Two infinitely long parallel wires carrying currents I = I_0 sin(omega t) in opposite directions are

Q: Why do the magnetic fields from the two wires add up instead of canceling out?

By using the right-hand grip rule, you can see that if the left wire has a downward current, its magnetic field on its right side points outwards. If the right wire has an upward current, its magnetic field on its left side also points outwards. Since both fields point in the same direction in the region between the wires, they add up.

Q: Why can't we just multiply the magnetic field at the center of the loop by the area of the loop?

The magnetic field produced by a straight wire is inversely proportional to the distance from the wire ($B \propto 1/r$). Because the field is non-uniform and changes continuously across the width of the loop, we must use integration to calculate the exact total magnetic flux.

Q: Why is the charge on the capacitor given by $q = Ce$?

The square loop is stated to have negligible resistance. In an ideal LC or purely capacitive circuit driven by an induced EMF, Kirchhoff's voltage law tells us that the induced EMF $e$ must exactly equal the potential difference across the capacitor, which is $V_C = q/C$.

Q: How do we determine the direction of the induced current at $t=0$?

At $t=0$, the current $I = I_0 \sin \omega t$ is zero, but its rate of change is maximum and positive. This means the outward magnetic flux is increasing. According to Lenz's Law, the induced current will try to oppose this increase by creating an inward magnetic field, which requires a clockwise current.

Q: Why does the maximum current depend on $\omega^2$?

The magnetic flux is proportional to $\sin \omega t$. Differentiating it once to find the EMF brings out one factor of $\omega$. The charge on the capacitor is proportional to this EMF. To find the current, we differentiate the charge with respect to time, which brings out a second factor of $\omega$. Hence, $i_{\max} \propto \omega^2$.

Question

Two infinitely long parallel wires carrying currents $I = I_0 \sin \omega t$ in opposite directions are placed a distance $3a$ 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.

Sigma Percentile · Accepted Answer

The problem of finding the induced current in a loop placed between two current-carrying wires is a classic application of Faraday's Law of Induction. It beautifully combines the concepts of magnetic fields, integration, and AC circuits. Let's break down the solution step-by-step.

### Analyzing the Setup

Imagine you are looking at two infinitely long parallel wires. They are carrying alternating currents, $I = I_0 \sin \omega t$, but in **opposite directions**. Right in the middle of these wires, at a distance $a$ from the left wire, sits a square loop of side $a$. This loop has a capacitor $C$ connected to it and has negligible resistance.

Our ultimate goal is to find the maximum induced current in this square loop. To do this, we first need to understand the magnetic field in the region where the loop is located.

### The Master Equation for Magnetic Flux

The magnetic field produced by a long straight wire at a distance $r$ is given by:

$$B = \frac{\mu_0 I}{2\pi r}$$

Because the currents in the two wires are flowing in opposite directions, their magnetic fields in the space between them will actually point in the **same direction**. Using the right-hand grip rule, if the left wire's current is downwards and the right wire's current is upwards, both magnetic fields will point outwards from the page.

At a distance $x$ from the left wire, the distance to the right wire is $(3a - x)$. Therefore, the total magnetic field at this position is the sum of the fields from both wires:

$$B = \frac{\mu_0 I}{2\pi x} + \frac{\mu_0 I}{2\pi (3a - x)}$$

### Integrating the Elemental Flux

Since the magnetic field is not uniform across the square loop, we cannot simply multiply the field by the area. We must use integration! Let's consider a thin vertical strip of width $dx$ at a distance $x$ from the left wire. The area of this elemental strip is $dS = a \, dx$.

The small magnetic flux $d\Phi$ passing through this strip is:

$$d\Phi = B \, dS = \frac{\mu_0 I a}{2\pi} \left( \frac{1}{x} + \frac{1}{3a - x} \right) dx$$

To find the total magnetic flux $\Phi$ through the entire loop, we integrate this expression from the left edge of the loop ($x = a$) to the right edge ($x = 2a$):

$$\Phi = \int_a^{2a} \frac{\mu_0 I a}{2\pi} \left( \frac{1}{x} + \frac{1}{3a - x} \right) dx$$

The integral of $1/x$ is $\ln x$. Evaluating this definite integral gives:

$$\Phi = \frac{\mu_0 I a}{2\pi} \left[ \ln x - \ln(3a - x) \right]_a^{2a}$$

When we substitute the limits, the terms simplify elegantly:

$$\Phi = \frac{\mu_0 I a}{\pi} \ln 2$$

### Faraday's Law and Induced EMF

Now that we have the total flux, we can substitute the alternating current $I = I_0 \sin \omega t$ into our expression:

$$\Phi = \frac{\mu_0 a \ln 2}{\pi} I_0 \sin \omega t$$

According to **Faraday's Law of Induction**, the induced EMF $e$ is the negative rate of change of magnetic flux:

$$e = -\frac{d\Phi}{dt}$$

Differentiating our flux expression with respect to time $t$, we get:

$$e = -\frac{\mu_0 a I_0 \omega \ln 2}{\pi} \cos \omega t$$

Let's define the constant amplitude of this EMF as $e_0 = \frac{\mu_0 a I_0 \omega \ln 2}{\pi}$. So, the induced EMF simplifies to $e = -e_0 \cos \omega t$.

### The Capacitor's Role and Final Calculation

The square loop has negligible resistance, which means the entire induced EMF appears across the capacitor. The charge $q$ stored on the capacitor is directly proportional to this voltage:

$$q = C |e| = C e_0 \cos \omega t$$

To find the current $i$ flowing in the loop, we differentiate the charge with respect to time:

$$i = \frac{dq}{dt} = -C e_0 \omega \sin \omega t$$

We are looking for the **maximum current** $i_{\max}$, which is simply the amplitude of this current expression:

$$i_{\max} = C e_0 \omega$$

Substituting the value of $e_0$ back into this equation, we arrive at our final answer:

$$i_{\max} = \frac{\mu_0 a C I_0 \omega^2 \ln 2}{\pi}$$

### The Charge Variation Graph

The charge on the capacitor varies as a cosine function, $q = q_0 \cos \omega t$. At $t=0$, the magnetic flux is zero but is increasing at its maximum rate. According to **Lenz's Law**, the induced current will flow in a direction to oppose this increase. This results in a clockwise current that pushes positive charge onto the upper plate of the capacitor. Consequently, the charge graph starts from a positive maximum and follows a standard cosine wave for one complete cycle.

Visualized Solution

$S y s t e m S e t u p$

$Magnetic Field & Flux$

$E l e m e n t a l F l ux$

$T o t a l M a g n e t i c F l ux$

$I n d u ce d E M F$

$C ha r g e an d C u r r e n t$

$M a x im u m C u r r e n t$

$C ha r g e V a r ia t i o n G r a p h$

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

Analyzing the Setup

The Master Equation for Magnetic Flux

Integrating the Elemental Flux

Faraday's Law and Induced EMF

The Capacitor's Role and Final Calculation

The Charge Variation Graph

Similar Questions

A planar loop of wire rotates in a uniform magnetic field. Initially at $t = 0$ , the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of $10 s$ about an axis in its plane, then the magnitude of induced emf will be maximum and minimum respectively at

Visualized Solution

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