A wheel having metal spokes of 1m long between its axle and rim is rotating in a magnetic field of flux density `5 xx 10^(-5)T` normal to the plane of the wheel. An e.m.f of 22/7 mV is produced between the rim and the axle of the wheel. The rate of rotation of the wheel in revolutions per seconds is

Options

A. 10
B. 20
C. 30
D. 4

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
We are given:
- Length of the spokes (r) = 1 m
- Magnetic flux density (B) = \(5 \times 10^{-5} \, T\)
- E.M.F (E) = \(\frac{22}{7} \, mV = \frac{22}{7} \times 10^{-3} \, V\)

### Step 2: Write down the formula for E.M.F
The formula for the induced E.M.F in a rotating wheel is given by:
\[ E = \frac{1}{2} B \omega r^2 \]
where:
- \(E\) is the induced E.M.F,
- \(B\) is the magnetic flux density,
- \(\omega\) is the angular velocity in radians per second,
- \(r\) is the radius (length of the spokes).

### Step 3: Rearrange the formula to find \(\omega\)
Rearranging the formula to solve for \(\omega\):
\[ \omega = \frac{2E}{B r^2} \]

### Step 4: Substitute the known values into the equation
Substituting the values into the equation:
\[ \omega = \frac{2 \times \left(\frac{22}{7} \times 10^{-3}\right)}{5 \times 10^{-5} \times (1)^2} \]

### Step 5: Calculate \(\omega\)
Calculating the numerator:
\[ 2 \times \left(\frac{22}{7} \times 10^{-3}\right) = \frac{44}{7} \times 10^{-3} \]
Calculating the denominator:
\[ 5 \times 10^{-5} \times 1^2 = 5 \times 10^{-5} \]
Now substituting these into the equation:
\[ \omega = \frac{\frac{44}{7} \times 10^{-3}}{5 \times 10^{-5}} = \frac{44 \times 10^{-3}}{7 \times 5 \times 10^{-5}} = \frac{44 \times 10^{-3}}{35 \times 10^{-5}} = \frac{44}{35} \times 10^{2} = 1.2571 \times 10^{2} \, rad/s \]
Thus,
\[ \omega \approx 125.71 \, rad/s \]

### Step 6: Convert angular velocity to revolutions per second
We know that:
\[ \omega = 2 \pi n \]
where \(n\) is the number of revolutions per second. Rearranging gives:
\[ n = \frac{\omega}{2\pi} \]
Substituting the value of \(\omega\):
\[ n = \frac{125.71}{2\pi} \approx \frac{125.71}{6.28} \approx 20 \, revolutions/second \]

### Step 7: Conclusion
The rate of rotation of the wheel is approximately 20 revolutions per second.

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