Magnetic effect of a current

1. Overview

Whenever an electric current flows through a conductor, a magnetic field is created around it. This phenomenon, known as electromagnetism, is fundamental to modern technology as it allows us to create magnets that can be turned on and off and to convert electrical energy into physical movement.

Key Definitions

• Magnetic Field: A region of space around a magnet or a current-carrying wire where a magnetic pole experiences a force.
• Solenoid: A long coil of wire, often wrapped into a cylindrical shape, which produces a magnetic field similar to a bar magnet when current passes through it.
• Electromagnet: A temporary magnet consisting of a coil of wire (solenoid) wrapped around a soft iron core; it is only magnetic when current flows.
• Right-Hand Grip Rule: A rule used to determine the direction of the magnetic field lines around a straight wire (Thumb = Current, Fingers = Field direction).

Core Content

Magnetic Field Patterns

• Straight Wire: The magnetic field forms concentric circles around the wire. The circles get further apart as you move away from the wire, indicating the field is getting weaker.
• Solenoid: The field inside a solenoid is strong and uniform (parallel lines). Outside the solenoid, the field pattern is identical to that of a bar magnet, with field lines emerging from the North pole and entering the South pole.
• 📊A vertical straight wire with circular field lines around it, labeled with arrows. A solenoid diagram showing field lines looping through the center and around the outside like a bar magnet.

Experiment: Identifying the Pattern and Direction

1. To find the pattern: Pass a wire through a piece of stiff cardboard. Sprinkle iron filings onto the card and turn on the current. Gently tap the card; the filings will align themselves with the magnetic field lines to show the pattern.
2. To find the direction: Place small plotting compasses on the cardboard around the wire or solenoid. The "North" end of the compass needle will point in the direction of the magnetic field at that point.

Applications of Electromagnetism

• Relays: A relay uses a small current in one circuit to switch on a much larger current in a second circuit.
  • How it works: Current flows through a coil $\rightarrow$ coil becomes an electromagnet $\rightarrow$ it pulls a magnetic armature $\rightarrow$ the armature closes contacts in a separate high-current circuit.
  • Example: Using a low-voltage dashboard switch to start a high-current car starter motor.
• Loudspeakers: Converts electrical signals into sound waves.
  • How it works: An alternating current (AC) flows through a coil attached to a speaker cone. This coil is placed near a permanent magnet. The changing magnetic field of the coil interacts with the permanent magnet, causing the coil (and the cone) to vibrate back and forth, creating sound waves.

Extended Content (Extended Only)

Field Strength Variation

• Straight Wire: The strength of the magnetic field is strongest closest to the wire and decreases as the distance from the wire increases.
• Solenoid: The field is strongest inside the coil. Increasing the number of turns per unit length increases the field strength.

Factors Affecting the Magnetic Field

1. Magnitude of Current: Increasing the current increases the strength of the magnetic field around both straight wires and solenoids.
2. Direction of Current: Reversing the direction of the current reverses the direction of the magnetic field lines (and swaps the North/South poles of a solenoid).
3. Soft Iron Core (Solenoid only): Placing a soft iron core inside a solenoid significantly increases the strength of the magnetic field because the iron becomes magnetized itself.

Key Equations

Note: While Topic 4.5.3 focuses on the effect of current, the following equations are essential for the related applications in transformers (Topic 4.5.4):

• The Transformer Equation: $\frac{V_p}{V_s} = \frac{N_p}{N_s}$
  • $V_p / V_s$: Voltage in primary/secondary coils (Volts, V)
  • $N_p / N_s$: Number of turns in primary/secondary coils
• Power in an Ideal Transformer: $I_p V_p = I_s V_s$
  • $I_p / I_s$: Current in primary/secondary coils (Amps, A)
• Efficiency: $\text{Efficiency} = \frac{\text{Power Output}}{\text{Power Input}} \times 100%$

Common Mistakes to Avoid

• ❌ Wrong: Forgetting that current and voltage have an inverse relationship in transformers.
• ✓ Right: If a transformer steps the voltage up, the current must step down (assuming 100% efficiency).
• ❌ Wrong: Dividing the input voltage by the turns ratio when you should multiply.
• ✓ Right: Always use the formula $\frac{V_p}{V_s} = \frac{N_p}{N_s}$ and rearrange carefully. If $N_s > N_p$, then $V_s$ must be greater than $V_p$.
• ❌ Wrong: Assuming all transformers are 100% efficient in real-world descriptions.
• ✓ Right: In reality, energy is lost as heat; $I_s V_s$ is usually slightly less than $I_p V_p$.
• ❌ Wrong: Using the Left-Hand Rule for field direction around a wire.
• ✓ Right: Use the Right-Hand Grip Rule for field direction around a wire, and the Left-Hand Motor Rule only when calculating the direction of a force (thrust).

