A straight conductor of length 0.4 m is moved with a speed of 7 m/s perpendicular to the magnetic field of intensity of `0.9 Wb//m^(2)`. The induced e.m.f. across the conductor will be

Options

A. 5.04 V
B. 1.26 V
C. 2.52 V
D. 25.2 V

Explanation

To solve the problem of finding the induced e.m.f. across a straight conductor moving in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Length of the conductor (L) = 0.4 m
- Speed of the conductor (V) = 7 m/s
- Magnetic field intensity (B) = 0.9 Wb/m²

2. **Understand the Formula for Induced e.m.f.:**
The induced e.m.f. (E) in a conductor moving through a magnetic field can be calculated using the formula:
\[ E = L \times V \times B \times \sin(\theta) \]
where:
- \(E\) is the induced e.m.f.
- \(L\) is the length of the conductor.
- \(V\) is the speed of the conductor.
- \(B\) is the magnetic field intensity.
- \(\theta\) is the angle between the direction of the magnetic field and the direction of motion of the conductor.

3. **Determine the Angle:**
In this problem, the conductor is moved perpendicular to the magnetic field, which means:
\[ \theta = 90^\circ \]
Therefore, \(\sin(90^\circ) = 1\).

4. **Substitute the Values into the Formula:**
Now, substituting the known values into the formula:
\[ E = 0.4 \, \text{m} \times 7 \, \text{m/s} \times 0.9 \, \text{Wb/m}^2 \times \sin(90^\circ) \]
\[ E = 0.4 \times 7 \times 0.9 \times 1 \]

5. **Calculate the Induced e.m.f.:**
Performing the multiplication:
\[ E = 0.4 \times 7 = 2.8 \]
\[ E = 2.8 \times 0.9 = 2.52 \, \text{V} \]

6. **Final Answer:**
The induced e.m.f. across the conductor is:
\[ E = 2.52 \, \text{V} \]