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Fields - Induction (HL)

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 strength BB and the area AA through which the field lines pass, mathematically given by Φ=BAcos⁡θ\Phi = BA \cos \theta, where θ\theta is the angle between the magnetic field and the normal to the area. It is measured in Webers (WbWb).

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Magnetic Flux Linkage (NΦN\Phi): For a coil with NN turns, the flux linkage is the product of the number of turns and the magnetic flux through each turn.

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Faraday’s Law: The magnitude of the induced electromotive force (EMF) ϵ\epsilon is equal to the rate of change of magnetic flux linkage through the circuit.

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Lenz’s Law: The direction of the induced EMF (and the resulting induced current) is such that it opposes the change in magnetic flux that produced 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 velocity vv perpendicular to a uniform magnetic field BB, an EMF is induced across its ends given by ϵ=BvL\epsilon = BvL.

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Alternating Current (AC): Induced by a coil rotating at a constant angular frequency ω\omega in a uniform magnetic field. The output voltage is sinusoidal: ϵ=ϵ0sin⁡(ωt)\epsilon = \epsilon_0 \sin(\omega t).

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Root Mean Square (RMS): Since AC varies over time, RMS values represent the equivalent DC value that would dissipate the same power in a resistor. Vrms=V02V_{rms} = \frac{V_0}{\sqrt{2}} and Irms=I02I_{rms} = \frac{I_0}{\sqrt{2}}.

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Transformers: Devices that increase or decrease AC voltage through mutual induction. An ideal transformer follows the relation VpVs=NpNs\frac{V_p}{V_s} = \frac{N_p}{N_s}, where pp and ss denote primary and secondary coils.

📐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)

Irms=I02I_{rms} = \frac{I_0}{\sqrt{2}}

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

Pavg=IrmsVrms=12I0V0P_{avg} = I_{rms} V_{rms} = \frac{1}{2} I_0 V_0

VpVs=NpNs=IsIp\frac{V_p}{V_s} = \frac{N_p}{N_s} = \frac{I_s}{I_p}

💡Examples

Problem 1:

A square coil of side length 0.10 m0.10 \text{ m} has 50 turns50 \text{ turns}. It is placed in a uniform magnetic field of 0.20 T0.20 \text{ T} such that the plane of the coil is perpendicular to the field. The magnetic field is reduced to zero in 0.50 s0.50 \text{ s}. Calculate the average induced EMF in the coil.

Solution:

Initial Flux Φi=B×A=0.20×(0.10)2=0.002 Wb\text{Initial Flux } \Phi_i = B \times A = 0.20 \times (0.10)^2 = 0.002 \text{ Wb} Final Flux Φf=0 Wb\text{Final Flux } \Phi_f = 0 \text{ Wb} Change in flux linkage Δ(NΦ)=N(Φf−Φi)=50×(0−0.002)=−0.10 Wb-turns\text{Change in flux linkage } \Delta(N\Phi) = N(\Phi_f - \Phi_i) = 50 \times (0 - 0.002) = -0.10 \text{ Wb-turns} ϵ=−Δ(NΦ)Δt=−−0.100.50=0.20 V\epsilon = -\frac{\Delta(N\Phi)}{\Delta t} = -\frac{-0.10}{0.50} = 0.20 \text{ V}

Explanation:

We first calculate the area and the initial magnetic flux. Since the field is perpendicular to the plane, θ=0∘\theta = 0^{\circ} and cos⁡0=1\cos 0 = 1. We then apply Faraday's law using the change in flux linkage over the given time interval.

Problem 2:

An AC generator produces a peak voltage of 311 V311 \text{ V}. Calculate the RMS voltage and the average power dissipated if this generator is connected to a 100 Ω100 \text{ } \Omega resistor.

Solution:

Vrms=V02=3111.414≈220 VV_{rms} = \frac{V_0}{\sqrt{2}} = \frac{311}{1.414} \approx 220 \text{ V} Pavg=Vrms2R=2202100=48400100=484 WP_{avg} = \frac{V_{rms}^2}{R} = \frac{220^2}{100} = \frac{48400}{100} = 484 \text{ W}

Explanation:

The RMS voltage is found by dividing the peak voltage by 2\sqrt{2}. Average power in an AC circuit is calculated using RMS values, similar to the DC power formula P=V2RP = \frac{V^2}{R}.