Question

A circular coil of one turn of radius \[5.0cm\] is rotated about a diameter with a constant angular speed of 80 revolution per minute. A uniform magnetic \[B = 0.010T\] exists in a direction perpendicular to the axis of rotation. Find the maximum emf induced

Answer 1

Hint: We can recall that the magnitude of the emf induced on a conductor is directly proportional to the rate of change of magnetic flux passing through the conductor. The magnetic flux, recall, in its fundamental form is the dot product of the magnetic and the area of the loop of the conductor.
Formula used: In this solution we will be using the following formulae;
 \[E = - \dfrac{{d\phi }}{{dt}}\] where \[E\] is the emf induced in a conductor, \[\phi \] is the magnetic flux passing through the conductor and \[t\] is time. \[\dfrac{{d\phi }}{{dt}}\] signifies the instantaneous rate of change of the magnetic flux with time.
 \[\phi = \vec B \cdot \vec A = BA\cos \theta \] where \[\vec B\] is the magnetic field passing through the conductor, \[\vec A\] is the area of the loop of the conductor, and \[\theta \] is the angle between \[\vec B\] and \[\vec A\] . The symbol \[ \cdot \] signifies the dot product.

Complete Step-by-Step solution:
To calculate the emf, we recall that the emf can be given by faraday’s law as in
 \[E = - \dfrac{{d\phi }}{{dt}}\] where \[\phi \] is the magnetic flux passing through the conductor and \[t\] is time and \[\dfrac{{d\phi }}{{dt}}\] signifies the instantaneous rate of change of the magnetic flux with time.
But \[\phi = \vec B \cdot \vec A = BA\cos \theta \] where \[\vec B\] is the magnetic field passing through the conductor, \[\vec A\] is the area of the loop of the conductor, and \[\theta \] is the angle between \[\vec B\] and \[\vec A\] .
Then, the emf will be
 \[E = - \dfrac{{d\left( {BA\cos \theta } \right)}}{{dt}}\]
Now, according to the question, the magnetic field is constant, and the area is not changing either, only that the conductor is moving in a circle. This means that the angle between the area of the loop and the magnetic field lines is constantly changing, hence, we have
 \[E = - BA\dfrac{{d\left( {\cos \theta } \right)}}{{dt}}\]
 \[\theta = \omega t\]
Hence, we have
 \[E = - BA\dfrac{{d\left( {\cos \omega t} \right)}}{{dt}}\] , which by differentiating we have
 \[E = BA\omega \sin \omega t\]
Then, the maximum emf would be
 \[{E_{\max }} = BA\omega \]
Area of a the loop of the conductor is
 \[A = \pi {r^2}\] where \[r\] is the radius. Hence, by inserting all known values into the above equation we have
 \[{E_{\max }} = 0.01\left( {\pi \times {{0.05}^2}} \right)\left( {80 \times \dfrac{{2\pi }}{{60}}} \right)\] (converting the cm to metres, and rev per minute to radian per second)
By computation, we get
 \[{E_{\max }} = 0.00065797...{\text{V}}\]
 \[ \Rightarrow {{\text{E}}_{\max }} \approx 0.7{\text{mV}}\]

Note: For clarity, we conclude that the next maximum emf is \[{E_{\max }} = BA\omega \] from the general equation \[E = BA\omega \sin \omega t\] because the value of the sine of an angle can only be less than or equal to 1. Hence, the maximum is when it is equal to 1. In that case
 \[{E_{\max }} = BA\omega \] , meaning the equation can be written as
 \[E = {E_{\max }}\sin \omega t\]