A conducting bar of length L is free to slide on two parallel conducting rails as shown in the figur

Q: Why does the current flow downwards in the central bar?

Because the Lorentz force $q(\mathbf{v} \times \mathbf{B})$ pushes positive charges downwards. The bar acts like a battery with its positive terminal at the bottom.

Q: Would the answer change if the magnetic field was pointing outwards?

Yes! If the magnetic field pointed outwards, the flux changes would be reversed, and the Lorentz force would push charges upwards. $I_1$ would be anti-clockwise and $I_2$ would be clockwise.

Q: Do the values of $R_1$ and $R_2$ affect the direction of the currents?

No, the resistance values only affect the *magnitude* of the induced currents, not their direction. The direction is purely determined by the velocity and the magnetic field.

Q: Can I solve this without using Lenz's Law?

Absolutely. You can rely entirely on the Lorentz force $\mathbf{F} = q(\mathbf{v} \times \mathbf{B})$ to find the polarity of the moving bar, which then dictates the current direction in both loops.

Q: What happens to the energy in this system?

The external agent pulling the bar does mechanical work. This work is converted into electrical energy (the induced EMF), which is ultimately dissipated as heat in the resistors $R_1$ and $R_2$.

Question

A conducting bar of length $L$ is free to slide on two parallel conducting rails as shown in the figure. Two resistors $R_1$ and $R_2$ are connected across the ends of the rails. There is a uniform magnetic field $\mathbf{B}$ pointing into the page. An external agent pulls the bar to the left at a constant speed $v$. The correct statement about the directions of induced currents $I_1$ and $I_2$ flowing through $R_1$ and $R_2$ respectively is

Options:

(A) both $I_1$ and $I_2$ are in anti-clockwise direction.

(B) both $I_1$ and $I_2$ are in clockwise direction.

(C) $I_1$ is in clockwise direction and $I_2$ is in anti-clockwise direction.

(D) $I_1$ is in anticlockwise direction and $I_2$ is in clockwise direction.

Sigma Percentile · Accepted Answer

### The Physical Setup

Imagine a conducting bar sliding smoothly along two parallel rails. The entire setup is immersed in a uniform magnetic field that points directly into your screen. On the left end, the rails are connected by a resistor $R_1$, forming a closed loop. On the right end, another resistor $R_2$ forms a second closed loop.

As an external agent pulls the bar to the left with a constant velocity $\mathbf{v}$, we are tasked with finding the direction of the induced currents $I_1$ and $I_2$ in the respective loops. This is a classic scenario of motional EMF, and we have two powerful tools at our disposal: Lenz's Law and the Lorentz Force.

### Applying Lenz's Law to the Left Loop

Let's focus our attention on the left loop first. As the bar moves to the left, the physical area of this loop is continuously shrinking.

Since the magnetic flux $\Phi$ is the product of the magnetic field $B$ and the area $A$, a decreasing area means the total magnetic flux pointing into the page is also decreasing.

Nature, as described by Lenz's Law, absolutely abhors a change in flux. To fight this decrease, the induced current $I_1$ will flow in a direction that creates its own magnetic field pointing **into** the page, trying to replenish the lost flux.

If you use the Right-Hand Grip Rule and point your thumb into the screen, your fingers will naturally curl in a **clockwise** direction. Therefore, the induced current $I_1$ must flow clockwise.

### Applying Lenz's Law to the Right Loop

Now, let's shift our gaze to the right loop. The situation here is the exact opposite. As the bar slides to the left, the area of the right loop is expanding.

This expanding area causes the magnetic flux pointing into the page to increase. Once again, Lenz's Law kicks in to oppose this change. To fight the increasing inward flux, the induced current $I_2$ must generate a magnetic field pointing **out** of the page (towards you).

Using the Right-Hand Grip Rule again, point your thumb towards yourself. Your fingers will curl in an **anti-clockwise** direction. Thus, the induced current $I_2$ must flow anti-clockwise.

### The Lorentz Force Perspective

We can beautifully verify our conclusions using the microscopic perspective of the Lorentz force. Inside the moving conducting bar, free positive charges are being dragged to the left with velocity $\mathbf{v}$ through the inward magnetic field $\mathbf{B}$.

The magnetic force on these charges is given by $\mathbf{F}_m = q(\mathbf{v} 	imes \mathbf{B})$.

Since velocity is to the left ($-\hat{i}$) and the magnetic field is inwards ($-\hat{k}$), their cross product points downwards ($-\hat{j}$). This means positive charges are pushed to the bottom of the bar, making the bottom end positive and the top end negative.

The bar acts exactly like a battery driving current downwards. If you trace a downward current in the central bar, it naturally splits to go UP through $R_1$ (which is a clockwise path for the left loop) and UP through $R_2$ (which is an anti-clockwise path for the right loop).

Everything aligns perfectly! The current $I_1$ is clockwise, and $I_2$ is anti-clockwise.

Select Answer:

Visualized Solution

The Physical Setup

Lenz's Law

Flux Change in Left Loop

Direction of $I_{1}$

Flux Change in Right Loop

Direction of $I_{2}$

Lorentz Force Perspective

Final Answer

The Sigma Insight: Motional EMF

The Physical Setup

Applying Lenz's Law to the Left Loop

Applying Lenz's Law to the Right Loop

The Lorentz Force Perspective

Similar Questions

A metal rod moves at a constant velocity in a direction perpendicular to its length. A constant uniform magnetic field exists in space in a direction perpendicular to the rod as well as its velocity. Select the correct statement (s) from the following.

A rectangular loop has a sliding connector $P Q$ of length $l$ and resistance $R Ω$ and it is moving with a speed $v$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $I_{1}, I_{2}$ and $I$ are

A metallic square loop $A B C D$ is moving in its own plane with velocity $v$ in a uniform magnetic field perpendicular to its plane as shown in the figure. Electric field is induced

Select Answer:

Visualized Solution

The Physical Setup

Lenz's Law

Flux Change in Left Loop

Direction of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Flux Change in Right Loop

Direction of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

Lorentz Force Perspective

Final Answer

The Sigma Insight: Motional EMF

The Physical Setup

Applying Lenz's Law to the Left Loop

Applying Lenz's Law to the Right Loop

The Lorentz Force Perspective

Similar Questions

A metal rod moves at a constant velocity in a direction perpendicular to its length. A constant uniform magnetic field exists in space in a direction perpendicular to the rod as well as its velocity. Select the correct statement (s) from the following.

Direction of $I_{1}$

Direction of $I_{2}$