Exam Tips

1. Check your poles: In a solenoid, use the "Right-Hand Grip Rule" where your fingers curl in the direction of the current, and your thumb points to the North Pole.
2. Draw carefully: When drawing field lines for a straight wire, ensure the circles are centered on the wire and the spacing increases as you move outward.
3. Relay questions: When describing a relay, always mention that the two circuits are electrically isolated, which allows a safe low-voltage circuit to control a dangerous high-voltage one.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

(a) A student investigates the magnetic field around a straight, current-carrying wire. Describe how they could use a compass and iron filings to map the magnetic field. [3]

(b) State two ways to increase the strength of the magnetic field around the wire. [2]

Worked Solution:

(a)

1. Place the wire vertically through a piece of card. Sets up the experiment.
2. Pass a current through the wire. Indicates the wire is active.
3. Sprinkle iron filings onto the card around the wire. Indicates how the field is visualized.
4. Gently tap the card. Allows filings to align.
5. Observe the pattern of the iron filings, which shows circular field lines around the wire. Describes observation.
6. Place a compass on the card and note the direction of the needle at different points around the wire. Describes how to determine direction.

How to earn full marks:

• Mention placing the wire through the card.
• Mention passing a current through the wire.
• Describe observing the pattern of the field (circular) using iron filings.

(b)

1. Increase the current flowing through the wire. Relates current to field strength.
2. Decrease the distance from the wire. Relates distance to field strength.

How to earn full marks:

• State increasing the current.
• State decreasing the distance.

Common Pitfall: Students often forget to mention the card when describing the iron filings method. Also, remember that the magnetic field strength decreases as you move further away from the wire.

Exam-Style Question 2 — Short Answer [6 marks]

Question:

(a) Draw the magnetic field pattern around a solenoid carrying a current. Indicate the direction of the magnetic field lines. [4]

(b) State two differences between the magnetic field pattern inside a solenoid and the magnetic field pattern outside the solenoid. [2]

Worked Solution:

(a)

1. 📊A solenoid is drawn as a coil of wire with at least 5 turns. Magnetic field lines are drawn both inside and outside the solenoid. Inside, the field lines are parallel, equally spaced, and run straight from one end of the solenoid to the other. Outside, the field lines loop around the solenoid, resembling the field of a bar magnet. Arrows on the field lines indicate the direction of the field, emerging from the north pole and entering the south pole of the solenoid. The N and S poles of the solenoid should be clearly indicated.

How to earn full marks:

• Draw the solenoid as a coil of wire (at least 5 turns).
• Draw field lines inside the solenoid as parallel and equally spaced.
• Draw field lines outside the solenoid looping around.
• Indicate the direction of the field with arrows AND label the N and S poles.

(b)

1. Inside the solenoid, the magnetic field is strong and uniform; outside the solenoid, the magnetic field is weaker and non-uniform. Compares strength and uniformity.
2. Inside the solenoid, the field lines are parallel; outside the solenoid, the field lines are curved. Compares field line shape.

How to earn full marks:

• State the difference in strength OR uniformity of the field.
• State the difference in the shape of the field lines.

Common Pitfall: Many students forget to draw the field lines as parallel and equally spaced inside the solenoid. Also, be sure to indicate the direction of the magnetic field lines with arrows and label the N and S poles.

Exam-Style Question 3 — Extended Response [8 marks]

Question:

A relay is used to control a high-current circuit with a low-current circuit.

(a) Explain how a relay works, including a diagram of a typical relay circuit. [5]

(b) State two advantages of using a relay in a circuit. [2]

(c) Suggest one application of a relay in a car. [1]

Worked Solution:

(a)

1. 📊A diagram of a relay circuit is drawn. The diagram shows two circuits: a low-current control circuit and a high-current output circuit. The low-current circuit includes a switch, a low-voltage power supply (e.g., 6V), and a coil of wire (with at least 20 turns) wrapped around an iron core (the electromagnet). The high-current circuit includes a high-voltage power supply (e.g., 12V), a lamp (or other high-power device), and a switch (armature) controlled by the electromagnet. The armature is shown as a pivoting metal bar. When the switch in the low-current circuit is closed, current flows through the coil, creating a magnetic field. This magnetic field attracts the metal armature, closing the high-current circuit and turning on the lamp. When the switch in the low-current circuit is opened, the magnetic field collapses, the armature returns to its original position (due to a spring), and the high-current circuit is opened, turning off the lamp. Label all components clearly.
2. When a small current flows through the coil in the low-current circuit, it creates a magnetic field. Describes the initial action.
3. This magnetic field attracts an iron armature (or switch). Describes the magnetic attraction.
4. The movement of the armature closes (or opens) a separate, high-current circuit. Describes the switching action.
5. When the current in the coil is switched off, the magnetic field collapses, and a spring returns the armature to its original position, breaking (or making) the high-current circuit. Describes the reverse action.

