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Electricity and Magnetism - Simple phenomena of magnetism

Grade 12IGCSEPhysics

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

🔑Concepts

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Magnetic poles: Every magnet has two poles, North (NN) and South (SS). Like poles repel, while unlike poles attract.

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Magnetic materials: Ferromagnetic materials include Iron, Nickel, Cobalt, and their alloys such as Steel. Non-magnetic materials include Copper, Aluminum, and Plastic.

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Induced Magnetism: When a piece of unmagnetized magnetic material (e.g., a soft iron nail) is placed near a magnet, it becomes magnetized. The end nearest the magnet's pole acquires the opposite polarity.

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Magnetic Field Lines: These represent the region where a magnetic force is exerted. Lines travel from the North pole to the South pole. The density of lines indicates the magnetic flux density (BB), measured in Tesla (TT).

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Hard and Soft Magnetic Materials: 'Soft' iron is easily magnetized and demagnetized (used in electromagnets). 'Hard' steel is difficult to magnetize but retains its magnetism (used for permanent magnets).

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Demagnetization methods: Heating the magnet to high temperatures, hammering it while oriented East-West, or placing it inside a solenoid carrying an alternating current (A.C.A.C.) and slowly withdrawing it.

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Magnetization methods: Stroking a magnetic material with a permanent magnet or placing it inside a solenoid carrying a direct current (D.C.D.C.).

📐Formulae

Magnetic Flux (\Phi)=B⋅A⋅cos⁡(θ)\text{Magnetic Flux (\Phi)} = B \cdot A \cdot \cos(\theta)

F=q(v×B)sin⁡(θ)F = q(v \times B) \sin(\theta)

F=BIlsin⁡(θ)F = BIl \sin(\theta)

💡Examples

Problem 1:

A student has two identical-looking metal bars, XX and YY. One is a permanent magnet and the other is a piece of unmagnetized soft iron. Describe how the student can identify which is which using only the two bars.

Solution:

The student should bring the end of bar XX near both ends of bar YY. If there is repulsion at any point, then both bars are permanent magnets. If there is only attraction regardless of which ends are brought together, then one is a magnet and the other is magnetic material (iron). To be certain, if bar XX attracts both ends of bar YY, XX is likely the magnet and YY is the iron.

Explanation:

Magnetic attraction occurs between a magnet and a magnetic material, but magnetic repulsion only occurs between two like poles of two permanent magnets.

Problem 2:

Calculate the magnetic force FF acting on a wire of length L=0.5 mL = 0.5\,m carrying a current I=2.0 AI = 2.0\,A placed perpendicularly in a uniform magnetic field of B=0.15 TB = 0.15\,T.

Solution:

Using the formula F=BIlsin⁡(θ)F = BIl \sin(\theta), where θ=90∘\theta = 90^\circ: F=0.15×2.0×0.5×sin⁡(90∘)F = 0.15 \times 2.0 \times 0.5 \times \sin(90^\circ) F=0.15 NF = 0.15\,N

Explanation:

Since the wire is perpendicular to the field, sin⁡(90∘)=1\sin(90^\circ) = 1, resulting in the maximum possible force for that field and current.

Simple phenomena of magnetism - Revision Notes & Key Formulas | IGCSE Grade 12 Physics