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Electromagnetic Induction — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsElectromagnetic Induction5 Marks
Question
(a) Derive the expression for the magnetic energy stored in a evil.
(b) Derive the expression for the mutual inductance between two plane coils.
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Answer

(a) Magnetic Energy Stored in a coil: When a current is established in an inductor (coil) by connected, a source of emf across its ends then an emf is induced in it due to self induction. This emf opposses the rise of current through the coil. Hence work has to be done by the current to get its steady value against the induced emf.
The energy for this work is obtained from the source of emf connected across the coil and this energy is stored as magnetic energy in the coil.
Let dW be the work done in establishing a current I in the coil in time dt.
Then, work $dW = eIdt$, where e is the induced emf
$dW = [L(\frac{dI}{dt})]Idt = LIdI$ ($\because$ $e = L\frac{dI}{dt}$)
Total work done in maintaining a steady current $I_{0}$ in the coil is given by
$W = \int dW = \int_{0}^{I_{0}} LIdI = L[\frac{I^{2}}{2}]_{0}^{I_{0}} = \frac{1}{2}LI_{0}^{2}$
Magnetic Energy stored in the coil :
$U=W \Rightarrow U=\frac{1}{2}LI_{0}^{2}$
(b) Mutual Inductance between two plane coils:
In adjoining figure two plane coils C1 and C2 are kept coaxially near each other. Let N1 be the number of turns in primary and N2 in secondary coil. If I1 current is flowing in primary coil, the magnetic field at the centre of the coil is given by :
Image
$B _1=\frac{\mu_0 N_1 I _1}{2 r_1}$
where $r_1$ is the radius of the primary coil. The magnetic field lines are treating uniformerly into the secondary coil. Hence magnetic flux passing through each turn of secondary coil will be $\Phi_2= B _1 \times A _2$
where $A_2$ = Area of cross-section of secondary coil =
$\pi r_2^2$ in which $r_2$ is the radius of the secondary coil
$\therefore \quad \Phi_2=\left(\frac{\mu_0 N_1 I _1}{2 r_1}\right) \times \pi r_2^2$
or $\Phi_2=\left(\frac{\mu_0 \pi r_2^2}{2 r_1}\right) N _1 I _1$
But coefficient inductance between these two coils is given by
$M =\frac{ N _2 \Phi_2}{ I _1}$
$\therefore \quad M =\frac{ N _2\left(\frac{\mu_0 \pi r_2^2}{2 r_1}\right) \cdot N _1 I _1}{ I _1}$
$\Rightarrow \quad M =\left(\frac{\mu_0 \pi r_2^2}{ 2 r_1}\right) N _1 N _2$ ...(1)
If the two coils are placed concentrically as shown in adjoining figure and $I_1$ be the current passed through coil C1 then magnetic field at the centre of this coil will be given by :
Image
$B _1=\frac{\mu_0 N_1 I _1}{2 r_1}$
where, N1= Number of turns in this coil C1
r₁ = Radius of this coil C1
Since coil C2 is small, the magnetic field B1 may be considered uniform throughout the area of cross-section of C2.
∴ Magnetic flux linked with the coil $C _2$ is given by :
$\Phi_2= B _1 A_2=\frac{\mu_0 N_1 I _1}{2 r_1} \times \pi r_2^2$
$\Rightarrow \quad \Phi_2=\left(\frac{\mu_0 \pi r_2^2}{2 r_1}\right) N _1 I _1$
where, r2 = Radius of inner coil C2.
But coefficient of mutual induction between these two coils is given by :
$M =\frac{ N _2 \Phi_2}{ I _1}=\frac{ N _2\left(\frac{\mu_0 \pi r_2^2}{2 r_1}\right) N _1 I _1}{ I _1}$
$\Rightarrow \quad M =\left(\frac{\mu_0 \pi r_2^2}{ 2 r _1}\right) N _1 N _2$ ...(2)
where, N2 = Number of turns in the inner coil. Thus formula for M is the same as formula (1).
In this way formula for M in both the situation is the same.

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