Week 8 · Chapter 3 Electromagnetic Induction (A): Faraday’s experiment and the induced emf, Faraday’s law emf = −N ΔΦ/Δt, Lenz’s rule, Fleming’s right-hand rule, mutual induction (M, the henry as V·s/A) and the two-coil experiments · 3 lessons
1
Faraday’s experiment and the law of induction
Physics Sec 3 Term 1, Chapter 3: pp. 60–67
GoalPredict when an emf is induced in a coil and calculate it from emf = −N ΔΦ ÷ Δt.
New words
induced emf · the emf that appears in a conductor or coil while the magnetic flux through it is changingmagnetic flux (Φ) · Φ = BA for a face at right angles to the field; unit weber (Wb), 1 Wb = 1 T·m²induced current · the current that the induced emf drives round a closed circuit
Explanation
After Oersted showed that a current makes a magnetic field, Faraday asked the opposite question: can a magnetic field make a current? In his experiment (a prediction task here, no equipment needed), an insulated coil with no battery is joined to a very sensitive galvanometer with its zero in the middle. While a magnet is pushed into the coil the pointer deflects; when the magnet stops, even inside the coil, the pointer returns to zero; when the magnet is pulled out the pointer deflects the other way. So an emf is induced only while the flux through the coil changes, not while it is large and steady. The emf grows with the rate of change of flux and with the number of turns N: emf = −N ΔΦ ÷ Δt, where ΔΦ = Φ(final) − Φ(initial) is measured along a positive direction of flux that you choose. The size is N|ΔΦ| ÷ Δt; the minus sign is explained by Lenz’s rule in the next lesson.
What happens to the magnet
Galvanometer
Why
pushed into the coil
deflects one way while it moves
flux through the coil is rising
at rest in or near the coil
zero
flux is not changing
pulled out of the coil
deflects the other way
flux is falling
moved faster
larger deflection
ΔΦ ÷ Δt is larger
Worked examples
A coil of N = 150 turns has the flux through it rise from 2.0 × 10⁻³ Wb to 8.0 × 10⁻³ Wb in 0.03 s (positive direction: the direction of the starting flux). ΔΦ = 6.0 × 10⁻³ Wb, so emf = −150 × 6.0 × 10⁻³ ÷ 0.03 = −30 V. The size is 30 V.
A coil of N = 80 turns and area 15 cm × 10 cm = 0.015 m² faces a field that falls from 0.80 T to 0.20 T in 0.06 s (positive direction: the direction of the field). ΔΦ = (0.20 − 0.80) × 0.015 = −9 × 10⁻³ Wb, so emf = −80 × (-9 × 10⁻³) ÷ 0.06 = +12 V. The size is 12 V.
A coil of N = 10 turns, area 0.03 m², is in a field of 0.4 T (Φ = 0.012 Wb) and is pulled out of the field. If this takes 0.5 s the size of the emf is 10 × 0.012 ÷ 0.5 = 0.24 V; if it takes 0.05 s it is 10 × 0.012 ÷ 0.05 = 2.4 V. Ten times faster gives ten times the emf.
Class activity
Predict and explain, on paper. A coil is joined to a zero-centre galvanometer. For each of these, say whether the pointer moves and why: (1) a magnet is held still inside the coil; (2) the magnet is dropped through the coil; (3) the coil is moved towards the still magnet instead; (4) the coil is turned so that its face is no longer at right angles to the field of a still magnet. A real coil, magnet and galvanometer belong in a school lab with a teacher.
Practice
1
A coil of 200 turns has the flux through it grow from 0 to 3.0 × 10⁻³ Wb in 0.05 s. Find the size of the induced emf.
The flux through a coil of 75 turns changes from +5.0 × 10⁻³ Wb to −3.0 × 10⁻³ Wb in 0.04 s (positive direction: the direction of the starting flux). Find emf = −N ΔΦ ÷ Δt with its sign.
