A square loop of side L, resistance R placed in a uniform magnetic field B acting normally to the plane of the loop. If we attempt to pull it out of the field with a constant velocity v, then the power needed is

Options

A. BRlv
B. `(B^(2)l^(2)v^(2))/(R)`
C. `(Bl^(2)v^(2))/(R)`
D. `(Bvl)/(R)`

Explanation

To solve the problem, we need to determine the power required to pull a square loop of side \( L \) and resistance \( R \) out of a uniform magnetic field \( B \) at a constant velocity \( v \). Here’s a step-by-step solution:

### Step 1: Determine the induced EMF
When the loop is pulled out of the magnetic field, an electromotive force (EMF) is induced in the loop due to the change in magnetic flux. The induced EMF (\( \mathcal{E} \)) can be calculated using Faraday's law of electromagnetic induction:

\[ \mathcal{E} = -\frac{d\Phi}{dt} \]

Where \( \Phi \) is the magnetic flux. The magnetic flux through the loop is given by:

\[ \Phi = B \cdot A = B \cdot (L^2) \]

As the loop is pulled out of the magnetic field, the area \( A \) that is in the magnetic field changes. The rate of change of flux can be expressed as:

\[ \frac{d\Phi}{dt} = B \cdot L \cdot v \]

Thus, the induced EMF is:

\[ \mathcal{E} = B \cdot L \cdot v \]

### Step 2: Calculate the induced current
Using Ohm's law, the induced current \( I \) in the loop can be calculated as:

\[ I = \frac{\mathcal{E}}{R} = \frac{B \cdot L \cdot v}{R} \]

### Step 3: Calculate the force on the loop
The force \( F \) acting on the loop due to the magnetic field can be calculated using the formula:

\[ F = I \cdot L \cdot B \]

Substituting the expression for \( I \):

\[ F = \left(\frac{B \cdot L \cdot v}{R}\right) \cdot L \cdot B = \frac{B^2 \cdot L^2 \cdot v}{R} \]

### Step 4: Calculate the power required
The power \( P \) required to pull the loop out of the magnetic field at constant velocity \( v \) is given by:

\[ P = F \cdot v \]

Substituting the expression for \( F \):

\[ P = \left(\frac{B^2 \cdot L^2 \cdot v}{R}\right) \cdot v = \frac{B^2 \cdot L^2 \cdot v^2}{R} \]

### Final Answer
Thus, the power needed to pull the loop out of the magnetic field is:

\[ P = \frac{B^2 \cdot L^2 \cdot v^2}{R} \]