Electromagnetic Induction and Alternating Current 7 Question 26
29. Two infinitely long parallel wires carrying currents $I=I _0 \sin \omega t$ in opposite directions are placed a distance $3 a$ apart. A square loop of side $a$ of negligible resistance with a capacitor of capacitance $C$ is placed in the plane of wires as shown. Find the maximum current in the square loop. Also, sketch the graph showing the variation of charge on the upper plate of the capacitor as a function of time for one complete cycle taking anti-clockwise direction for the current in the loop as positive.
$(2003,4$ M)
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Answer:
Correct Answer: 29. $i _{\max }=\frac{\mu _0 a C I _0 \omega^{2} \ln (2)}{\pi}$
Solution:
- (a) For an elemental strip of thickness $d x$ at a distance $x$ from left wire, net magnetic field (due to both wires)
$$ \begin{aligned} B & =\frac{\mu _0}{2 \pi} \frac{I}{x}+\frac{\mu _0}{2 \pi} \frac{I}{3 a-x} \\ & =\frac{\mu _0 I}{2 \pi} \frac{1}{x}+\frac{1}{3 a-x} \end{aligned} $$
Magnetic flux in this strip,
$$ d \varphi=B d S=\frac{\mu _0 I}{2 \pi} \frac{1}{x}+\frac{1}{3 a-x} a d x $$
$\therefore$ Total flux, $\varphi=\int _a^{2 a} d \varphi=\frac{\mu _0 I a}{2 \pi} \int _a^{2 a} \frac{1}{x}+\frac{1}{3 a-x} d x$
or
$$ \begin{aligned} & \varphi=\frac{\mu _0 I a}{\pi} \ln (2) \\ & \varphi=\frac{\mu _0 a \ln (2)}{\pi}\left(I _0 \sin \omega t\right) \end{aligned} $$
Magnitude of induced emf,
$$ e=-\frac{d \varphi}{d t}=\frac{\mu _0 a I _0 \omega \ln (2)}{\pi} \cos \omega t=e _0 \cos \omega t $$
where, $e _0=\frac{\mu _0 a I _0 \omega \ln (2)}{\pi}$
Charge stored in the capacitor,
$$ q=C e=C e _0 \cos \omega t $$
and current in the loop
$$ \begin{aligned} i & =\frac{d q}{d t}=C \omega e _0 \sin \omega t \\ i _{\max } & =C \omega e _0=\frac{\mu _0 a I _0 \omega^{2} C \ln (2)}{\pi} \end{aligned} $$
(b) Magnetic flux passing through the square loop
$$ \varphi \propto \sin \omega t $$
[From Eq. (i)]
i.e. $U$ magnetic field passing through the loop is increasing at $t=0$. Hence, the induced current will produce $\otimes$ magnetic field (from Lenz’s law). Or the current in the circuit at $t=0$ will be clockwise (or negative as per the given convention). Therefore, charge on upper plate could be written as,
$$ q=+q _0 \cos \omega t $$
[From Eq. (ii)]
Here, $\quad q _0=C e _0=\frac{\mu _0 a C I _0 \omega \ln (2)}{\pi}$
The corresponding $q-t$ graph is shown in figures,
