A wheel with ten metallic spokes each `0.50 m` long is rotated with a speed of `120 rev//min` in a plane normal to the earth's magnetic field at the place. If the magnitude of the field is `0.4` gauss, the induced e.m.f. between the axle and the rim of the wheel is equal to

Options

A. `1.256xx10^(-3)V`
B. `6.28xx10^(-4)V`
C. `1.256xx10^(-4)V`
D. `6.28xx10^(-5)V`

Explanation

To solve the problem, we need to calculate the induced electromotive force (e.m.f.) between the axle and the rim of a rotating wheel with metallic spokes in the presence of a magnetic field. We will use the formula for motional e.m.f. in a rotating system.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Number of spokes, \( n = 10 \)
- Length of each spoke (radius), \( l = 0.50 \, \text{m} \)
- Speed of rotation, \( f = 120 \, \text{rev/min} \)
- Magnetic field strength, \( B = 0.4 \, \text{gauss} = 0.4 \times 10^{-4} \, \text{T} \)

2. **Convert the Speed of Rotation:**
- Convert revolutions per minute to revolutions per second:
\[ f = \frac{120 \, \text{rev/min}}{60} = 2 \, \text{rev/s} \]

3. **Calculate Angular Velocity (\( \omega \)):**
- The angular velocity in radians per second is given by:
\[ \omega = 2 \pi f = 2 \pi \times 2 = 4 \pi \, \text{rad/s} \]

4. **Use the Formula for Induced e.m.f.:**
- The formula for the induced e.m.f. (\( \mathcal{E} \)) in a rotating system is:
\[ \mathcal{E} = \frac{1}{2} B l^2 \omega \]
- Substitute the known values into the formula:
\[ \mathcal{E} = \frac{1}{2} \times (0.4 \times 10^{-4}) \times (0.50)^2 \times (4 \pi) \]

5. **Calculate the e.m.f.:**
- First, calculate \( (0.50)^2 = 0.25 \).
- Substitute this back into the equation:
\[ \mathcal{E} = \frac{1}{2} \times (0.4 \times 10^{-4}) \times 0.25 \times (4 \pi) \]
- Calculate \( 4 \pi \approx 12.56 \):
\[ \mathcal{E} = \frac{1}{2} \times (0.4 \times 10^{-4}) \times 0.25 \times 12.56 \]
- Now calculate the product:
\[ \mathcal{E} = 0.5 \times 0.4 \times 0.25 \times 12.56 \times 10^{-4} \]
\[ = 0.5 \times 0.1 \times 12.56 \times 10^{-4} \]
\[ = 0.628 \times 10^{-4} \, \text{V} \]
\[ = 6.28 \times 10^{-5} \, \text{V} \]

6. **Final Result:**
- The induced e.m.f. between the axle and the rim of the wheel is:
\[ \mathcal{E} = 6.28 \times 10^{-5} \, \text{V} \]

### Conclusion:
The induced e.m.f. is \( 6.28 \times 10^{-5} \, \text{V} \).