A square loop of side 20 cm and resistance 1 ohm is moved towards right with a constant speed v_0. T

Q: Why is only the right arm considered for motional EMF?

The left arm is completely outside the magnetic field, so it cuts no flux. The top and bottom arms are moving parallel to their own length, so the cross product in the motional EMF formula $(\vec{v} \times \vec{B}) \cdot \vec{l}$ evaluates to zero for them. Only the right arm is perpendicular to both the velocity and the magnetic field.

Q: How did we get $4\Omega$ for the resistor network?

Based on the problem's intended circuit logic, the network acts as two parallel branches connected across the loop. Each branch has two $4\Omega$ resistors in series, making $8\Omega$ per branch. Two $8\Omega$ resistors in parallel give an equivalent resistance of $4\Omega$.

Q: Why do we add the $1\Omega$ resistance of the loop?

The loop itself forms part of the closed electrical circuit. The motional EMF acts like a battery inside the right arm, and the rest of the loop's wire acts as an internal resistance in series with the external resistor network.

Q: What does the negative sign in Faraday's law mean here?

The negative sign (Lenz's Law) gives the direction of the induced current, which opposes the change in magnetic flux. However, since we are only asked to relate the magnitudes of the current, velocity, and resistance, we can drop the negative sign and use $e = Blv$.

Q: Why did we convert the current to Amperes?

To maintain consistency with SI units (Tesla for magnetic field, meters for length, Ohms for resistance), the current must be in Amperes. If you leave it in milliamps, your resulting velocity will be off by a factor of 1000.

Question

A square loop of side $20	ext{ cm}$ and resistance $1\Omega$ is moved towards right with a constant speed $v_0$. The right arm of the loop is in a uniform magnetic field of $5	ext{ T}$. The field is perpendicular to the plane of the loop and is going into it. The loop is connected to a network of resistors each of value $4\Omega$. What should be the value of $v_0$, so that a steady current of $2	ext{ mA}$ flows in the loop ?

Options:

(A) $1	ext{ m/s}$

(B) $1	ext{ cm/s}$

(C) $10^2	ext{ m/s}$

(D) $10^{-2}	ext{ cm/s}$

Sigma Percentile · Accepted Answer

### The Moving Loop and Motional EMF

Imagine a square conducting loop being pulled steadily out of a region containing a uniform magnetic field. As the loop moves to the right with a constant speed $v_0$, its right arm cuts through the magnetic field lines. According to Faraday's Law of Electromagnetic Induction, this cutting of flux generates an electromotive force (EMF) across the ends of the moving arm.

We can think of this right arm as a virtual battery driving current through the entire circuit. The magnitude of this motional EMF is given by the elegant formula:

$$e = B l v_0$$

Here, $B$ is the magnetic field strength ($5	ext{ T}$), $l$ is the length of the arm inside the field ($20	ext{ cm} = 0.2	ext{ m}$), and $v_0$ is the unknown velocity we need to find.

### Untangling the Resistor Network

Now, let's turn our attention to the external resistor network connected to the loop. At first glance, the diagram might look like a complex web, but the intended circuit logic simplifies it beautifully. The network is designed to act as two parallel branches connected across the terminals of the loop.

Each branch consists of two $4\Omega$ resistors in series, giving a branch resistance of $4\Omega + 4\Omega = 8\Omega$. Because these two $8\Omega$ branches are in parallel, we can calculate the equivalent resistance of the external network ($R_{PQ}$) as:

$$R_{PQ} = \frac{8 	imes 8}{8 + 8} = 4 \Omega$$

**Don't fall into a common trap here!** The square loop itself is made of conducting wire and has its own internal resistance of $1\Omega$. Since the loop is in series with the external network, the total resistance of the entire circuit is:

$$R_{total} = R_{PQ} + R_{loop} = 4 \Omega + 1 \Omega = 5 \Omega$$

### Bringing It All Together

With the total resistance and the EMF expression in hand, we can apply Ohm's Law to find the steady current flowing through the loop:

$$I = \frac{e}{R_{total}} = \frac{B l v_0}{R_{total}}$$

We are given that a steady current of $2	ext{ mA}$ (which is $2 	imes 10^{-3}	ext{ A}$) flows through the circuit. Let's substitute all our known values into the equation:

$$2 	imes 10^{-3} = \frac{5 	imes 0.2 	imes v_0}{5}$$

The numerator simplifies beautifully since $5 	imes 0.2 = 1$. This leaves us with:

$$2 	imes 10^{-3} = \frac{v_0}{5}$$

Multiplying both sides by $5$, we isolate $v_0$:

$$v_0 = 10 	imes 10^{-3}	ext{ m/s} = 10^{-2}	ext{ m/s}$$

Finally, looking at our multiple-choice options, we need to convert this speed into centimeters per second. Since $10^{-2}	ext{ m}$ is exactly $1	ext{ cm}$, the required speed is:

$$v_0 = 1	ext{ cm/s}$$

And there we have it! By carefully breaking down the motional EMF and the circuit's equivalent resistance, we've arrived at the perfect solution.

Select Answer:

Visualized Solution

Visual Anchor & Problem Setup

Motional EMF Formula

Equivalent Resistance of the Network

Total Resistance of the Circuit

Using Ohm's Law

Substituting the Values

Solving for Velocity

Final Conversion

The Sigma Insight: Motional EMF

The Moving Loop and Motional EMF

Untangling the Resistor Network

Bringing It All Together

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