Magnetic fields due to currents

1. Overview

The fundamental principle of electromagnetism is that moving electric charges (electric currents) generate magnetic fields in the surrounding space. Unlike the static electric fields produced by stationary charges, magnetic fields are inherently linked to the motion of electrons through conductors. The geometry of the conductor—whether it is a straight wire, a circular loop, or a multi-turn solenoid—determines the specific shape, direction, and strength of the resulting magnetic field. This topic focuses on visualizing these fields, understanding how materials like iron can amplify them, and explaining the mechanical forces that arise when two current-carrying conductors interact.

Key Definitions

• Magnetic Flux Density ($B$): The force acting per unit length on a straight conductor carrying unit current placed perpendicular to the direction of the magnetic field. It is a vector quantity.
• Tesla (T): The SI unit of magnetic flux density. One Tesla is defined as the magnetic flux density that produces a force of 1 Newton per metre on a conductor carrying a current of 1 Ampere at right angles to the field ($1\text{ T} = 1\text{ N A}^{-1}\text{ m}^{-1}$).
• Permeability of Free Space ($\mu_0$): A physical constant that describes the ability of a vacuum to support the formation of a magnetic field. Its value is exactly $4\pi \times 10^{-7}\text{ H m}^{-1}$ (or $\text{T m A}^{-1}$).
• Solenoid: A long coil of wire consisting of many loops wound into a tightly packed helix. When current flows through it, the individual fields of each loop superpose to create a strong, uniform magnetic field inside the coil.
• Ferrous Core: A core made of iron or a similar ferromagnetic material inserted into a solenoid to concentrate and increase the magnetic flux density.

Content

3.1 Magnetic Field Patterns

To describe magnetic fields, we use magnetic field lines (flux lines). The density of these lines represents the magnitude of the magnetic flux density ($B$), and the tangent to the line at any point indicates the direction of the field.

1. Long Straight Current-Carrying Wire

• Field Shape: The field lines form concentric circles centered on the wire, lying in a plane perpendicular to the wire.
• Direction: Determined by the Right-Hand Grip Rule. If the right thumb points in the direction of conventional current ($I$), the curled fingers indicate the direction of the magnetic field ($B$).
• Field Strength: The field is strongest near the wire and weakens with distance. Mathematically, $B \propto \frac{1}{r}$.
• Visual Representation: When sketching, the spacing between the concentric circles must increase as you move further from the wire to demonstrate the decreasing field strength.
• Notation: Use a cross ($\times$) to represent current flowing into the page (like the feathers of an arrow moving away) and a dot ($\cdot$) to represent current flowing out of the page (like the tip of an arrow approaching).

2. Flat Circular Coil

• Field Shape: Near the wire itself, the field lines are circular. As you move toward the center of the loop, the lines become less curved. At the exact center of the coil, the field line is a straight line perpendicular to the plane of the loop.
• Polarity: A circular coil acts like a thin disc magnet. One face acts as a North pole and the other as a South pole.
• Right-Hand Grip Rule (Alternative): Curl your fingers in the direction of the current around the loop; your thumb points toward the North pole of the coil.

3. Long Solenoid

• Internal Field: Inside the solenoid (away from the ends), the field lines are parallel, straight, and equally spaced. This indicates that the magnetic field is uniform (constant magnitude and direction).
• External Field: Outside the solenoid, the field pattern is identical to that of a bar magnet. The lines emerge from the North pole, loop around, and enter the South pole.
• Field Strength: The strength depends on the current ($I$) and the number of turns per unit length ($n$).
• Determining Poles: Use the Right-Hand Grip Rule where fingers curl with the current in the coils, and the thumb points to the end of the solenoid that acts as the North pole.

3.2 The Effect of a Ferrous Core

The magnetic field of a solenoid can be made significantly stronger (often by a factor of hundreds or thousands) by placing a ferrous core (such as soft iron) inside the coils.

