A wire of length 1 m moving with velocity 8 m/s at right angles to a magnetic field of 2T. The magnitude of induced emf, between the ends of wire will be ______

Options

A. 20 V
B. 16 V
C. 8 V
D. 12 V

Answer

To solve the problem of finding the induced electromotive force (emf) in a wire moving through a magnetic field, we can use the formula for induced emf:

\[ \text{emf} (E) = B \cdot L \cdot v \]

where:
- \( E \) is the induced emf,
- \( B \) is the magnetic field strength,
- \( L \) is the length of the wire,
- \( v \) is the velocity of the wire.

### Step-by-Step Solution:

1. **Identify the given values:**
- Length of the wire, \( L = 1 \, \text{m} \)
- Velocity of the wire, \( v = 8 \, \text{m/s} \)
- Magnetic field strength, \( B = 2 \, \text{T} \)

2. **Substitute the values into the formula:**
\[ E = B \cdot L \cdot v \]
\[ E = 2 \, \text{T} \cdot 1 \, \text{m} \cdot 8 \, \text{m/s} \]

3. **Calculate the induced emf:**
\[ E = 2 \cdot 1 \cdot 8 = 16 \, \text{V} \]

4. **Conclusion:**
The magnitude of the induced emf between the ends of the wire is \( 16 \, \text{V} \).

### Final Answer:
The magnitude of the induced emf between the ends of the wire will be **16 V**.

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