A conducting square loop of side l and resistance R moves in its plane with a uniform velocity v perpendicular ot one of its sides. A uniform and constant magnetic field B exists along the perpendicualr ot the plane of the loop as shown in . The current induced in the loop is
(##HCV_VOL2_C38_E01_025_Q01##)

Options

A. `(Blv)/(R)`clockwise
B. `(Blv)/(R)` anti-clockwise
C. `(2Blv)/(R)` anti-clockwise
D. zero

Explanation

To solve the problem of the induced current in a conducting square loop moving in a magnetic field, we can follow these steps:

### Step 1: Understand the setup
We have a square loop of side length \( l \) and resistance \( R \). The loop is moving with a uniform velocity \( v \) perpendicular to one of its sides. There is a uniform magnetic field \( B \) that is perpendicular to the plane of the loop.

### Step 2: Determine the area of the loop
The area \( A \) of the square loop is given by:
\[ A = l^2 \]

### Step 3: Analyze the magnetic flux
The magnetic flux \( \Phi \) through the loop is given by the product of the magnetic field \( B \) and the area \( A \):
\[ \Phi = B \cdot A = B \cdot l^2 \]

### Step 4: Check for change in magnetic flux
Since the loop is moving with a constant velocity \( v \) and the magnetic field \( B \) is uniform and constant, the area of the loop that is within the magnetic field does not change. Therefore, the magnetic flux \( \Phi \) remains constant over time.

### Step 5: Apply Faraday's law of electromagnetic induction
According to Faraday's law, the induced electromotive force (EMF) \( \mathcal{E} \) in the loop is given by the negative rate of change of magnetic flux:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Since \( \Phi \) is constant, we have:
\[ \frac{d\Phi}{dt} = 0 \]
Thus, the induced EMF \( \mathcal{E} \) is:
\[ \mathcal{E} = 0 \]

### Step 6: Determine the induced current
Using Ohm's law, the induced current \( I \) in the loop can be calculated as:
\[ I = \frac{\mathcal{E}}{R} \]
Since \( \mathcal{E} = 0 \), we find:
\[ I = \frac{0}{R} = 0 \]

### Conclusion
The current induced in the loop is zero.