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

(b): Amplitude of alternating voltage = 20.942 Volts.

Explanation:

Given:

• Area of the coil = A.
• Number of turns of coil = N.
• Magnetic field = B
• Rotation frequency = f.

(a):

The magnetic flux through the coil is given by

$\phi = \vec B \cdot \vec A=BA\cos\theta$

where,

$\vec A$ = area vector of the coil directed along the normal to the plane of the coil.

$\theta$ = angle between $\vec B$ and $\vec A$.

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

At any time t, the angle which magnetic field makes with the normal to the plane of the coil is $2\pi ft$

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

$\phi(t) = NBA\cos(2\pi ft)$

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

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

(b):

The amplitude of the alternating voltage is the maximum value of the emf and emf is maximum when $\sin(2\pi ft)=1.$

Therefore, the amplitude of the alternating voltage is given by

$e_o = 2\pi ft NBA.$

We have,

$N=100\\A=10^{-2}\ m^2\\B=0.1\ T\\f=2000\ rev/ min = 2000\times \dfrac{1}{60}\ rev/sec=33.33\ rev/sec.$

Putting all these values,

$e_o = 2\pi \times 33.33\times 100\times 0.1\times 10^{-2}=20.942\ Volts.$