A wheel with 20 metallic spokes each of length 8.0 m long is rotated with a speed of 120 revolution per minute in a plane normal to the horizontal component of earth magnetic field H at a place. If `H=0.4xx10^(-4)` T at the place, then induced emf between the axle the rim of the wheel is

Options

A. `2.3xx10^(-4)V`
B. `3.1xx10^(-4)V`
C. `2.9xx10^(-4)V`
D. `1.61xx10^(-4)V`

Explanation

To find the induced EMF between the axle and the rim of the wheel, we can use the formula for induced EMF in a rotating system in a magnetic field. The formula is:

\[ \epsilon = \frac{1}{2} B \omega L^2 \]

Where:
- \(\epsilon\) is the induced EMF,
- \(B\) is the magnetic field strength,
- \(\omega\) is the angular frequency in radians per second,
- \(L\) is the length of each spoke.

### Step 1: Convert the speed of rotation from revolutions per minute to radians per second.

Given that the wheel rotates at 120 revolutions per minute (rpm), we need to convert this to radians per second.

\[ \text{Revolutions per second} = \frac{120 \text{ revolutions}}{60 \text{ seconds}} = 2 \text{ revolutions/second} \]

Since one revolution is \(2\pi\) radians, we can convert this to radians per second:

\[ \omega = 2 \text{ revolutions/second} \times 2\pi \text{ radians/revolution} = 4\pi \text{ radians/second} \]

### Step 2: Identify the values for the magnetic field and the length of the spokes.

From the problem statement:
- The magnetic field \(H = 0.4 \times 10^{-4} \text{ T}\)
- The length of each spoke \(L = 8.0 \text{ m}\)

### Step 3: Substitute the values into the induced EMF formula.

Now we can substitute the values into the induced EMF formula:

\[ \epsilon = \frac{1}{2} B \omega L^2 \]

Substituting \(B = H = 0.4 \times 10^{-4} \text{ T}\), \(\omega = 4\pi \text{ radians/second}\), and \(L = 8.0 \text{ m}\):

\[ \epsilon = \frac{1}{2} \times (0.4 \times 10^{-4}) \times (4\pi) \times (8.0)^2 \]

### Step 4: Calculate the induced EMF.

Calculating \(L^2\):

\[ L^2 = (8.0)^2 = 64 \text{ m}^2 \]

Now substituting this back into the equation:

\[ \epsilon = \frac{1}{2} \times (0.4 \times 10^{-4}) \times (4\pi) \times 64 \]

Calculating the numerical values:

\[ \epsilon = \frac{1}{2} \times (0.4 \times 10^{-4}) \times (4 \times 3.14) \times 64 \]

\[ \epsilon = \frac{1}{2} \times (0.4 \times 10^{-4}) \times (12.56) \times 64 \]

Calculating further:

\[ \epsilon = \frac{1}{2} \times (0.4 \times 10^{-4}) \times 804.16 \]

\[ \epsilon = 0.2 \times 10^{-4} \times 804.16 \]

\[ \epsilon = 1.60832 \times 10^{-4} \text{ V} \approx 1.61 \times 10^{-4} \text{ V} \]

### Final Answer:

The induced EMF between the axle and the rim of the wheel is approximately:

\[ \epsilon \approx 1.61 \times 10^{-4} \text{ V} \]