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  1. >>Class 12
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  3. >>Electromagnetic Induction
Problem solving tips
2 min read

Electromagnetic Induction

- Want to score better marks in exams? Have a look at some problem solving tips.
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Tip 1: Finding mutual inductance of two inductors of different shapes. There is a general equality between mutual inductance of inductor 1 w.r.t inductor 2 () and that of inductor 2 w.r.t. inductor 1 ()
So for a given pair of inductors, we can either calculate or , whichever is easier to calculate. But how do we decide that? We follow the given step-by-step strategy.
  • Usually, one of the two inductors (say coil 1) is dimensionally much much smaller than the other inductor (say coil 2). In such cases, we can assume the magnetic field by the large inductor to be uniform throughout the area/ volume of the small inductor. Thus it becomes much easier to write the flux linkage for the small inductor due to the large inductor.
  • We write the flux linkage of coil 1 due to current in coil 2,
  • Following are some of the examples in which we will find calculating much easier than
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Here the inner solenoid is effectively immersed in a uniform magnetic field due to the outer solenoid. Thus, the calculation of would be easy. However, it would be extremely difficult to calculate the flux linkage with the outer solenoid as the magnetic field due to the inner solenoid would vary across the length as well as cross-section of the outer solenoid.
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Similarly, we can easily find here when as we have to find magnetic field at the centre of the outer square loop and consider it to be uniform throughout the area of the inner square loop.
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Same approach can be used for this situation where we can write easily by writing the magnetic field due to ring 1 on its axis.
Find an expression of mutual inductance for shown concentric co-planner circular and regular hexagonal loops
A rectangular loop of N closely packed turns is positioned near a long straight wire as shown in Figure . What is the mutual inductance M for the loopwire combination if N = 100, a = 1.0 cm, b = 8.0 cm, and l = 30 cm?
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Tip 2: Finding motional emf in loops or conductors rotating perpendicular to the plane of magnetic field.
For a conducting disc rotating in a plane perpendicular to the plane of magnetic field, the EMF is induced across its centre and its circumference.
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The magnetic flux through it might seem constant but EMF is still induced. The induced EMF is a result of "the rate of cutting lines of magnetic flux".
Now let's understand the breakdown of any complex problem of this nature.
For a conducting rod of length l moving with velocity v perpendicular to its length and in a plane perpendicular to the plane of magnetic field B, the motional EMF across its ends is given by

We can extend this result to a rotating rod where we have to consider an element of length dx at distance X from the centre which moves at right angles to the magnetic field. Motional EMF across it is given by:

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We can integrate this result from x= 0 to x= l
  • Now we can further use this result for a rotating conducting disc. The disc can be considered as a collection of closely packed rods rotating with .
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All the points on the circumference will be equipotential. So the emf between centre O and any point C on circumference is effectively a parallel combination of infinite batteries each of EMF . So the effective EMF will also be same as .
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  • In case of an annular disc rotating in a plane perpendicular to the magnetic field, we have to again use the elementary rod. But in that, we integrate the result from to
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A metal rod of resistance 20 is fixed along a diameter of conducting ring of radius and lies in x-y plane. There is a magnetic field   . The ring rotates with an angular velocity about its axis. An external resistance of 10 is connected across the centre of the ring and rim. The current through external resistance is

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B
C
D
zero