Problem

The back e.m.f. in a motor is $120 \, \mathrm{V}$ when the motor is turning at $1680 \, \mathrm{rev/min}$. What is the back e.m.f. when the motor turns at $3360 \, \mathrm{rev/min}$?

Data

• Initial back e.m.f.: $\varepsilon_i = 120 \, \mathrm{V}$
• Initial angular velocity: $\omega_i = 1680 \, \mathrm{rev/min}$
• Final angular velocity: $\omega_f = 3360 \, \mathrm{rev/min}$

To find:

• Final back e.m.f. ($\varepsilon_f$).

Prerequisite Concepts

1. Back e.m.f. is proportional to the angular velocity of the motor:

$$\varepsilon = N B A \omega,$$

where:

• $\varepsilon$ is the back e.m.f.,
• $N$ is the number of turns,
• $B$ is the magnetic field strength,
• $A$ is the area of the loop,
• $\omega$ is the angular velocity.

2. For two states:

$$\frac{\varepsilon_f}{\varepsilon_i} = \frac{\omega_f}{\omega_i}.$$

Solution

1. Use the proportional relationship for back e.m.f.:

$$\varepsilon_f = \varepsilon_i \frac{\omega_f}{\omega_i}.$$

2. Substitute the given values:

$$\varepsilon_f = 120 \cdot \frac{3360}{1680}.$$

3. Simplify:

$$\varepsilon_f = 120 \cdot 2 = 240 \, \mathrm{V}.$$

Answer

The back e.m.f. when the motor turns at $3360 \, \mathrm{rev/min}$ is:

$$\varepsilon_f = 240 \, \mathrm{V}.$$