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Physics - Magnetism and Electromagnets

Grade 8IGCSE

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

🔑Concepts

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

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

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

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

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

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.

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

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.

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

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

\frac{V_p}{V_s} = rac{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 - Revision Notes & Key Formulas | IGCSE Grade 8 Science