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Fields - Electromagnetic Induction

Grade 12IBPhysics

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 cross-sectional area (AA) perpendicular to the field, measured in Webers (WbWb). Formally, Φ=BAcos⁡θ\Phi = BA \cos \theta, where θ\theta is the angle between the magnetic field and the normal to the area.

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

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Faraday’s Law: The magnitude of the induced electromotive force (emf, ϵ\epsilon) is directly proportional to the rate of change of magnetic flux linkage.

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Lenz’s Law: The direction of the induced current (and emf) is such that it creates a magnetic field that opposes the change in magnetic flux that produced it. This is a statement 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 magnetic field BB, the induced emf is given by ϵ=BvL\epsilon = BvL.

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Alternating Current (AC): An electric current that periodically reverses direction. In IB Physics, we use root mean square (rmsrms) values for voltage (VrmsV_{rms}) and current (IrmsI_{rms}) to represent the equivalent DC values that would dissipate the same power.

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Transformers: Devices used to change the voltage of an alternating current. They consist of a primary and secondary coil. An ideal transformer assumes 100%100\% efficiency, where Pin=PoutP_{in} = P_{out}.

📐Formulae

Φ=BAcos⁡θ\Phi = BA \cos \theta

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

ϵ=BvL\epsilon = BvL

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

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

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

Pavg=IrmsVrmsP_{avg} = I_{rms} V_{rms}

💡Examples

Problem 1:

A circular coil with 250250 turns and a radius of 0.050.05 m is placed in a uniform magnetic field of 0.400.40 T. The coil is rotated from a position where its plane is perpendicular to the field to a position where its plane is parallel to the field in 0.200.20 s. Calculate the average induced emf.

Solution:

  1. Initial flux: Φi=BA=0.40×(π×0.052)≈0.00314\Phi_i = BA = 0.40 \times (\pi \times 0.05^2) \approx 0.00314 Wb.
  2. Final flux: Φf=0\Phi_f = 0 (since the plane is parallel to the field, θ=90∘\theta = 90^\circ).
  3. Change in flux: ΔΦ=0−0.00314=−0.00314\Delta \Phi = 0 - 0.00314 = -0.00314 Wb.
  4. Using Faraday's Law: ϵ=−NΔΦΔt=−250×−0.003140.20=3.925\epsilon = -N \frac{\Delta \Phi}{\Delta t} = -250 \times \frac{-0.00314}{0.20} = 3.925 V.

Explanation:

The induced emf is determined by the rate at which the magnetic flux linkage changes. By rotating the coil, the effective area perpendicular to the field changes from maximum to zero.

Problem 2:

An ideal transformer has 500500 turns on the primary coil and 2525 turns on the secondary coil. If the primary voltage is 240240 V AC, calculate the secondary voltage and the current in the secondary if the primary current is 0.100.10 A.

Solution:

  1. Secondary Voltage: VpVs=NpNs  ⟹  Vs=VpNsNp=240×25500=12\frac{V_p}{V_s} = \frac{N_p}{N_s} \implies V_s = V_p \frac{N_s}{N_p} = 240 \times \frac{25}{500} = 12 V.
  2. Secondary Current: IsIp=NpNs  ⟹  Is=IpNpNs=0.10×50025=2.0\frac{I_s}{I_p} = \frac{N_p}{N_s} \implies I_s = I_p \frac{N_p}{N_s} = 0.10 \times \frac{500}{25} = 2.0 A.

Explanation:

This is a step-down transformer. Because it is ideal, power is conserved (Pp=PsP_p = P_s), meaning the decrease in voltage leads to a proportional increase in current.

Electromagnetic Induction - Revision Notes & Key Formulas | IB Grade 12 Physics