The magnetic induction in the region between the pole faces of an electromagnet is 0.7 weber/m 2. The induced e.m.f. in a straight conductor 10 cm long, perpendicular to B and moving perpendicular both to magnetic induction and its own length with a velocity 2 m/sec is

Options

A. `0.08`V
B. `0.14`V
C. `0.35`V
D. `0.07`V

Answer

To solve the problem step by step, we will use the formula for induced electromotive force (e.m.f.) in a conductor moving in a magnetic field. The formula is given by:

\[ \text{Induced e.m.f.} (E) = B \cdot v \cdot L \]

where:
- \( E \) is the induced e.m.f. (in volts),
- \( B \) is the magnetic induction (in weber/m² or tesla),
- \( v \) is the velocity of the conductor (in meters/second),
- \( L \) is the length of the conductor (in meters).

### Step 1: Identify the given values
From the problem statement, we have:
- Magnetic induction, \( B = 0.7 \, \text{W/m}^2 \)
- Length of the conductor, \( L = 10 \, \text{cm} = 0.1 \, \text{m} \) (convert centimeters to meters)
- Velocity of the conductor, \( v = 2 \, \text{m/s} \)

### Step 2: Substitute the values into the formula
Now, we substitute the values into the formula for induced e.m.f.:

\[ E = B \cdot v \cdot L \]
\[ E = 0.7 \, \text{W/m}^2 \cdot 2 \, \text{m/s} \cdot 0.1 \, \text{m} \]

### Step 3: Perform the calculation
Now we calculate the values step by step:

1. First, calculate \( B \cdot v \):
\[ 0.7 \cdot 2 = 1.4 \]

2. Next, multiply the result by \( L \):
\[ 1.4 \cdot 0.1 = 0.14 \]

### Step 4: Write the final result
Thus, the induced e.m.f. is:
\[ E = 0.14 \, \text{V} \]

### Conclusion
The induced e.m.f. in the conductor is **0.14 volts**.

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