The square armature coil of an alternating current generator has 200 turns and is 20.0 cm on side. When it rotates at 3600 rpm, its peak output voltage is 120 V. A) What is the frequency of the output voltage?
B) What is the strength of the magnetic field in which the coil is turning?

A) 60 Hz

B) 0.04 T

Explanation:

Given that.

Number of turns, N = 200

Length of the side, l = 20 cm = 0.2 m

Speed if rotation, w = 3600 rpm

Voltage, V = 120 V

First, we try to convert the speed from rpm to rad/s

3600 * (2π/60)

3600 * 0.10473

3600 rpm = 377 rad/s

Now, we use that as our w, speed of rotation

Frequency of output, f =

w/2π

f = 377 / 6.284

f = 59.99 Hz or approximately, 60 Hz.

B

Strength of the magnetic field in which the coil is turning

E• = NABw

Where, A = l² = 0.2² = 0.04, on substituting the values to the equation, we have

120 = 200 * 0.04 * 377 * B

120

Making B subject of formula,

B = 120/ 3016

B = 0.04 T..

The frequency of the output voltage is 60 Hz and the strength of the magnetic field is 0.04 T