A bicycle is resting on its stand in the east-west direction and the rear wheel is rotated at an angular speed of `50` revolutions per minute.If the length of each spoke is `30.0 cm` and the horizontal component of the earth's magnetic field is `4xx10^(-5) T`,find the `emf` induced between the axis and the outer end of a spoke Neglect centripetal force acting on the free electrons of the spoke.

Explanation

To solve the problem, we need to find the induced electromotive force (emf) between the axis and the outer end of a spoke of a bicycle wheel that is rotating in the presence of a magnetic field. Let's break down the solution step by step.

### Step-by-Step Solution:

**Step 1: Convert Angular Speed to Radians per Second**
The angular speed is given as 50 revolutions per minute (rpm). We need to convert this to radians per second.

\[ \text{Angular speed} (\omega) = 50 \text{ revolutions/minute} \times \frac{2\pi \text{ radians}}{1 \text{ revolution}} \times \frac{1 \text{ minute}}{60 \text{ seconds}} \]

Calculating this gives:

\[ \omega = 50 \times \frac{2\pi}{60} = \frac{100\pi}{60} = \frac{5\pi}{3} \text{ radians/second} \]

**Step 2: Calculate the Length of the Spoke**
The length of each spoke is given as 30.0 cm. We need to convert this to meters:

\[ L = 30.0 \text{ cm} = 0.3 \text{ m} \]

**Step 3: Calculate the Velocity of the Spoke's End**
The velocity \(v\) of the outer end of the spoke can be calculated using the formula:

\[ v = L \cdot \omega \]

Since the spoke rotates about the axis, we use half the length of the spoke for the calculation:

\[ v = \frac{L}{2} \cdot \omega = \frac{0.3 \text{ m}}{2} \cdot \frac{5\pi}{3} = 0.15 \cdot \frac{5\pi}{3} = \frac{0.75\pi}{3} \text{ m/s} \]

**Step 4: Calculate the Induced EMF**
The induced emf (E) can be calculated using the formula:

\[ E = B \cdot L \cdot v \]

Where:
- \(B\) is the magnetic field strength, given as \(4 \times 10^{-5} \text{ T}\).
- \(L\) is the length of the spoke, which we have as \(0.3 \text{ m}\).
- \(v\) is the velocity we calculated.

Substituting the values:

\[ E = (4 \times 10^{-5} \text{ T}) \cdot (0.3 \text{ m}) \cdot \left(\frac{0.75\pi}{3}\right) \]

Calculating this gives:

\[ E = (4 \times 10^{-5}) \cdot (0.3) \cdot \left(\frac{0.75 \cdot 3.14}{3}\right) \]

\[ E = (4 \times 10^{-5}) \cdot (0.3) \cdot (0.785) \]

\[ E = (4 \times 10^{-5}) \cdot (0.3) \cdot (0.785) \approx 9.42 \times 10^{-5} \text{ V} \]

Thus, the induced emf is approximately:

\[ E \approx 9.42 \times 10^{-5} \text{ V} \]

### Final Answer:
The induced emf between the axis and the outer end of a spoke is approximately \(9.42 \times 10^{-5} \text{ V}\).

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