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

Explanation

To solve the problem, we need to find the rate at which work is being done to keep the wire moving at a constant speed in a magnetic field. This can be calculated using the concepts of electromagnetic induction and power.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Length of the wire (L) = 1 m
- Speed of the wire (v) = 2 m/s
- Magnetic field (B) = 0.5 T
- Resistance of the circuit (R) = 6 Ω

2. **Calculate the Electromotive Force (EMF):**
The induced EMF (E) in the wire moving in a magnetic field can be calculated using the formula:
\[ E = B \cdot L \cdot v \]
Substituting the values:
\[ E = 0.5 \, \text{T} \cdot 1 \, \text{m} \cdot 2 \, \text{m/s} = 1 \, \text{V} \]

3. **Calculate the Current (I) in the Circuit:**
Using Ohm's law, the current (I) can be calculated as:
\[ I = \frac{E}{R} \]
Substituting the values:
\[ I = \frac{1 \, \text{V}}{6 \, \Omega} = \frac{1}{6} \, \text{A} \]

4. **Calculate the Power (P):**
The power (P) can be calculated using the formula:
\[ P = I^2 \cdot R \]
Substituting the values:
\[ P = \left(\frac{1}{6} \, \text{A}\right)^2 \cdot 6 \, \Omega = \frac{1}{36} \cdot 6 = \frac{1}{6} \, \text{W} \]

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

### Final Answer:
The rate at which work is being done is \( \frac{1}{6} \, \text{W} \).

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