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 find the power needed to pull a square loop of side L and resistance R out of a uniform magnetic field B with a constant velocity v, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We have a square loop of side L and resistance R placed in a magnetic field B. The magnetic field is acting normally (perpendicular) to the plane of the loop. We want to pull the loop out of the magnetic field with a constant velocity v.

2. **Identify the Relevant Equations**:
- Power (P) is given by the formula:
\[ P = F \cdot v \]
where F is the force and v is the velocity.
- The force (F) acting on the loop in a magnetic field can be expressed as:
\[ F = I \cdot L \cdot B \]
where I is the current flowing through the loop, L is the length of the side of the loop, and B is the magnetic field strength.

3. **Relate Current to EMF**:
- The electromotive force (EMF) induced in the loop when it moves through the magnetic field is given by:
\[ \text{EMF} = B \cdot v \cdot L \]
- According to Ohm's law, the current (I) can be expressed as:
\[ I = \frac{\text{EMF}}{R} = \frac{B \cdot v \cdot L}{R} \]

4. **Substitute Current into the Power Equation**:
- Now, substitute the expression for current (I) into the power equation:
\[ P = (I \cdot L \cdot B) \cdot v \]
- Replacing I with \(\frac{B \cdot v \cdot L}{R}\):
\[ P = \left(\frac{B \cdot v \cdot L}{R}\right) \cdot L \cdot B \cdot v \]

5. **Simplify the Expression**:
- This simplifies to:
\[ P = \frac{B^2 \cdot L^2 \cdot v^2}{R} \]

6. **Final Result**:
- Thus, the power needed to pull the loop out of the magnetic field is:
\[ P = \frac{B^2 \cdot L^2 \cdot v^2}{R} \]