Question

The rectangular coils in a 360-turn generator are 10 cm by 19 cm Part A What is the maximum emf produced by this generator when it rotates with an angular speed of 525 rpm in a magnetic field of 0.40 T? Express your answer using two significant figures. IVO ΑΣΦ ? Emax kV

Answer 1

We have,
Number of turns in coil
N= 360

Area of the coil
+A= 10 cm x 19 cm 190 cm2 = 190 x 10-4 m

Angular velocity of the coil
— w = 525 трт

Magnetic field $\rightarrow B=0.40\ T$

The angular frequency $\omega$ in SI unit:

$\omega= \frac{525\ rev}{min}\cdot \frac{1\ min}{60\ sec}\cdot \frac{2\pi \ rad}{1\ rev}$

$\therefore \omega= 54.98\ rad/sec$

The induced emf is given as:

$e=e_{0}\ sin\ \omega t$

Where $e_{0}$ is the maximum value of induced emf when $sin \ \omega t=\pm 1$

$\therefore e=e_{0}$

$\therefore e=NAB\ \omega$

$\therefore e=900\times190\times10^{-4}\times0.40\times54.98$

$\therefore e=376.0632\ V$

$\therefore e=376.0632\times10^{-3}\ kV$

${\color{Blue} \therefore e=0.3761\ kV}$

${\color{Blue}\therefore e=0.4\ kV}$

Therefore, the maximum induced emf in the coil is ${\color{Blue} e=0.4\ kV}$ which in two significant figures.