Published by Patrick Mutisya · 14 days ago
State the factors that affect the magnitude of an induced e.m.f.
Faraday’s law of electromagnetic induction states that the induced e.m.f. (\$\mathcal{E}\$) in a coil is proportional to the rate of change of magnetic flux (\$\Phi\$) linking the coil:
\$\mathcal{E}= -N\frac{d\Phi}{dt}\$
where
The magnitude of the induced e.m.f. depends on how quickly the magnetic flux through the coil changes. The following factors influence this rate of change:
More turns increase the total change in flux, giving a larger e.m.f.
A stronger field produces a greater flux for a given area, so a change in \$B\$ yields a larger e.m.f.
Larger area intercepts more field lines; a change in area (e.g., by moving a rectangular loop) leads to a larger change in flux.
Faster changes give a greater \$\frac{d\Phi}{dt}\$ and thus a larger e.m.f.
When a straight conductor of length \$l\$ moves with velocity \$v\$ perpendicular to a uniform field \$B\$, the induced e.m.f. is \$\mathcal{E}=Blv\$. Faster motion or longer conductor increases the e.m.f.
Since \$\Phi = BA\cos\theta\$, changing \$\theta\$ changes the flux. Rotating the coil more rapidly (larger \$\frac{d\theta}{dt}\$) increases the induced e.m.f.
| Factor | Effect on Induced e.m.f. |
|---|---|
| Number of turns (\$N\$) | Proportional – double \$N\$, double \$\mathcal{E}\$ |
| Magnetic field strength (\$B\$) | Proportional – stronger \$B\$ gives larger \$\mathcal{E}\$ |
| Coil area (\$A\$) | Proportional – larger \$A\$ increases flux change |
| Rate of change of \$B\$ or \$A\$ | Directly proportional – faster change → larger \$\mathcal{E}\$ |
| Length of moving conductor (\$l\$) in a uniform field | Proportional – \$\mathcal{E}=Blv\$ |
| Velocity of motion (\$v\$) | Proportional – faster motion → larger \$\mathcal{E}\$ |
| Angle \$\theta\$ between \$B\$ and coil normal | Flux varies as \$\cos\theta\$; rapid change of \$\theta\$ increases \$\mathcal{E}\$ |
A rectangular coil of 20 turns has an area of \$0.04\ \text{m}^2\$ and is placed in a uniform magnetic field of \$0.5\ \text{T}\$. The coil is rotated from \$\theta = 0^\circ\$ to \$\theta = 90^\circ\$ in \$0.2\ \text{s}\$. Calculate the average induced e.m.f.
Solution:
\$\Phi_i = BA\cos0^\circ = (0.5)(0.04)(1)=0.02\ \text{Wb}\$
\$\Phi_f = BA\cos90^\circ = (0.5)(0.04)(0)=0\ \text{Wb}\$
\$\Delta\Phi = \Phif-\Phii = -0.02\ \text{Wb}\$
\$\mathcal{E}_{\text{avg}} = -N\frac{\Delta\Phi}{\Delta t}= -20\frac{-0.02}{0.2}=2\ \text{V}\$