An infinitely long straight wire carrying current I, one side opened rectangular loop and a conducto

Q: Why do we use the formula for a long straight wire's magnetic field?

The problem states the wire is 'infinitely long'. For an infinitely long straight wire carrying current $I$, the magnetic field at a perpendicular distance $r$ is derived from Ampere's Law as $B = \frac{\mu_0 I}{2\pi r}$.

Q: Why is the EMF only induced in the sliding connector and not the horizontal wires?

Motional EMF is given by $\vec{v} \times \vec{B} \cdot \vec{l}$. For the horizontal wires, their length vector $\vec{l}$ is parallel to the velocity vector $\vec{v}$. The cross product $\vec{v} \times \vec{B}$ is perpendicular to $\vec{l}$, so the dot product is zero. Only the vertical sliding connector cuts the magnetic field lines effectively.

Q: Does the magnetic field vary along the length of the connector?

No, the connector is vertical and parallel to the straight wire. Every point on the connector is at the exact same perpendicular distance $r$ from the straight wire. Therefore, the magnetic field $B$ is uniform across the entire length $l$ of the connector.

Q: What if the connector was moving to the left instead of the right?

If it moved to the left, the magnitude of the velocity $v$ would be the same, so the magnitude of the induced EMF and current would remain exactly the same. However, the *direction* of the induced current would reverse (it would become clockwise) because the magnetic flux would be decreasing instead of increasing.

Q: Why did we ignore the self-inductance of the loop?

The problem explicitly states to assume the self-inductance is negligible. If we didn't ignore it, the changing current would induce an additional back-EMF ($-L \frac{di}{dt}$), which would make the current a time-dependent differential equation rather than a simple steady-state algebraic expression.

Question

An infinitely long straight wire carrying current $I$, one side opened rectangular loop and a conductor $C$ with a sliding connector are located in the same plane, as shown in the figure. The connector has length $l$ and resistance $R$. It slides to the right with a velocity $v$. The resistance of the conductor and the self-inductance of the loop are negligible. The induced current in the loop, as a function of separation $r$ between the connector and the straight wire is

Options:

(A) $\frac{\mu_0}{4\pi} \frac{Ivl}{Rr}$

(B) $\frac{\mu_0}{\pi} \frac{Ivl}{Rr}$

(C) $\frac{2\mu_0}{\pi} \frac{Ivl}{Rr}$

(D) $\frac{\mu_0}{2\pi} \frac{Ivl}{Rr}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are observing a classic electromagnetic induction setup. We have a very long, straight wire carrying a steady current $I$. Placed nearby in the same plane is a rectangular conducting loop. This isn't just any loop; one of its sides is open, and a conducting rod (or connector) $C$ of length $l$ bridges the gap.

This connector isn't stationary. It is sliding to the right with a constant velocity $v$. The loop itself has a resistance $R$, while the resistance of the sliding connector and the self-inductance of the loop are negligible. Our mission is to determine the induced current in this loop as a function of the separation distance $r$ between the straight wire and the sliding connector.

### The Magnetic Field Environment

Before we talk about induction, we need to understand the environment the connector is moving through. The long straight wire acts as the source of a magnetic field.

According to Ampere's Law, the magnetic field $B$ produced by a long straight wire at a perpendicular distance $r$ is given by:

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

Using the Right-Hand Thumb Rule—pointing your thumb in the direction of the current $I$ (upwards)—your fingers curl into the plane of the screen on the right side of the wire. Therefore, the magnetic field $\vec{B}$ at the location of the sliding connector is directed **into the plane of the paper**.

### The Master Equation: Motional EMF

Now, let's focus on the sliding connector. It is a conductor of length $l$ moving with velocity $v$ through a perpendicular magnetic field $B$. As it moves, the free electrons inside the conductor experience a magnetic Lorentz force, which pushes them to one end of the rod. This separation of charge creates a potential difference across the ends of the connector.

This phenomenon is known as **Motional EMF**. The magnitude of this induced EMF ($\varepsilon$) is given by the elegant formula:

$$\varepsilon_{	ext{induced}} = Bvl$$

Let's substitute the magnetic field $B$ we found earlier into this equation. We get:

$$\varepsilon_{	ext{induced}} = \left(\frac{\mu_0 I}{2\pi r}ight) vl$$

This induced EMF acts exactly like a virtual battery placed in the arm of the sliding connector, ready to drive a current through the closed circuit.

### Final Calculation: The Induced Current

We have our virtual battery, and we know the total resistance of the circuit is $R$. To find the current, we simply call upon our old friend, Ohm's Law. The induced current $I_{	ext{induced}}$ is the induced EMF divided by the resistance:

$$I_{	ext{induced}} = \frac{\varepsilon_{	ext{induced}}}{R}$$

Plugging in our expression for the EMF, we arrive at the final answer:

$$I_{	ext{induced}} = \frac{\mu_0 I v l}{2\pi r R}$$

This perfectly matches option (d).

**A Quick Note on Direction:** If you want to take it a step further, think about Lenz's Law. As the connector moves to the right, the area of the loop increases, which increases the magnetic flux pointing *into* the page. To oppose this increase, the induced current will generate a magnetic field pointing *out* of the page. Using the Right-Hand Rule again, this requires the induced current to flow in a **counter-clockwise** direction!

Select Answer:

Visualized Solution

Visualizing the Setup

Magnetic Field due to Straight Wire

Motional EMF in the Connector

Substituting the Magnetic Field

Calculating Induced Current

Final Expression

Direction of Induced Current (Lenz's Law)

The Sigma Insight: Motional EMF

Analyzing the Setup

The Magnetic Field Environment

The Master Equation

Final Calculation

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