Problem

A wheel with 12 metal spokes, each 0.6 m long, is rotated at a speed of 180 r.p.m. in a plane normal to Earth’s magnetic field. If the magnitude of the magnetic field is $0.6 \, \mathrm{G}$, calculate the induced e.m.f. between the axle and the rim of the wheel.

Data

• Length of a spoke: $l = 0.314 \, \mathrm{m}$
• Radius of the wheel: $r = 0.6 \, \mathrm{m}$
• Angular velocity:

$$\omega = 180 \, \mathrm{r.p.m.} = \frac{180 \times 2\pi}{60} = 6\pi \, \mathrm{rad/s}$$

• Magnetic field: $B = 0.6 \, \mathrm{G} = 0.6 \times 10^{-4} \, \mathrm{T}$

To find:

• Induced e.m.f. ($\varepsilon$).

Prerequisite Concepts

1. The motional e.m.f. is given by:

$$\varepsilon = B l v$$

2. The relation between linear velocity ($v$) and angular velocity ($\omega$) is:

$$v = r \omega$$

Solution

1. Using the formula for motional e.m.f.:

$$\varepsilon = B l r \omega$$

2. Substituting the values:

$$\varepsilon = (0.6 \times 10^{-4}) \times 0.314 \times 0.6 \times (6 \pi)$$

3. Simplifying step-by-step:
   • $\varepsilon = (0.6 \times 10^{-4}) \times 0.314 \times 0.6 \times 18.84$
   • $\varepsilon = 2.129 \times 10^{-4} \, \mathrm{V}$.

Answer

The induced e.m.f. between the axle and the rim of the wheel is:

$$\varepsilon = 2.129 \times 10^{-4} \, \mathrm{V}.$$

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