25. A rectangular frame ABCD, made of a uniform metal wire, has a straight connection between E and

Q: Why do we treat the two loops separately when calculating EMF?

Because the magnetic flux passing through each loop depends on its specific area. Even though they are part of the same frame, the rate of change of flux (and thus the induced EMF) is proportional to the area enclosed by each individual loop.

Q: How do we know the direction of the batteries in the equivalent circuit?

Lenz's Law tells us that the induced current must oppose the change in magnetic flux. Since the inward magnetic field is increasing, the induced current must flow counter-clockwise to produce an outward field. We place the batteries such that they drive current in this counter-clockwise direction.

Q: Why is the resistance of segment BE 0.5 ohms?

The problem states that the resistance per unit length is $1 \, \Omega/\text{m}$. Since the length of segment BE is given as 0.5 meters, its resistance is simply $0.5 \text{ m} \times 1 \, \Omega/\text{m} = 0.5 \, \Omega$.

Q: What happens to the shared segment EF?

Segment EF is part of both loops. Therefore, the net current flowing through it is the superposition of the currents from both loops. If we define $i_1$ and $i_2$ as the counter-clockwise currents in the left and right loops respectively, the net current in EF is $i_1 - i_2$ (flowing upwards).

Q: Can I solve this using nodal analysis instead of loop analysis?

Absolutely! You can define the potential at the bottom wire as 0V and set up a nodal equation for the top wire. However, because the EMF is distributed along the loops, you have to be very careful about where you place the equivalent batteries. Loop analysis is generally less prone to sign errors in these types of Faraday's Law problems.

Question

A rectangular frame $ABCD$, made of a uniform metal wire, has a straight connection between $E$ and $F$ made of the same wire, as shown in figure. $AEFD$ is a square of side 1 m and $EB = FC = 0.5$ m. The entire circuit is placed in a steadily increasing uniform magnetic field directed into the plane of the paper and normal to it.
The rate of change of the magnetic field is 1 T/s. The resistance per unit length of the wire is 1 $\Omega$/m. Find the magnitudes and directions of the currents in the segments $AE$, $BE$ and $EF$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a rectangular metal frame placed in a uniform magnetic field. The field is directed into the plane of the paper and is steadily increasing at a rate of $1 	ext{ T/s}$. The frame is divided into two adjacent loops: a square loop $AFED$ on the left with a side length of $1 	ext{ m}$, and a smaller rectangular loop $EBCF$ on the right with a width of $0.5 	ext{ m}$. The resistance of the wire is $1 \, \Omega/	ext{m}$. Our goal is to find the exact currents flowing through specific segments of this frame.

### The Master Equation

According to **Faraday's Law of Induction**, a changing magnetic flux induces an electromotive force (EMF) in a closed loop. The magnitude of this EMF is given by:

$$e = \left| \frac{d\phi}{dt} ight| = A \frac{dB}{dt}$$

Since the rate of change of the magnetic field $\frac{dB}{dt}$ is constant at $1 	ext{ T/s}$, the induced EMF in each loop is simply numerically equal to its area.

For the larger loop $AFED$:
$$A_1 = 1 	ext{ m} 	imes 1 	ext{ m} = 1 	ext{ m}^2 \implies e_1 = 1 	ext{ V}$$

For the smaller loop $EBCF$:
$$A_2 = 0.5 	ext{ m} 	imes 1 	ext{ m} = 0.5 	ext{ m}^2 \implies e_2 = 0.5 	ext{ V}$$

Now, we must apply **Lenz's Law** to determine the direction of these induced EMFs. The external magnetic field is pointing *into* the page and is *increasing*. To oppose this change, the induced current must create a magnetic field pointing *out* of the page. Using the right-hand grip rule, this means the induced currents must flow in a **counter-clockwise** direction in both loops.

### Constructing the Equivalent Circuit

To solve for the currents, we can replace the physical frame with an equivalent electrical circuit. We place a $1 	ext{ V}$ battery in the left loop and a $0.5 	ext{ V}$ battery in the right loop, oriented such that they drive current counter-clockwise.

Next, we calculate the resistance of each segment based on the $1 \, \Omega/	ext{m}$ specification:
- $R_{AE} = R_{DF} = R_{EF} = R_{AD} = R_{BC} = 1 \, \Omega$
- $R_{EB} = R_{FC} = 0.5 \, \Omega$

Let $i_1$ be the counter-clockwise current in loop $AFED$ and $i_2$ be the counter-clockwise current in loop $EBCF$. The shared segment $EF$ will carry a net current of $i_1 - i_2$ flowing upwards (from $F$ to $E$).

### Final Calculation

We now apply **Kirchhoff's Voltage Law (KVL)** to both loops.

**For Loop 1 ($AFED$):**
Starting from $A$ and moving counter-clockwise:
$$-i_1(1) + 1 - i_1(1) - (i_1 - i_2)(1) - i_1(1) = 0$$
$$4i_1 - i_2 = 1 \quad \dots (i)$$

**For Loop 2 ($EBCF$):**
Starting from $E$ and moving counter-clockwise:
$$(i_1 - i_2)(1) - i_2(0.5) - i_2(1) + 0.5 - i_2(0.5) = 0$$
$$i_1 - 3i_2 = -0.5 \quad \dots (ii)$$

We have a simple system of linear equations. From equation (ii), we can express $i_1$ as:
$$i_1 = 3i_2 - 0.5$$

Substituting this into equation (i):
$$4(3i_2 - 0.5) - i_2 = 1$$
$$12i_2 - 2 - i_2 = 1$$
$$11i_2 = 3 \implies i_2 = \frac{3}{11} = \frac{6}{22} 	ext{ A}$$

Plugging $i_2$ back to find $i_1$:
$$i_1 = 3\left(\frac{6}{22}ight) - \frac{11}{22} = \frac{18}{22} - \frac{11}{22} = \frac{7}{22} 	ext{ A}$$

Finally, we map these loop currents back to the physical segments requested:
- **Current in $AE$:** This is simply $i_1$, flowing from $E$ to $A$. Magnitude: **$\frac{7}{22} 	ext{ A}$**.
- **Current in $BE$:** This is simply $i_2$, flowing from $B$ to $E$. Magnitude: **$\frac{6}{22} 	ext{ A}$**.
- **Current in $EF$:** This is the shared segment, carrying $i_1 - i_2$ from $F$ to $E$. Magnitude: $\frac{7}{22} - \frac{6}{22} =$ **$\frac{1}{22} 	ext{ A}$**.

Visualized Solution

$Visualizing the Setup$

$Faraday’s Law of Induction$

$Calculating Induced EMFs$

$Lenz’s Law and Equivalent Circuit$

$Kirchhoff’s Voltage Law: Loop 1$

$Kirchhoff’s Voltage Law: Loop 2$

$Solving the Equations$

$Final Currents in Segments$

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

Analyzing the Setup

The Master Equation

Constructing the Equivalent Circuit

Final Calculation

Similar Questions

In the given figure, the magnetic flux through the loop increases according to the relation $ϕ_{B} (t) = 10 t^{2} + 20 t$ , where $ϕ_{B}$ is in milliwebers and $t$ is in seconds. The magnitude of current through $R = 2Ω$ resistor at $t = 5$ s is ...... mA.

At time $t = 0$ magnetic field of 1000 gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to 500 gauss, in the next 5 s, then induced emf in the loop is

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Faraday’s Law of Induction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Induced EMFs

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lenz’s Law and Equivalent Circuit

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kirchhoff’s Voltage Law: Loop 1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kirchhoff’s Voltage Law: Loop 2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving the Equations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Currents in Segments