-5}) \times (1200) \times (50) \times \sin(90^\circ)$$ $$F = 3.0\text{ N}$$

Step 3: Determine direction using FLHR

• First finger (Field): North.
• Second finger (Current): East.
• Thumb (Force): Points vertically upwards.

Answer: $3.0\text{ N}$ acting vertically upwards.

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Worked Example 2 — Determining $B$ from a Balance

A $10\text{ cm}$ length of wire is placed perpendicular to a magnetic field between two poles. The magnets are on a top-pan balance. When a current of $2.5\text{ A}$ passes through the wire, the balance reading increases from $150.00\text{ g}$ to $151.25\text{ g}$. Calculate the magnetic flux density $B$.

Step 1: Calculate the additional force $\Delta m = 151.25 - 150.00 = 1.25\text{ g} = 1.25 \times 10^{-3}\text{ kg}$ $F = \Delta m g = (1.25 \times 10^{-3}) \times 9.81$ $F = 0.01226\text{ N}$

Step 2: Identify other variables $L = 10\text{ cm} = 0.10\text{ m}$ $I = 2.5\text{ A}$ $\theta = 90^\circ$

Step 3: Rearrange and solve for $B$ $$B = \frac{F}{IL}$$ $$B = \frac{0.01226}{2.5 \times 0.10}$$ $$B = \frac{0.01226}{0.25} = 0.04904\text{ T}$$

Answer: $B = 4.9 \times 10^{-2}\text{ T}$ (to 2 s.f.)

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Worked Example 3 — Wire at an Angle

A wire of length $0.40\text{ m}$ carries a current of $3.5\text{ A}$. It is placed in a uniform magnetic field of $0.15\text{ T}$. The wire is oriented at an angle of $60^\circ$ to the field lines. Calculate the force on the wire.

Step 1: Identify variables $B = 0.15\text{ T}$ $I = 3.5\text{ A}$ $L = 0.40\text{ m}$ $\theta = 60^\circ$

Step 2: Substitution $$F = BIL \sin \theta$$ $$F = (0.15) \times (3.5) \times (0.40) \times \sin(60^\circ)$$ $$F = 0.21 \times 0.866$$ $$F = 0.1818\dots\text{ N}$$

Answer: $0.18\text{ N}$ (to 2 s.f.)

Key Equations

Equation	Description	Data Sheet?
$F = BIL \sin \theta$	Force on a current-carrying conductor in a magnetic field.	Yes
$B = \frac{F}{IL}$	Definition of Magnetic Flux Density (for $\theta = 90^\circ$).	No
$F = \Delta m g$	Force calculation for top-pan balance experiments.	No
$1\text{ T} = 1\text{ N A}^{-1}\text{ m}^{-1}$	Unit consistency for the Tesla.	No

Common Mistakes to Avoid

• ❌ Using the Right Hand: Students often use the Right-Hand Rule for force.
  • ✓ Right: Use the Left Hand for the Motor Effect (Force). Use the Right Hand only for Induced Current (Topic 21).
• ❌ Confusing $\sin \theta$ and $\cos \theta$: Using the angle between the wire and the normal to the field.
  • ✓ Right: $\theta$ is the angle between the wire and the field lines. If the wire is perpendicular, $\theta = 90^\circ$ and $\sin(90^\circ) = 1$.
• ❌ Neglecting Unit Conversions: Using $L$ in cm or $m$ in grams.
  • ✓ Right: Always convert to metres (m), kilograms (kg), and Amperes (A).
• ❌ Misinterpreting Balance Readings: Assuming the mass reading is the force.
  • ✓ Right: The balance measures mass. You must multiply the change in mass ($\Delta m$) by $g$ ($9.81\text{ m s}^{-2}$) to find the force in Newtons.
• ❌ Electron Flow vs. Conventional Current:
  • ✓ Right: If a question says "electrons flow South," your second finger (Current) must point North.

Exam Tips

1. The "Perpendicular" Keyword: In any definition of $B$ or the Tesla, you will lose the mark if you do not mention that the wire/current is perpendicular or at right angles to the field.
2. 3D Notation:
   • $\odot$ (Dot): Field or current coming out of the page (like the tip of an arrow).
   • $\otimes$ (Cross): Field or current going into the page (like the feathers of an arrow).
3. Newton's Third Law in Magnetism: If a question asks why the balance reading changes, explain that the magnetic field exerts a force on the wire, and the wire exerts an equal and opposite force on the magnets.
4. Zero Force Situations: Always check if the wire is parallel to the field. Examiners often include a "trick" section of a circuit that is parallel to the $B$-field; the force on that specific section is $0\text{ N}$.
5. Deriving Units: You may be asked for the base units of Tesla.
   • $B = \frac{F}{IL} \rightarrow \frac{\text{N}}{\text{A m}}$
   • Since $1\text{ N} = 1\text{ kg m s}^{-2}$
   • Base units of $B = \frac{\text{kg m s}^{-2}}{\text{A m}} = \mathbf{kg s^{-2} A^{-1}}$.