A wire of length `1 m` is moving at a speed of `2 ms^(-1)` perpendicular to its length and a homogeneous magnetic field of `0.5T`. The ends of the wire are joined to a circuit of resistance `6 Omega`. The rate at which work is being done to keep the wire moving at constant speed is

Options

A. `(1)/(12)W`
B. `(1)/(6)W`
C. `(1)/(3)W`
D. `1W`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify the given values
- Length of the wire, \( L = 1 \, \text{m} \)
- Speed of the wire, \( V = 2 \, \text{m/s} \)
- Magnetic field strength, \( B = 0.5 \, \text{T} \)
- Resistance of the circuit, \( R = 6 \, \Omega \)

### Step 2: Calculate the induced electromotive force (emf)
The induced emf (\( \mathcal{E} \)) in the wire moving through a magnetic field can be calculated using the formula:
\[ \mathcal{E} = B \cdot L \cdot V \]
Substituting the known values:
\[ \mathcal{E} = 0.5 \, \text{T} \cdot 1 \, \text{m} \cdot 2 \, \text{m/s} = 1 \, \text{V} \]

### Step 3: Calculate the current flowing through the circuit
Using Ohm's law, the current (\( I \)) can be calculated as:
\[ I = \frac{\mathcal{E}}{R} \]
Substituting the values:
\[ I = \frac{1 \, \text{V}}{6 \, \Omega} = \frac{1}{6} \, \text{A} \]

### Step 4: Calculate the power (rate of work done)
The power (\( P \)) can be calculated using the formula:
\[ P = I^2 \cdot R \]
Substituting the value of current:
\[ P = \left(\frac{1}{6} \, \text{A}\right)^2 \cdot 6 \, \Omega = \frac{1}{36} \cdot 6 = \frac{1}{6} \, \text{W} \]

### Conclusion
The rate at which work is being done to keep the wire moving at constant speed is:
\[ \boxed{\frac{1}{6} \, \text{W}} \]