Question

An alternator consists of a coil of area A with N turns that rotates in a uniform field B around a diameter perpendicular to the field with a rotation frequency f.
a)find the fem in the coil

b)what is the amplitude of the alternating voltage if N= 100 turns, A = 10 ^ -2 m ^ 2, B = 0.1T, f = 2000 rev/min

Answer 1

Answer:

(a): $\rm 2\pi f\ NBA\ \sin(2\pi ft).$.

(b): 20.94 Volts.

Explanation:

Given:

• Area of the coil = A.
• Number of turns of the coil = N.
• Magnetic field in which the coil is placed = B.
• The frequency with which the coil is rotating = f.

(a):

The magnetic flux linked with a coil is defined as

$\phi = N\vec B \cdot \vec A=\rm NBA\cos\theta$

where,

• $\vec A$ is the area vector of the coil, directed along the normal to the plane of the coil.

• $\theta$ = angle between the magnetic field and the area vector of the coil.

Assuming that the magnetic field is along the normal to the plane of the coil, initially.

Therefore, at any later time t, the angle which the magnetic field makes with the normal to the plane of the coil is is given by

$\rm \theta = 2\pi ft.$

Therefore, the magnetic flux linked with the coil at any time t is given by

$\rm \phi = NBA\cos(2\pi ft).$

According to the Faraday's law of electromagnetic induction, the emf induced in the coil is given by

$\rm e=-\dfrac{d\phi}{dt}\\=-\dfrac{d(NBA\cos(2\pi ft))}{dt}\\=-NBA\dfrac{d(\cos(2\pi ft))}{dt}\\=-NBA(-2\pi f\sin(2\pi ft))\\=2\pi f\ NBA\ \sin(2\pi ft).$

(b):

The amplitude of the alternating voltage is the maximum value of the induced emf in the coil, the induced emf in the coil is maximum when $\rm \sin(2\pi ft)=1$.

Therefore, the amplitude of the alternating voltage is given by

$\rm e_o=(2\pi f)NBA\ 1\\=2\pi ft \ NBA.$

Given values are:

• N = 100 turns.
• A = $\rm 10^{-2}\ m^2$.
• B = 0.1 T.
• f = 2000 rev/min = $\rm 2000\times \dfrac1{60}\ rev/s = 33.33\ rev/s.$

Putting all these values,

$\rm e_o=2\pi \times 33.33\times (100)\times (0.1)\times (10^{-2})=20.94\ Volts.$