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Fields - Electromagnetic Induction

Grade 11IBPhysics

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

🔑Concepts

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Magnetic Flux (Φ\Phi): Defined as the product of the magnetic field BB and the area AA through which the field lines pass, given by Φ=BAcos⁡θ\Phi = BA \cos \theta, measured in Webers (WbWb).

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Magnetic Flux Linkage: For a coil with NN turns, the total flux linkage is NΦN\Phi.

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Faraday’s Law: The magnitude of the induced electromotive force (emf) ϵ\epsilon is proportional to the rate of change of magnetic flux linkage, expressed as ϵ=NΔΦΔt\epsilon = N \frac{\Delta \Phi}{\Delta t}.

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Lenz’s Law: The direction of the induced emf is such that it will oppose the change in magnetic flux that created it. This is a consequence of the law of conservation of energy.

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Motional emf: When a straight conductor of length LL moves with a velocity vv perpendicular to a uniform magnetic field BB, the induced emf is ϵ=BvL\epsilon = BvL.

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Alternating Current (AC): In a rotating coil within a magnetic field, the induced emf varies sinusoidally as ϵ=ϵ0sin⁡(ωt)\epsilon = \epsilon_0 \sin(\omega t), where ω\omega is the angular frequency.

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Root Mean Square (rms): For AC circuits, IrmsI_{rms} and VrmsV_{rms} represent the equivalent DC values that would dissipate the same power: Irms=I02I_{rms} = \frac{I_0}{\sqrt{2}} and Vrms=V02V_{rms} = \frac{V_0}{\sqrt{2}}.

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Transformers: Devices that change the voltage of an alternating current via mutual induction. For an ideal transformer, VpVs=NpNs=IsIp\frac{V_p}{V_s} = \frac{N_p}{N_s} = \frac{I_s}{I_p}.

📐Formulae

Φ=BAcos⁡θ\Phi = BA \cos \theta

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

ϵ=BvL\epsilon = BvL

ϵ=NBAωsin⁡(ωt)\epsilon = NBA\omega \sin(\omega t)

Vrms=V02V_{rms} = \frac{V_0}{\sqrt{2}}

Paverage=IrmsVrmsP_{average} = I_{rms}V_{rms}

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

💡Examples

Problem 1:

A circular coil of N=200N = 200 turns and radius r=0.05 mr = 0.05\text{ m} is placed in a magnetic field that decreases from 0.8 T0.8\text{ T} to 0.2 T0.2\text{ T} in a time interval of 0.1 s0.1\text{ s}. The field is perpendicular to the plane of the coil. Calculate the average induced emf.

Solution:

First, calculate the area: A=πr2=π(0.05)2≈7.85×10−3 m2A = \pi r^2 = \pi (0.05)^2 \approx 7.85 \times 10^{-3}\text{ m}^2. The change in flux is ΔΦ=A(Bfinal−Binitial)=7.85×10−3×(0.2−0.8)=−4.71×10−3 Wb\Delta \Phi = A(B_{final} - B_{initial}) = 7.85 \times 10^{-3} \times (0.2 - 0.8) = -4.71 \times 10^{-3}\text{ Wb}. Using Faraday's Law: ϵ=−NΔΦΔt=−200×−4.71×10−30.1=9.42 V\epsilon = -N \frac{\Delta \Phi}{\Delta t} = -200 \times \frac{-4.71 \times 10^{-3}}{0.1} = 9.42\text{ V}.

Explanation:

The negative sign in Lenz's law indicates opposition, but the magnitude of the induced emf is determined by the rate of change of flux linkage over time.

Problem 2:

An ideal transformer has 500500 turns on the primary coil and 2525 turns on the secondary coil. If the primary voltage is 240 V240\text{ V} (rms), find the secondary voltage.

Solution:

Using the transformer equation: VpVs=NpNs  ⟹  Vs=Vp(NsNp)\frac{V_p}{V_s} = \frac{N_p}{N_s} \implies V_s = V_p \left( \frac{N_s}{N_p} \right). Substituting the values: Vs=240×(25500)=240×0.05=12 VV_s = 240 \times \left( \frac{25}{500} \right) = 240 \times 0.05 = 12\text{ V}.

Explanation:

This is a step-down transformer because the number of turns in the secondary coil is less than in the primary coil, resulting in a lower output voltage.

Electromagnetic Induction - Revision Notes & Key Formulas | IB Grade 11 Physics