Question

A 1000-turns coil of cross-sectional area 30cm^2 rotates at a frequency of 120 Hz in a magnetic field 0.1 T. Calculate the peak value of the induced emf.

a. 346.4V
b. 112.3V
c. 97.6V
d. 226.2V

Answer 1

Question 32839

$$N = 1000, \quad B_0 = 0.1\,T, \quad v = 120\,Hz, \quad S = 30\,cm^2.$$

The induced emf, according to Faraday's law is $\varepsilon = \frac{-\partial\Phi}{\partial t}$ .

The magnetic flux might be expressed as $\Phi(t) = NBS$ . It is more convenient to let the magnetic field rotate (not the coil). Thus, magnetic field as the function of time is

$$B(t) = B_0 \cos(\omega t) = B_0 \cos(2\pi v t).$$

Then, EMF as a function of time is $\varepsilon(t) = 2\pi B_0 \nu N S \cdot \sin(2\pi \nu t)$ .

Obviously, the peak value reaches when sine function is equal to one (the amplitude).

Hence, the peak value of induced EMF is $\varepsilon_{max} = 2\pi B_0 N S v = 226.195 V \approx 226.2 V$ .

The answer is d)