A wire of length 10cm translates in a direction making an angle of `60^@` with its length. The plane of motion is perpendicular ot a uniform magnetic field of 1.0 T that exists in the space. Find the emf induced between the ends of the rod if the speed of translation of `20 cm s^(-1)`.

Options

A. `1.7xx10^(-3)V`
B. `17xx10^(-3)V`
C. `0.17xx10^(-3)V`
D. `1.7xx10^(-4)V`

Explanation

To solve the problem of finding the induced EMF in a wire translating in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Parameters:**
- Length of the wire, \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Angle with the length of the wire, \( \theta = 60^\circ \)
- Magnetic field strength, \( B = 1.0 \, \text{T} \)
- Speed of translation, \( v = 20 \, \text{cm/s} = 0.2 \, \text{m/s} \)

2. **Use the Formula for Induced EMF:**
The formula for the induced EMF (\( E \)) in a wire moving through a magnetic field is given by:
\[ E = L \cdot v \cdot B \cdot \sin(\theta) \]

3. **Calculate \( \sin(60^\circ) \):**
We know that:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

4. **Substitute the Values into the Formula:**
Now, substituting the known values into the EMF formula:
\[ E = 0.1 \, \text{m} \cdot 0.2 \, \text{m/s} \cdot 1.0 \, \text{T} \cdot \sin(60^\circ) \]
\[ E = 0.1 \cdot 0.2 \cdot 1.0 \cdot \frac{\sqrt{3}}{2} \]

5. **Calculate the Result:**
\[ E = 0.1 \cdot 0.2 \cdot 1.0 \cdot \frac{\sqrt{3}}{2} = 0.01 \cdot \sqrt{3} \]
Using \( \sqrt{3} \approx 1.732 \):
\[ E \approx 0.01 \cdot 1.732 = 0.01732 \, \text{V} = 17.32 \, \text{mV} \]

6. **Final Answer:**
Rounding to two significant figures, the induced EMF is approximately:
\[ E \approx 17 \times 10^{-3} \, \text{V} = 17 \, \text{mV} \]

### Conclusion:
The induced EMF between the ends of the rod is \( 17 \, \text{mV} \).

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