electromagnetic-induction Question 33
Question: Q. 6. Define mutual inductance of a pair of coils and write on which factors does it depend.
A square loop of side $20 \mathrm{~cm}$ is initially kept $30 \mathrm{~cm}$ away from a region of uniform magnetic field of $0.1 \mathrm{~T}$ as shown in the figure. It is then moved towards the right with a velocity of $10 \mathrm{~cm}$ $\mathrm{s}^{-1}$ till it goes out of the field.
Plot a graph showing the variation of
(i) magnetic flux $(\phi)$ through the loop with time $(\tau)$.
(ii) induced emf $(\varepsilon)$ in the loop with time $t$.
(iii) induced current in the loop if it has resistance of $0.1 \Omega$.
R A [O.D. I, II, III 2015]
Show Answer
Solution:
Ans. Mutual Inductance : The mutual inductance, of a pair of coils, equals the magnetic flux linked with one of them due to a unit current in the other. $\mathbf{1}$ Alternatively, The mutual inductance, of a pair of coils, equals the emf induced in one of them when the rate of change of current in the other is unity.
Factors affecting the mutual inductance of a pair of coils :
(i) The sizes of the two coils
(ii) The shape of the two coils
(iii) The distance of separation between the two coils
(iv) The nature of the medium between the two coils
(v) The relative orientation of the two coils.
[Any two] 1
From $t=0$ to $t=3 \mathrm{~s}\left(=\frac{30 \mathrm{~cm}}{10 \mathrm{~cm} / \mathrm{s}}\right)$, the flux through the coil is zero.
From $t=3 \mathrm{~s}$ to $t=5 \mathrm{~s}$, the flux through the coil increases from 0 to $\left[0.1 \times\left(\frac{20}{100}\right)^{2}\right] \mathrm{Wb}$. i.e. $0.004 \mathrm{~Wb}$.
From $t=5 \mathrm{~s}$ to $t=11 \mathrm{~s}$, the flux remains constant at the value $0.004 \mathrm{~Wb}$.
From $t=11 \mathrm{~s}$ to $t=13 \mathrm{~s}$, the flux decreases through the will from $0.004 \mathrm{~Wb}$ to $0 \mathrm{~Wb}$.
(i) The plot of $\phi$ against $t$ is, therefore, as shown :
(ii) $\varepsilon=-\frac{d \phi}{d t}$
The plot of $\varepsilon$ against $t$ is, therefore, as shown :
(iii)
$$ i=\frac{\varepsilon}{R}=\frac{2 \times 10^{-3}}{0.1 \Omega}=20 \mathrm{~mA} $$
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