Question

Two identical conductors 1 and 2 are placed on two frictionless conducting rails R and S in a uniform magnetic field directed vertically downward into the plane of the page. If conductor 1 is moved with a constant velocity in the direction as shown in figure, the force on conductor 2 will be along :

• A. \( -\hat{i} \)
• B. \( - \hat{j} \)
• C. \( \hat{k} \)
• D. \( \hat{j} \)

Hint:

Save time by analyzing this via energy conservation and Lenz's Law directly: Lenz's Law states that any induced effect always attempts to destroy the cause that created it. Here, the cause of the induction is the expansion of the loop's area due to conductor 1 moving left. To counteract this growth and try to decrease the area, the free-to-move conductor 2 will experience an electromagnetic force pushing it inward to contract the loop. Since it is on the right side, it must move to the left ($-\hat{i}$) to shrink the area! No cross products are required.

Answer 1

The Correct Option is A

Concept: This problem can be comprehensively analyzed using the principles of Electromagnetic Induction (Faraday's Law and Lenz's Law) and the Magnetic Lorentz Force acting on a current-carrying conductor.
• Motional Electromotive Force (EMF): When a conducting rod of length $L$ moves with a velocity $\vec{v}$ perpendicular to a uniform magnetic field $\vec{B}$, an EMF ($\varepsilon$) is induced across its ends, given by: \[ \varepsilon = (\vec{v} \times \vec{B}) \cdot \vec{L} \] The direction of the driving force on positive charge carriers inside the moving conductor is given by the vector cross product $\vec{v} \times \vec{B}$.
• Lenz's Law: The direction of an induced current is always such that it will oppose the change in magnetic flux that produced it.
• Magnetic Force on a Straight Conductor: A wire carrying a current $I$ with a length vector $\vec{L}$ placed inside a uniform external magnetic field $\vec{B}$ experiences a mechanical deflecting force expressed by: \[ \vec{F} = I (\vec{L} \times \vec{B}) \]

Step 1: Define the coordinate system and given vectors.
From the reference coordinate axes provided in the diagram:
• The positive $x$-axis is directed to the right, represented by the unit vector $\hat{i}$.
• The positive $y$-axis is directed vertically upwards within the plane of the paper, represented by the unit vector $\hat{j}$.
• The positive $z$-axis points perpendicularly outwards from the plane of the paper towards the reader, represented by the unit vector $\hat{k}$. The uniform magnetic field $\vec{B}$ is directed vertically downward into the plane of the page, so it can be vectorially written as: \[ \vec{B} = -B\hat{k} \] Conductor 1 is forced to move horizontally to the left with a constant velocity. Therefore, its velocity vector $\vec{v}_1$ is written as: \[ \vec{v}_1 = -v\hat{i} \]

Step 2: Determine the direction of the induced current using motional EMF.
As conductor 1 moves through the uniform magnetic field, the free charges inside it experience a magnetic force. The direction of this force on positive charge carriers is given by the cross product: \[ \vec{v}_1 \times \vec{B} = (-v\hat{i}) \times (-B\hat{k}) \] Using the scalar property of vector multiplication: \[ \vec{v}_1 \times \vec{B} = vB (\hat{i} \times \hat{k}) \] According to the standard cross product rules for unit vectors ($\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, and $\hat{k} \times \hat{i} = \hat{j}$), we know that: \[ \hat{i} \times \hat{k} = -\hat{j} \] Substituting this back into our expression: \[ \vec{v}_1 \times \vec{B} = vB (-\hat{j}) = -vB\hat{j} \] Since the direction of the force on positive charges is along $-\hat{j}$ (downwards), the induced current flows downwards within conductor 1. Tracing this current through the closed loop formed by conductor 1, rail S, conductor 2, and rail R:
• The current flows downwards in conductor 1 ($-\hat{j}$ direction).
• It moves to the right along the bottom conducting rail S ($+\hat{i}$ direction).
• It travels upwards through conductor 2 ($+\hat{j}$ direction).
• It travels back to the left along the top conducting rail R ($-\hat{i}$ direction). Thus, the induced current circles in a counter-clockwise direction around the loop. For conductor 2, the current vector points vertically upwards: \[ \vec{I}_2 = I\hat{j} \]

Step 3: Alternative Approach using Lenz's Law (Flux Change Analysis).
Let us cross-verify the current direction using Lenz's Law to ensure absolute certainty:
• The initial magnetic field lines are pointing straight into the page ($\times$).
• As conductor 1 is pulled towards the left, the enclosed surface area of the rectangular loop bounded by rails R, S and conductors 1, 2 is continuously increasing.
• Because the area increases, the total inward magnetic flux ($\Phi_B = B \cdot A$) piercing through the loop increases.
• According to Lenz's law, the system will establish an induced current to generate an opposing magnetic field pointing out of the page ($\cdot$) to mitigate this flux increase.
• By applying the right-hand grip rule, a counter-clockwise current is required to produce a magnetic field pointing outwards. In a counter-clockwise loop, the current must flow from bottom to top through the rightmost vertical branch, which is conductor 2. This perfectly confirms that the current flows in the $+\hat{j}$ direction in conductor 2.

Step 4: Calculate the magnetic force acting on conductor 2.
Now that we have established that conductor 2 carries an upward current in the presence of an inward magnetic field, we can compute the mechanical force $\vec{F}_2$ acting on it: \[ \vec{F}_2 = I (\vec{L}_2 \times \vec{B}) \] Here, the length vector $\vec{L}_2$ is in the direction of the current, so $\vec{L}_2 = L\hat{j}$. The magnetic field is $\vec{B} = -B\hat{k}$. Substituting these vectors into the formula: \[ \vec{F}_2 = I \left( L\hat{j} \times (-B\hat{k}) \right) \] Factoring out the scalar constants: \[ \vec{F}_2 = -ILB (\hat{j} \times \hat{k}) \] Using the cross product rule for the unit vectors where $\hat{j} \times \hat{k} = \hat{i}$: \[ \vec{F}_2 = -ILB (\hat{i}) = -ILB\hat{i} \] The negative sign attached to the unit vector $\hat{i}$ indicates that the net magnetic force on conductor 2 is directed horizontally to the left. Therefore, the force acts along the $-\hat{i}$ direction, matching Option (A).