Motional E.m.f. | CAIE 9702 Paper 4 Notes

Paper 4

Physics9702

Motional E.m.f.

A straight conductor moving across a magnetic field sweeps out area. Faraday's law turns the swept flux per second into an e.m.f.

Derive the moving-wire equation

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In one second the wire sweeps area Lv. The flux cut each second is therefore BLv, so the induced e.m.f. has magnitude BLv when B, L and v are mutually perpendicular.

In time $\Delta t$ , the wire moves distance $v\Delta t$ .
A wire of length $L$ therefore sweeps area:
$\Delta A=Lv\Delta t$
The flux cut is $B\Delta A$ . Divide by time to obtain the relation below.

|\mathcal E|=BLv

Symbols

$B$ = magnetic flux density (T)
$L$ = wire length inside the field (m)
$v$ = speed perpendicular to the field (m s⁻¹)

The wire, velocity and magnetic field must be mutually perpendicular for this form.
Charge separation gives an e.m.f. even if the circuit is open. A current flows only after a complete path is connected.
Aircraft wings can form the moving conductor; a recent question linked e.m.f., swept area, wingspan and speed (9702/41/O/N/25 Q7(b)(i)–(iii)).

Your turncalculation check

A 0.25 m wire moves at 3.0 m s⁻¹ across a 0.80 T field. The wire, motion and field are mutually perpendicular. Calculate the induced e.m.f.

Show worked answer

|\mathcal E|=(0.80)(0.25)(3.0)

Answer

|\mathcal E|=0.60\ \mathrm{V}