Question

A rectangular coil of N turns, length a, width b is rotated at a frequency f in a uniform magnetic field, B. The ends of the coil are connected to sliding rings that keep the same end of the wire coil connected to the probes of a voltage measuring device. (a) Determine the expression for the voltage from the system as a function of time. (b) What is the magnitude of the voltage maximum? (c) If the following values are given determine the area of the coils if the maximum voltage is 60 V. N = 500, f = 60 rev/s, B = 0.200 T

Answer 1

In order to solve the problem it is necessary to take into account the concepts of Magnetic Flow, Magnetic Field, Angular frequency and induced voltage.

By definition we know that the magnetic flux is defined as

$\Phi = B*ACos\theta$

Where,

B = Magnetic Field

A = Cross-sectional area

$\theta =$ Angle between the axis of the lopp and the field direction

The angle is given as a function of frequency and angular velocity, therefore

$\theta = \omega t$

In this way we can rewrite the equation we had as,

$\phi = BAcos(\omega t)$

PART A) By Faraday's law we know that the induced voltage emf is given by the expression,

$\epsilon_{emf} = -N\frac{d\Phi}{dt}$

Therefore replacing we have,

$\epsilon_{emf} = -N \frac{d(BAcos\theta)}{dt}$

$\epsilon_{emf} = -NBA \frac{d(cos(\omega t))}{dt}$

The final expression would be,

$\epsilon_{emf} = NBA\omega sin(\omega t)$

Where,

$\omega =$ Angular Velocity

N = Number of loops

A = Cross-sectional area

B = Magnetic Field

t = time

PART B) To find the maximum voltage $\omega t = 90$

Therefore $sin(90) = 1$ as well

$\epsilon_{emfMAX} = NAB*2\omega$

PART C) With the values given then the maximum area is,

$\epsilon_{max} = 500 * A * 0.2 * 2*3.14 * 60$

Re-arrange to find A,

$A = \frac{60}{(500 * 0.2* 2*3.14 * 60)}$

$A = 1.59*10^{-3} m^2$