How to earn full marks:

• Draw a complete and correctly labelled diagram of a relay circuit.
• Explain that a small current creates a magnetic field in the coil.
• Explain that the magnetic field attracts the armature.
• Explain that the armature controls a separate high-current circuit.
• Explain what happens when the current in the coil is switched off, including the role of the spring.

(b)

1. A relay allows a low-voltage circuit to control a high-voltage circuit, providing electrical isolation. Isolation of circuits.
2. A relay allows a circuit to be controlled remotely, from a distance. Remote control capability.

How to earn full marks:

• State that a relay allows a low-voltage circuit to control a high-voltage circuit (mentioning isolation).
• State that a relay allows remote control.

(c)

1. Controlling the starter motor $\boxed{\text{or}}$ switching headlights on/off. Example application.

How to earn full marks:

• State a valid application of a relay in a car.

Common Pitfall: Students often forget to include the spring in their relay diagram or explain its function. Also, remember that relays provide electrical isolation between the control circuit and the high-current circuit.

Exam-Style Question 4 — Extended Response [10 marks]

Question:

A student sets up the following experiment to investigate the magnetic field produced by a solenoid. The solenoid is connected to a variable power supply and an ammeter. A small Hall probe is placed near the center of the solenoid to measure the magnetic field strength. The student varies the current through the solenoid and records the magnetic field strength for each current value.

(a) Describe how the student could ensure that the Hall probe is measuring the magnetic field strength accurately at the center of the solenoid. [2]

(b) The student obtains the following data:

Current (A)	Magnetic Field Strength (mT)
0.5	2.5
1.0	5.0
1.5	7.5
2.0	10.0

Plot a graph of magnetic field strength (y-axis) against current (x-axis). [4]

(c) Determine the gradient of the graph. State the unit. [2]

(d) Explain how the number of turns per unit length of the solenoid and the permeability of free space affect the gradient of the graph. [2]

Worked Solution:

(a)

1. Ensure the Hall probe is positioned perpendicular to the axis of the solenoid. Correct orientation.
2. Place the Hall probe at the center of the solenoid along its axis. Correct location.

How to earn full marks:

• Mention the Hall probe is perpendicular to the solenoid's axis.
• Mention the Hall probe is placed at the center of the solenoid.

(b)

1. [GRAPH: A graph is drawn with Current (I / A) on the x-axis and Magnetic Field Strength (B / mT) on the y-axis. The scale on the x-axis ranges from 0 to 2.5 A, with major gridlines at 0.5 A intervals, and the scale on the y-axis ranges from 0 to 12.5 mT, with major gridlines at 2.5 mT intervals. The points (0.5, 2.5), (1.0, 5.0), (1.5, 7.5), and (2.0, 10.0) are plotted accurately using small crosses or dots. A straight line of best fit is drawn through the points, passing through (or very close to) the origin.]

How to earn full marks:

• Correctly label both axes with quantity AND unit (B / mT and I / A).
• Choose a suitable linear scale for both axes, using at least half of the grid.
• Plot all points accurately (within half a small square).
• Draw a single straight line of best fit through the points.

(c)

1. Calculate the gradient: $gradient = \frac{\Delta y}{\Delta x} = \frac{(10.0 - 0) \text{ mT}}{(2.0 - 0) \text{ A}} = 5.0 \text{ mT/A}$ Correct substitution from the graph.
2. The gradient is $\boxed{5.0 \times 10^{-3} \text{ T/A}}$ Correct answer with unit.

How to earn full marks:

• Show the working for calculating the gradient, including correct units in the calculation.
• Calculate the correct gradient value, with correct unit (T/A or mT/A).

(d)

1. The gradient is directly proportional to the number of turns per unit length of the solenoid. Increasing the number of turns per unit length increases the gradient, resulting in a stronger magnetic field for the same current. Effect of turns per length.
2. The gradient is also directly proportional to the permeability of free space. A higher permeability results in a stronger magnetic field for the same current, increasing the gradient. Effect of permeability.

How to earn full marks:

• State that the gradient is proportional to the number of turns per unit length.
• State that the gradient is proportional to the permeability of free space.

Common Pitfall: When plotting the graph, make sure to label the axes with both the quantity AND the unit. Also, remember to convert mT to T when calculating the gradient if you want the answer in SI units.