The spokes of a wheel are made of metal and their lengths are of one metre. On rotating the wheel about its own axis in a uniform magnetic field of `5xx10^(-5)` tesla normal to the plane of the wheel, a potential difference of `3.14 mV` is generated between the rim and the axis. The rotational velocity of the wheel is-

Options

A. 63 rev/s
B. 50 rev/s
C. 31.4 rev/s
D. 20 rev/s

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- Length of the spokes (L) = 1 meter
- Magnetic field (B) = \(5 \times 10^{-5}\) Tesla
- Potential difference (V) = 3.14 mV = \(3.14 \times 10^{-3}\) V

### Step 2: Use the formula for EMF generated
The formula for the electromotive force (EMF) generated due to the rotation of a wheel in a magnetic field is given by:
\[ \text{EMF} = \frac{1}{2} B \omega L^2 \]
Where:
- EMF is the potential difference generated,
- \(B\) is the magnetic field strength,
- \(\omega\) is the angular velocity in radians per second,
- \(L\) is the length of the spokes.

### Step 3: Substitute the known values into the formula
We can rearrange the formula to solve for \(\omega\):
\[ \omega = \frac{2 \times \text{EMF}}{B \times L^2} \]
Now substituting the known values:
\[ \omega = \frac{2 \times (3.14 \times 10^{-3})}{(5 \times 10^{-5}) \times (1^2)} \]

### Step 4: Calculate \(\omega\)
Calculating the numerator:
\[ 2 \times (3.14 \times 10^{-3}) = 6.28 \times 10^{-3} \]
Calculating the denominator:
\[ (5 \times 10^{-5}) \times (1) = 5 \times 10^{-5} \]
Now substituting back:
\[ \omega = \frac{6.28 \times 10^{-3}}{5 \times 10^{-5}} = \frac{6.28}{5} \times 10^{2} = 1.256 \times 10^{2} \text{ rad/s} \]
\[ \omega \approx 125.6 \text{ rad/s} \]

### Step 5: Convert \(\omega\) to revolutions per second
To convert from radians per second to revolutions per second, we use the fact that \(1 \text{ revolution} = 2\pi \text{ radians}\):
\[ \text{Revolutions per second} = \frac{\omega}{2\pi} \]
Substituting the value of \(\omega\):
\[ \text{Revolutions per second} = \frac{125.6}{2\pi} \approx \frac{125.6}{6.28} \approx 20 \text{ revolutions per second} \]

### Final Answer
The rotational velocity of the wheel is approximately **20 revolutions per second**.

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