• Magnetic Domains: Ferromagnetic materials like iron consist of microscopic regions called domains. In each domain, the magnetic moments of atoms are aligned.
• Alignment: In an unmagnetized state, these domains are oriented randomly, so their individual magnetic fields cancel out. When the solenoid's magnetic field passes through the iron, it exerts a torque on these domains, aligning them with the external field.
• Resultant Field: The total magnetic field is the vector sum of the field produced by the current in the wire and the field produced by the aligned domains in the iron.
• Permeability: Iron has a much higher relative permeability than air, meaning it "concentrates" the magnetic flux lines, resulting in a much higher magnetic flux density ($B$).
• Soft Iron: "Soft" iron is typically used because it loses its induced magnetism quickly when the current is turned off (low retentivity), making it ideal for electromagnets.

3.3 Forces Between Current-Carrying Conductors

When two parallel wires carry current, they exert a force on each other. This is not a direct "current-to-current" interaction but a two-step magnetic process.

The Origin of the Force (Step-by-Step Explanation):

1. Field Generation: Wire 1 carries a current $I_1$ and creates a magnetic field $B_1$ in the space surrounding it. At the position of Wire 2 (distance $d$ away), the magnitude of this field is $B_1 = \frac{\mu_0 I_1}{2\pi d}$.
2. Interaction: Wire 2, which carries current $I_2$, is now a current-carrying conductor sitting inside an external magnetic field ($B_1$).
3. Force Application: According to the motor effect ($F = BIL \sin \theta$), Wire 2 experiences a magnetic force. Since the wires are parallel, the current $I_2$ is perpendicular to the field $B_1$ ($\theta = 90^\circ$), so the force per unit length is $F/L = B_1 I_2$.
4. Symmetry: By Newton's Third Law, Wire 2 creates a field $B_2$ that exerts an equal and opposite force on Wire 1.

Determining Direction:

• Parallel Currents (Same Direction): The wires attract each other.
• Anti-parallel Currents (Opposite Directions): The wires repel each other.
• Verification: Use the Right-Hand Grip Rule to find the direction of $B_1$ at Wire 2, then use Fleming's Left-Hand Rule (First finger = Field $B_1$, Second finger = Current $I_2$, Thumb = Force $F$) to find the direction of the force.

Worked Example 1 — Force Between Parallel Wires

Two long, vertical, parallel wires, X and Y, are separated by a distance of $8.0\text{ cm}$ in a vacuum. Wire X carries a current of $4.0\text{ A}$ upwards. Wire Y carries a current of $6.0\text{ A}$ downwards. Calculate the magnitude and direction of the force per unit length acting on Wire X.

Step 1: Identify the given values and convert to SI units.

• $I_X = 4.0\text{ A}$
• $I_Y = 6.0\text{ A}$
• $d = 8.0\text{ cm} = 0.080\text{ m}$
• $\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1}$

Step 2: State the formula for force per unit length. $$\frac{F}{L} = \frac{\mu_0 I_X I_Y}{2\pi d}$$

Step 3: Substitute the values into the equation. $$\frac{F}{L} = \frac{(4\pi \times 10^{-7}) \times 4.0 \times 6.0}{2\pi \times 0.080}$$

Step 4: Simplify and calculate. $$\frac{F}{L} = \frac{2 \times 10^{-7} \times 24.0}{0.080}$$ $$\frac{F}{L} = \frac{4.8 \times 10^{-6}}{0.080}$$ $$\frac{F}{L} = 6.0 \times 10^{-5}\text{ N m}^{-1}$$

Step 5: Determine the direction. Since the currents are in opposite directions (one up, one down), the wires repel each other. Therefore, the force on Wire X is directed away from Wire Y (to the left if Y is to the right of X).

Worked Example 2 — Resultant Magnetic Field

Two long straight wires are placed $10.0\text{ cm}$ apart. Wire A carries $5.0\text{ A}$ out of the page. Wire B carries $10.0\text{ A}$ into the page. Calculate the magnitude of the resultant magnetic flux density at a point $P$ exactly midway between the two wires.

