Electricity & Magnetism

Magnetism & Electromagnetism

5th Year · 6th Year (Leaving Cert)

  • By the end of this lesson students will be able to define magnetic flux density and state its unit.
  • By the end of this lesson students will be able to calculate the force on a current-carrying conductor in a magnetic field.
  • By the end of this lesson students will be able to explain electromagnetic induction and state Faraday's and Lenz's laws.
  • By the end of this lesson students will be able to solve problems involving induced e.m.f. using Faraday's law.
  • By the end of this lesson students will be able to apply the transformer equation to ideal transformers (HL).

Key concepts

Magnetic Flux Density (B)

Magnetic flux density is a measure of the strength of a magnetic field. It is a vector quantity. It is defined as the force per unit current per unit length on a conductor placed perpendicular to the field. The greater the magnetic flux density, the stronger the magnetic field.

B = F / (Il) (when θ = 90°)
Magnetic Flux (Φ)

Magnetic flux is the total number of magnetic field lines passing normally through a given area. It quantifies the amount of magnetic field passing through a surface.

Φ = BA cos θ
Force on a Conductor (F = BIl sin θ)

A current-carrying conductor placed in a magnetic field experiences a force. The magnitude of this force depends on the magnetic flux density (B), the current (I), the length of the conductor within the field (l), and the angle (θ) between the direction of the current and the magnetic field. The direction of the force is given by Fleming's Left-Hand Rule.

F = BIl sin θ
Electromagnetic Induction

Electromagnetic induction is the process of generating an electromotive force (e.m.f.) and, consequently, an electric current in a conductor by changing the magnetic flux through it. This change can be achieved by moving a magnet near a coil, moving a coil in a magnetic field, or changing the current in a nearby coil.

Faraday's Law of Electromagnetic Induction

Faraday's Law states that the magnitude of the induced e.m.f. is directly proportional to the rate of change of magnetic flux through the circuit. The negative sign in the formula indicates the direction of the induced e.m.f. as described by Lenz's Law.

E = -N (ΔΦ/Δt)
Lenz's Law

Lenz's Law states that the direction of the induced current (or e.m.f.) is always such as to oppose the change in magnetic flux that produced it. This law is a direct consequence of the principle of conservation of energy.

Transformer Equation (HL)

A transformer is a device that changes a high alternating voltage to a low alternating voltage, or vice versa, using the principle of mutual induction. For an ideal transformer (one with no energy losses), the ratio of voltages is equal to the ratio of the number of turns in the coils, and inversely proportional to the ratio of currents.

Vp/Vs = Np/Ns = Is/Ip

Key facts to remember

  • 1Magnetic flux density (B) is measured in Tesla (T).
  • 2Magnetic flux (Φ) is measured in Weber (Wb).
  • 3Fleming's Left-Hand Rule determines the direction of the force on a current-carrying conductor in a magnetic field.
  • 4Electromagnetic induction is the generation of an e.m.f. by a changing magnetic flux.
  • 5Faraday's Law quantifies the induced e.m.f.: E = -N (ΔΦ/Δt).
  • 6Lenz's Law states that the induced current opposes the change in magnetic flux that produced it, consistent with energy conservation.
  • 7Transformers operate on the principle of mutual induction and only work with alternating current (a.c.) (HL).
  • 8For an ideal step-up transformer, the secondary voltage is higher than the primary voltage, and the secondary current is lower than the primary current (HL).

Worked examples

Example 1

A straight conductor of length 25 cm carrying a current of 4.0 A is placed in a uniform magnetic field of flux density 0.60 T. Calculate the force on the conductor when it is (a) perpendicular to the field and (b) makes an angle of 30° with the field.

IGiven: l = 25 cm = 0.25 m, I = 4.0 A, B = 0.60 T.
II(a) When perpendicular, θ = 90°. sin 90° = 1.
IIIF = BIl sin θ = (0.60 T)(4.0 A)(0.25 m)(sin 90°)
IVF = 0.60 × 4.0 × 0.25 × 1 = 0.60 N.
V(b) When it makes an angle of 30° with the field, θ = 30°.
VIF = BIl sin θ = (0.60 T)(4.0 A)(0.25 m)(sin 30°)
VIIF = 0.60 × 4.0 × 0.25 × 0.5 = 0.30 N.

Answer

(a) 0.60 N, (b) 0.30 N

Remember to convert length from cm to m.

Example 2

A coil of 200 turns has a cross-sectional area of 0.03 m². It is placed in a uniform magnetic field of 0.4 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 0.2 s. Calculate the average induced e.m.f.

IGiven: N = 200, A = 0.03 m², B = 0.4 T, Δt = 0.2 s.
IIInitial position: Plane perpendicular to field means the normal to the plane is parallel to the field. So, θ₁ = 0°.
IIIInitial magnetic flux, Φ₁ = BA cos θ₁ = (0.4 T)(0.03 m²) cos 0° = 0.012 Wb.
IVFinal position: Plane parallel to field means the normal to the plane is perpendicular to the field. So, θ₂ = 90°.
VFinal magnetic flux, Φ₂ = BA cos θ₂ = (0.4 T)(0.03 m²) cos 90° = 0 Wb.
VIChange in magnetic flux, ΔΦ = Φ₂ - Φ₁ = 0 - 0.012 Wb = -0.012 Wb.
VIIApply Faraday's Law: E = -N (ΔΦ/Δt).
VIIIE = -200 (-0.012 Wb / 0.2 s) = -200 (-0.06 V) = 12 V.

Answer

12 V

The negative sign in Faraday's Law accounts for Lenz's Law, but when asked for magnitude, the absolute value is usually given.

Example 3

An ideal transformer (HL) has 300 turns in its primary coil and 1500 turns in its secondary coil. If the primary coil is connected to a 230 V a.c. supply, calculate the voltage across the secondary coil. If the current in the primary coil is 0.6 A, calculate the current in the secondary coil.

IGiven: Np = 300, Ns = 1500, Vp = 230 V, Ip = 0.6 A.
IITo find Vs, use the voltage-turns ratio: Vp/Vs = Np/Ns.
III230 V / Vs = 300 / 1500
IV230 V / Vs = 1 / 5
VVs = 230 V × 5 = 1150 V.
VITo find Is, use the current-turns ratio: Np/Ns = Is/Ip.
VII300 / 1500 = Is / 0.6 A
VIII1 / 5 = Is / 0.6 A
9Is = 0.6 A / 5 = 0.12 A.

Answer

Secondary voltage (Vs) = 1150 V, Secondary current (Is) = 0.12 A

This is a step-up transformer, so voltage increases and current decreases in the secondary coil.

Common mistakes

  • Confusing magnetic flux (Φ) with magnetic flux density (B) or their units.
  • Incorrectly applying Fleming's Left-Hand Rule, leading to errors in force direction.
  • Forgetting to convert units (e.g., cm to m, milliseconds to seconds) before performing calculations.
  • Not understanding the significance of the negative sign in Faraday's Law (which represents Lenz's Law).
  • Assuming transformers can operate with direct current (d.c.) instead of alternating current (a.c.) (HL).
  • Mixing up primary and secondary values in the transformer equations, especially for current (HL).

Exam tips

  • Always draw clear diagrams to visualise the relative directions of magnetic fields, current, and force when using Fleming's Left-Hand Rule.
  • Ensure all quantities are in SI units (metres, seconds, amperes, teslas, webers) before substituting them into formulae.
  • When explaining electromagnetic induction, clearly state both Faraday's Law and Lenz's Law, explaining their relationship.
  • For transformer problems, carefully identify primary and secondary values and use the correct ratios for voltage, turns, and current (HL).

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