(a) State the laws of electromagnetic induction.
(b) (i) Describe a simple experiment to show how an induced e.m.f, can be produced; (ii) State two factors on which the magnitude of the induced e.m.f. depends
(c) Explain what is meant by the r.m.s. value of an alternating current
(d) (i) If the alternating current is represented by \(I = l_{o} \sin \omega t\), state what the symbol \(I, I_{o}, \omega\) and \(\omega t\)represent.
(ii) Calculate the instantaneous value of such a current, if in a circuit it has r.m.s value of 15.0A when its phase angle is 30°.
Explanation
(a) The laws of electromagnetic induction states that
(1) whenever there is a change in the magnetic flux linked with a circuit, an e.m.f. is induced, the strength of which is proportional to the rate of change of the flux linked with the circuit. The above law is sometimes called faraday's law of electromagnetic induction.
(2) The direction of the induced current is always such as to oppose the change producing it. This is called Lenz's law.
b(i) Experiment to show how an induced e.m.f. can be produced.
A coil with many turns is connected to centre-scale zero galvanometer G. As the bar magnet is moved rapidly close and through the turns, an induced e.m.f. is produced. The galvanometer pointer is deflected showing a flow of current. The same effects could be observed if the coil is moving towards the magnet instead. Or as long as there is relative motion between the two.
(ii) The magnitude of the induced e.m.f. depends on:
(1) The magnitude of flux linking the coil. (2) The relative speed between the magnet and the coil.
(c) The root-mean-square (r.m.s.) value of an alternating current is the steady or direct current which produces the same heating effect per secoridin a given resistor. Its relationship with the peak value of alternating current \(I_{o}\) is given by
\(I_{r.m.s} = \frac{1}{\sqrt{2}} I_{o}\)
d(i) \(I\) is the instantaneous current
\(I_{o}\) is the peak value of the current
\(\alpha\) is the angular frequency
\(\alpha t\) is the phase or the current (since it is wave-like).
(ii) \(I = I_{o} \sin \alpha t\)
= \(I_{r.m.s} \sqrt{2} \sin 30°\)
= \(\sqrt{2} \times 15 \times \frac{1}{2} = 10.6A\)