Step 1: Calculate the field contribution from Wire A ($B_A$). Point $P$ is $5.0\text{ cm} = 0.050\text{ m}$ from Wire A. $$B_A = \frac{\mu_0 I_A}{2\pi r} = \frac{4\pi \times 10^{-7} \times 5.0}{2\pi \times 0.050} = \frac{2 \times 10^{-7} \times 5.0}{0.050} = 2.0 \times 10^{-5}\text{ T}$$ Using the Right-Hand Grip Rule for a current coming out of the page, the field at $P$ (to the right of A) points upwards.

Step 2: Calculate the field contribution from Wire B ($B_B$). Point $P$ is $5.0\text{ cm} = 0.050\text{ m}$ from Wire B. $$B_B = \frac{\mu_0 I_B}{2\pi r} = \frac{4\pi \times 10^{-7} \times 10.0}{2\pi \times 0.050} = \frac{2 \times 10^{-7} \times 10.0}{0.050} = 4.0 \times 10^{-5}\text{ T}$$ Using the Right-Hand Grip Rule for a current going into the page, the field at $P$ (to the left of B) also points upwards.

Step 3: Calculate the resultant field ($B_{total}$). Since both vectors point in the same direction (upwards): $$B_{total} = B_A + B_B = 2.0 \times 10^{-5} + 4.0 \times 10^{-5} = 6.0 \times 10^{-5}\text{ T}$$

Key Equations

Equation	Description	Status
$F = BIL \sin \theta$	Force on a conductor in an external field	On Data Sheet
$B = \frac{\mu_0 I}{2\pi d}$	Flux density at distance $d$ from a long straight wire	Memorise
$B = \mu_0 n I$	Flux density inside a long solenoid ($n = N/L$)	Memorise
$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$	Force per unit length between two parallel wires	Memorise
$1\text{ T} = 1\text{ N A}^{-1}\text{ m}^{-1}$	Definition of the Tesla in base units	Memorise

Common Mistakes to Avoid

• ❌ Confusing Attraction/Repulsion: Students often assume "like" currents repel (like charges).
  • ✓ Right: Parallel currents in the same direction attract. Parallel currents in opposite directions repel.
• ❌ Incorrect Field Line Spacing: Drawing equally spaced circles around a single wire.
  • ✓ Right: The distance between circles must increase as you move outward to show $B \propto 1/r$.
• ❌ Unit Errors: Using distance in centimeters ($cm$) or millimeters ($mm$) in the formula.
  • ✓ Right: Always convert to meters ($m$) before calculating.
• ❌ Solenoid Field Termination: Drawing field lines that stop at the ends of the solenoid.
  • ✓ Right: Magnetic field lines are continuous loops. They must pass through the center and return around the outside.
• ❌ Misapplying the Grip Rule: Using the left hand for field direction.
  • ✓ Right: Always use the Right Hand for field patterns (Grip Rule) and the Left Hand for force/motion (Fleming's Rule).

Exam Tips

1. The "Explain" Question: If asked to explain why two wires exert a force, use this logical chain:
   • Current in Wire 1 creates a magnetic field.
   • Wire 2 is a current-carrying conductor located within that field.
   • A magnetic force acts on Wire 2 according to $F=BIL$.
   • Mention Fleming's Left-Hand Rule to determine the specific direction.
2. Newton’s Third Law: Remember that the force on Wire 1 is always equal in magnitude to the force on Wire 2, even if the currents are different (e.g., $2\text{ A}$ and $100\text{ A}$).
3. Dot and Cross Diagrams: Practice drawing the field between two wires using dot/cross notation.
   • Between two wires with currents in the same direction, the fields oppose each other, creating a region of weak field (neutral point).
   • Between two wires with currents in opposite directions, the fields reinforce each other, creating a region of strong field.
4. Solenoid Uniformity: When sketching a solenoid, ensure the lines inside are perfectly parallel and equally spaced. This is a specific requirement for "uniform field" marks.
5. Defining the Tesla: If asked for the definition, you must include the word perpendicular or at right angles. "The force per unit length per unit current" is not enough; the orientation of the wire relative to the field is essential.