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Physics - Magnetism and Electromagnets

Grade 8Cambridge (IGCSE)

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Magnets have two poles: North (NN) and South (SS). Like poles repel, and opposite poles attract.

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Magnetic materials include Iron (FeFe), Nickel (NiNi), and Cobalt (CoCo). Steel is an alloy containing iron and is used to make permanent magnets.

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Magnetic field lines represent the direction and strength of a magnetic field. They always point from North (NN) to South (SS).

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The density of field lines indicates field strength; where lines are closest together (at the poles), the field is strongest.

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Induced magnetism occurs when a magnetic material is placed in a magnetic field and becomes a magnet itself. This magnetism is usually temporary.

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Hard magnetic materials (e.g., Steel) are difficult to magnetize but retain their magnetism. Soft magnetic materials (e.g., Iron) are easy to magnetize but lose it quickly.

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An electromagnet is a temporary magnet produced by an electric current flowing through a coil of wire (a solenoid).

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The strength of an electromagnet can be increased by: increasing the current (II), increasing the number of turns in the coil (NN), or adding a soft iron core.

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The Right-Hand Grip Rule: If you wrap your right hand around a solenoid with fingers in the direction of current (II), your thumb points toward the North pole (NN).

📐Formulae

Magnetic Field Strength∝I (Current)\text{Magnetic Field Strength} \propto I \text{ (Current)}

Magnetic Field Strength∝N (Number of turns)\text{Magnetic Field Strength} \propto N \text{ (Number of turns)}

VpVs=NpNs\frac{V_p}{V_s} = \frac{N_p}{N_s}

P=V×IP = V \times I

💡Examples

Problem 1:

A student wants to increase the strength of an electromagnet. If the current is currently I=2 AI = 2\text{ A} and the coil has N=50N = 50 turns, calculate the new relative strength if the current is doubled and the turns are tripled.

Solution:

The strength is proportional to I×NI \times N. Initial strength ∝2×50=100\propto 2 \times 50 = 100. New current Inew=4 AI_{new} = 4\text{ A} and new turns Nnew=150N_{new} = 150. New strength ∝4×150=600\propto 4 \times 150 = 600.

Explanation:

Since the strength of an electromagnet is directly proportional to both the current (II) and the number of turns (NN), the total factor of increase is 2×3=62 \times 3 = 6 times the original strength.

Problem 2:

A transformer is used to step up voltage. The primary coil has Np=100N_p = 100 turns and a voltage of Vp=12 VV_p = 12\text{ V}. If the secondary coil has Ns=500N_s = 500 turns, calculate the secondary voltage VsV_s.

Solution:

12Vs=100500\frac{12}{V_s} = \frac{100}{500} Vs=12×500100=60 VV_s = \frac{12 \times 500}{100} = 60\text{ V}

Explanation:

Using the transformer equation VpVs=NpNs\frac{V_p}{V_s} = \frac{N_p}{N_s}, we can solve for the unknown voltage by cross-multiplying and isolating VsV_s.

Problem 3:

Identify the polarity of an electromagnet if the current enters the solenoid and flows clockwise when viewed from the right end.

Solution:

The right end acts as a South (SS) pole.

Explanation:

According to the clock rule, if the current flows in a clockwise direction, that end of the solenoid behaves as a South pole. Conversely, an anti-clockwise current indicates a North pole.

Magnetism and Electromagnets Grade 8 Notes & Examples