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Electricity and Magnetism - Electromagnetic effects

Grade 12IGCSEPhysics

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

🔑Concepts

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Magnetic Field of a Current: A current-carrying conductor produces a magnetic field. For a straight wire, the field lines are concentric circles, the direction of which is determined by the Right-Hand Grip Rule.

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The Motor Effect: When a current II flows through a conductor of length LL placed in an external magnetic field BB, it experiences a force FF. This force is maximum when the conductor is perpendicular to the field.

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Fleming's Left-Hand Rule: Used to predict the direction of the force in the motor effect. The Thumb represents Force (FF), the First Finger represents Magnetic Field (BB), and the Second Finger represents Current (II).

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Electromagnetic Induction: The process of generating an e.m.f. (electromotive force) by moving a conductor through a magnetic field or by changing the magnetic field within a coil.

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Faraday's Law: The magnitude of the induced e.m.f. ϵ\epsilon is directly proportional to the rate of change of magnetic flux linkage NΦN\Phi.

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Lenz's Law: The direction of the induced e.m.f. is such that it creates a current whose magnetic field opposes the change in magnetic flux that produced it. This is a statement of the Law of Conservation of Energy.

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Transformers: Devices that increase or decrease alternating voltages. They consist of a primary coil and a secondary coil wound around a soft iron core. A Step-up transformer has Ns>NpN_s > N_p, while a Step-down transformer has Np>NsN_p > N_s.

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A.C. Generator: Uses a rotating coil in a magnetic field to produce alternating current. The e.m.f. is maximum when the coil is parallel to the field lines (cutting them at the highest rate).

📐Formulae

F=BILF = B I L

ϵ=−NΔΦΔt\epsilon = -N \frac{\Delta \Phi}{\Delta t}

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

IpVp=IsVsI_p V_p = I_s V_s

Efficiency=IsVsIpVp×100%\text{Efficiency} = \frac{I_s V_s}{I_p V_p} \times 100\%

Φ=BA\Phi = B A

💡Examples

Problem 1:

A step-down transformer is used to reduce the mains voltage from 240 V240\text{ V} to 12 V12\text{ V}. If the primary coil has 40004000 turns, calculate the number of turns required in the secondary coil.

Solution:

Using the transformer equation: VpVs=NpNs\frac{V_p}{V_s} = \frac{N_p}{N_s} Rearranging for NsN_s: Ns=Np×VsVpN_s = N_p \times \frac{V_s}{V_p} Ns=4000×12 V240 VN_s = 4000 \times \frac{12\text{ V}}{240\text{ V}} Ns=4000×0.05=200N_s = 4000 \times 0.05 = 200

Explanation:

The number of turns in the secondary coil must be 200200 to achieve the desired voltage reduction.

Problem 2:

A straight wire of length 0.5 m0.5\text{ m} carries a current of 2.0 A2.0\text{ A} at right angles to a uniform magnetic field of flux density 0.15 T0.15\text{ T}. Calculate the force acting on the wire.

Solution:

Using the formula for magnetic force: F=BILF = B I L F=0.15 T×2.0 A×0.5 mF = 0.15\text{ T} \times 2.0\text{ A} \times 0.5\text{ m} F=0.15 NF = 0.15\text{ N}

Explanation:

The force is calculated by multiplying the magnetic flux density, the current, and the length of the wire that is within the field.

Problem 3:

A coil with 5050 turns experiences a change in magnetic flux from 0.02 Wb0.02\text{ Wb} to 0.10 Wb0.10\text{ Wb} in a time interval of 0.4 s0.4\text{ s}. Determine the magnitude of the induced e.m.f.

Solution:

According to Faraday's Law: ϵ=NΔΦΔt\epsilon = N \frac{\Delta \Phi}{\Delta t} ΔΦ=0.10 Wb−0.02 Wb=0.08 Wb\Delta \Phi = 0.10\text{ Wb} - 0.02\text{ Wb} = 0.08\text{ Wb} ϵ=50×0.08 Wb0.4 s\epsilon = 50 \times \frac{0.08\text{ Wb}}{0.4\text{ s}} ϵ=50×0.2=10 V\epsilon = 50 \times 0.2 = 10\text{ V}

Explanation:

The induced e.m.f. is found by taking the product of the number of turns and the rate of change of magnetic flux.

Electromagnetic effects - Revision Notes & Key Formulas | IGCSE Grade 12 Physics