recall and use Φ = BA

Electromagnetic Induction – Magnetic Flux (Φ = B A cos θ)

1. Magnetic Flux

• Definition: The total magnetic field that passes through a surface of area A.

  \[

  \Phi = B A \cos\theta

  \]

  where

  • \(B\) = magnetic field strength (tesla, T)
  • \(A\) = area of the surface (m²)
  • \(\theta\) = angle between the field direction and the normal (perpendicular) to the surface.

• Units: 1 weber (Wb) = 1 T·m².
• Key point for the Cambridge syllabus: The expression \(\Phi = B A \cos\theta\) and the unit weber must be memorised.

2. Flux Linkage (N Φ)

• When a coil has N identical turns, the total flux linked with the coil is the product of the number of turns and the flux through one turn:

  \[

  N\Phi = N \times \Phi

  \]

  This quantity is called flux linkage. Some textbooks denote it by the symbol \(\Psi\); however, the Cambridge syllabus uses the notation N Φ. Both represent the same physical quantity (weber‑turns).

• Flux linkage appears directly in Faraday’s law for a multi‑turn coil.

3. Faraday’s Law of Electromagnetic Induction

The induced electromotive force (emf) in a coil is proportional to the rate of change of its flux linkage:

\[

\varepsilon = -\,\frac{d(N\Phi)}{dt}= -\,N\frac{d\Phi}{dt}

\]

• The minus sign embodies Lenz’s law: the induced emf always opposes the change that produces it.
• In most calculations we use the magnitude:

  \[

  |\varepsilon| = N\left|\frac{d\Phi}{dt}\right|

  \]

4. Lenz’s Law – Direction of the Induced emf / Current

• When the magnetic flux through a circuit changes, the induced current creates a magnetic field that opposes that change.
• Right‑hand rule for a coil: Curl the fingers in the direction of the induced current; the thumb points in the direction of the induced magnetic field.
• Examples:

• If the external field is increasing into the page, the induced field points out of the page, so the current flows anticlockwise.
• If the external field is decreasing, the induced field points into the page, giving a clockwise current.

5. Factors that Influence the Magnitude of the Induced emf

6. Practical Demonstrations

• Moving magnet through a coil: Slide a bar magnet into a solenoid connected to a galvanometer. The needle deflects while the magnet moves and returns to zero when the magnet stops.
• Rotating rectangular loop (hand‑crank generator): Rotate a coil of area A in a uniform field B. The induced emf varies sinusoidally; the peak value is

  \[

  \varepsilon_{\text{max}} = N B A \omega

  \]

  where \(\omega\) is the angular speed (rad s⁻¹).

• Simple AC generator with an LED: Connect a rotating coil to an LED. Each half‑turn the LED flashes, illustrating the alternating nature of the induced emf.

7. Application to Alternating Current (AC) Generation

For a coil of area A rotating with angular speed ω in a uniform magnetic field B, the instantaneous flux is

\[

\Phi(t)=B A \cos(\omega t)

\]

The induced emf (magnitude) is then

\[

\varepsilon(t)=N\frac{d\Phi}{dt}=N B A \omega \sin(\omega t)

\]

• The emf varies sinusoidally – the fundamental principle of AC generators.
• Peak (maximum) emf: \(\varepsilon_{\text{peak}} = N B A \omega\).

8. Worked Example – Flux Through a Square Loop

Problem: Calculate the flux through a square loop of side 0.10 m placed in a uniform magnetic field of 0.50 T that is perpendicular to the plane of the loop.

1. Area: \(A = (0.10\ \text{m})^{2}=0.010\ \text{m}^{2}\).
2. Because the field is perpendicular, \(\theta =0^{\circ}\) and \(\cos\theta =1\).
3. Flux: \(\displaystyle \Phi = B A = 0.50\ \text{T}\times0.010\ \text{m}^{2}=5.0\times10^{-3}\ \text{Wb}\).

9. Practice Problems (Cambridge‑style)

1. A rectangular coil \(0.20\ \text{m} \times 0.30\ \text{m}\) is in a magnetic field of 0.80 T that makes an angle of 30° with the normal. The coil has 50 turns.

   • (i) Find the flux linkage \(N\Phi\).
   • (ii) If the coil is rotated so that the angle changes from 30° to 90° in 0.10 s, calculate the average induced emf.

2. What is the flux through a circular loop of radius 0.05 m in a field of 0.25 T perpendicular to the loop?
3. If the flux through a single‑turn coil is 0.02 Wb and the area is 0.04 m², determine the magnitude of the magnetic field.
4. A hand‑crank generator has a coil of area 0.015 m², 200 turns, and rotates at 300 rev min⁻¹ in a 0.30 T field. Find the peak emf produced.

10. Summary

• Magnetic flux: \(\displaystyle \Phi = B A \cos\theta\) (Wb).
• Flux linkage (Cambridge notation): \(\displaystyle N\Phi\) (weber‑turns). Some texts use \(\Psi\); both are equivalent.
• Faraday’s law: \(\displaystyle \varepsilon = -N\,\frac{d\Phi}{dt}\). The negative sign reflects Lenz’s law.
• The magnitude of the induced emf depends on N, A, B, the rate of change of the field (or rotation speed), and the orientation (θ).
• Practical set‑ups (moving magnet, rotating coil) demonstrate these principles and form the basis of generators and transformers.