A horizontal straight wire 10m long extending from east to west is falling with a speed of 5.0m/s at right angles to the horizontal component of earths magnetic field `0.3xx10^(-4) tesla. The emf induced across the ends of wire is

Options

A. 1.5 mV with higher potential at eastern end
B. 1.5 mV with higher potential at western end
C. 0.75 mV with higher potential at eastern end
D. zero

Answer

To find the induced EMF across the ends of a horizontal straight wire falling in a magnetic field, we can use the formula for induced EMF (E):

\[ E = B \cdot L \cdot V \]

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, \( L = 10 \, \text{m} \)
- Speed of the wire, \( V = 5 \, \text{m/s} \)
- Magnetic field strength, \( B = 0.3 \times 10^{-4} \, \text{T} \)

2. **Substitute the values into the formula:**
\[ E = B \cdot L \cdot V \]
\[ E = (0.3 \times 10^{-4} \, \text{T}) \cdot (10 \, \text{m}) \cdot (5 \, \text{m/s}) \]

3. **Calculate the induced EMF:**
\[ E = 0.3 \times 10^{-4} \cdot 10 \cdot 5 \]
\[ E = 0.3 \times 10^{-4} \cdot 50 \]
\[ E = 15 \times 10^{-4} \, \text{V} \]
\[ E = 1.5 \times 10^{-3} \, \text{V} \]

4. **Convert the induced EMF to millivolts:**
\[ E = 1.5 \, \text{mV} \]

5. **Determine the direction of the induced EMF:**
According to Fleming's right-hand rule, the direction of the induced EMF is from west to east. Therefore, the western end of the wire is at a higher potential.

### Final Answer:
The induced EMF across the ends of the wire is **1.5 millivolts** with the higher potential at the **western end**.

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