Faraday's Law | CAIE 9702 Paper 4 Notes

Chapters

Paper 4

Physics9702

Faraday's Law

Faraday's law connects the size of an induced e.m.f. to how quickly the magnetic flux linkage changes.

Change per second sets the e.m.f.

Definition

2 marks

What is meant by Faraday's law of electromagnetic induction?

Model answer: The induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage.

Mark schemes repeatedly award one mark for “directly proportional to the rate” and one for “change of magnetic flux linkage” (9702/42/M/J/24 Q7(a)) (9702/42/F/M/25 Q7(a)) (9702/41/O/N/25 Q7(a)).

|\mathcal E|=\left|\dfrac{\Delta(N\Phi)}{\Delta t}\right|

Symbols

$|\mathcal E|$ = magnitude of the induced e.m.f. (V)
$\Delta(N\Phi)$ = change in magnetic flux linkage (Wb)
$\Delta t$ = time taken (s)

The same change completed in less time gives a larger e.m.f.
Constant flux linkage has zero rate of change, even if its value is large.
This magnitude form gives size. Lenz's law later gives direction or sign.

Calculate an average induced e.m.f.

Worked example

Field rises at a steady rate

A 3000-turn coil of area $3.4\times10^{-4}\ \mathrm{m^2}$ faces a field. The field rises from zero to 4.0 μT in 0.020 s (9702/42/O/N/23 Q7(b)(i)).

Find the change in flux linkage.
$\Delta(N\Phi)=NA\Delta B$
$\Delta(N\Phi)=4.1\times10^{-6}\ \mathrm{Wb}$
Divide by the time interval.
$|\mathcal E|=\dfrac{4.1\times10^{-6}}{0.020}$

Answer

|\mathcal E|=2.0\times10^{-4}\ \mathrm{V}

Your turnnumerical check

The flux through each turn of a 250-turn coil changes by 1.2 × 10⁻⁴ Wb in 0.040 s. Calculate the average induced e.m.f.

Show worked answer

\Delta(N\Phi)=3.0\times10^{-2}\ \mathrm{Wb}

|\mathcal E|=\dfrac{3.0\times10^{-2}}{0.040}

Answer

|\mathcal E|=0.75\ \mathrm{V}

Faraday's Law

Faraday's law connects the size of an induced e.m.f. to how quickly the magnetic flux linkage changes.

Change per second sets the e.m.f.

Definition

2 marks

What is meant by Faraday's law of electromagnetic induction?

Model answer: The induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage.

|\mathcal E|=\left|\dfrac{\Delta(N\Phi)}{\Delta t}\right|

Symbols

$|\mathcal E|$ = magnitude of the induced e.m.f. (V)
$\Delta(N\Phi)$ = change in magnetic flux linkage (Wb)
$\Delta t$ = time taken (s)

The same change completed in less time gives a larger e.m.f.
Constant flux linkage has zero rate of change, even if its value is large.
This magnitude form gives size. Lenz's law later gives direction or sign.

Calculate an average induced e.m.f.

Worked example

Field rises at a steady rate

A 3000-turn coil of area $3.4\times10^{-4}\ \mathrm{m^2}$ faces a field. The field rises from zero to 4.0 μT in 0.020 s (9702/42/O/N/23 Q7(b)(i)).

Find the change in flux linkage.
$\Delta(N\Phi)=NA\Delta B$
$\Delta(N\Phi)=4.1\times10^{-6}\ \mathrm{Wb}$
Divide by the time interval.
$|\mathcal E|=\dfrac{4.1\times10^{-6}}{0.020}$

Answer

|\mathcal E|=2.0\times10^{-4}\ \mathrm{V}

Your turnnumerical check

The flux through each turn of a 250-turn coil changes by 1.2 × 10⁻⁴ Wb in 0.040 s. Calculate the average induced e.m.f.

Show worked answer

\Delta(N\Phi)=3.0\times10^{-2}\ \mathrm{Wb}

|\mathcal E|=\dfrac{3.0\times10^{-2}}{0.040}

Answer

|\mathcal E|=0.75\ \mathrm{V}