(a) Explain mutual induction
(b) State four use of electromagnets
The diagram above illustrates two coils X an arranged so that their axes are collinear. X is connected to an a.c. supply and has an amme in series with it while Y is connected to a lamp Explain the following observations. The (i) lamp is lit when the a.c. supply is switched on;
(ii) brightness of the light from the lamp increases when distance between X. and Y is decreased;
(iii) filament of the lamp glows brighter whet bundle of insulatttriron wires is placed along ccmmon axis of the coils.
(d) Two cells, one have an emf of 2.0 V and an internal resistance of 0.4 and the other having an emf of 2.0 V and an internal resistance of 0.1 \(\Omega\), are connected in parallel. The combination is then connected in series with a resistor.
(i) Draw a circuit diagram of the arrangement.
(ii) Calculate the current through 5\(\Omega\) resistor.
Explanation
(a) Mutual Induction: Mutual Induction is a phenomenon which occurs when the current in a coil is changing, an e.m.f will be induced in a tear-by circuit due to some of the magnetic flux produced by the first circuit linking the second circuit.
(b) The uses of electromagnets are in:
- electric bells or telephone earpieces
- generators or electric motors
- industrial robots
- separation of iron scraps from debris
- electromagnetic switches or magnetic relays
- cranes fbr lifting or transporting heavy magnetic materials
- magnetic resonance scanners, i.e. in medical diagnosis.
(i) The changing magnetic flux in X induces changing magnetic flux in Y which induces the current that light up the bulb.
(ii) As the distance between X and Y decreases, the magnetic flux density linking coil Y increases, hence the induced e.m.f. increase hence the brightners of the lamp.
(iii) The insulated soft iron wires increases the magnetic flux through coil Y, hence increase in the induced current in Y, therefore, the filament of the lamp glows brighter.
(d)
Total e,m,f. E = 2V
Total internal resistance r
\(\frac{1}{r} + \frac{1}{r_2} = \frac{1}{0.4} + \frac{1}{0.1}\)
e = 0.08\(\Omega\)
E = I(R + r)
2 = I(5 + 0.08)
I = \(\frac{2}{5.08}\) = 0.394A