Question

An elliptical loop having resistance R, of semi-major axis a, and semi-minor axis b is placed in the magnetic field as shown in the figure. If the loop is rotated about the x-axis with angular frequency \[\omega \], Find the average power loss in the loop due to Joule heating.

A. \[\dfrac{{\pi abB\omega }}{R}\]
B. \[\dfrac{{{\pi ^2}{a^2}{b^2}{B^2}{\omega ^2}}}{R}\]
C. \[\dfrac{{{\pi ^2}{a^2}{b^2}{B^2}{\omega ^2}}}{{2R}}\]
D. Zero

Answer 1

Hint: Induced emf is defined as the generation of a potential difference in a coil due to the change in the magnetic flux through it. In other words, electromotive Force or EMF is said to be induced when the flux linking with a conductor or coil changes. We are going to use the expressions of magnetic flux and average power loss to answer this question.

Formula Used:
The formula to find the flux is,
\[\phi = \overrightarrow B \cdot \overrightarrow A \]
Where, \[\overrightarrow B \] is magnetic field and \[\overrightarrow A \] is area of loop.
The formula to find the average power loss is,
\[P = \dfrac{{{e^2}}}{R}\]
Where, e is induced emf and R is resistance.

Complete step by step solution:

Image: An elliptical loop having resistance R, of semi-major axis a, and semi-minor axis b is placed in the magnetic field.

Consider an elliptical loop having resistance R, of semi-major axis a, and semi-minor axis b which is placed in the magnetic field as shown in the figure. When the loop is rotated about the x-axis with angular frequency\[\omega \], then we need to find the average power loss in the loop due to Joule heating. We know the formula to find the flux is,
\[\phi = \overrightarrow B \cdot \overrightarrow A \]
\[\Rightarrow \phi = BA\cos \theta \]
Since \[\omega \] is angular frequency, in place of \[\theta \] we write\[\omega t\], then the above equation will become,
\[\phi = BA\cos \omega t\]

We know that, the induced emf is,
\[e = - \dfrac{{d\phi }}{{dt}}\]
\[\Rightarrow e = - \dfrac{{d\left( {BA\cos \omega t} \right)}}{{dt}}\]
\[\Rightarrow e = BA\sin \omega t \times \omega \]……… (1)
We know that, the area of ellipse is,
\[A = \pi ab\]
Substitute the area of ellipse in equation (1) we get,
\[e = \pi abB\sin \omega t \times \omega \]
The average power loss is,
\[P = \dfrac{{{e^2}}}{R}\]
\[\Rightarrow P = {\dfrac{{{\pi ^2}{a^2}{b^2}{B^2}{{\sin }^2}\left( {\omega t} \right) \times \omega }}{R}^2}\]
The average value of \[{\sin ^2}\theta = \dfrac{1}{2}\]
Then,
\[P = {\dfrac{{{\pi ^2}{a^2}{b^2}{B^2}\omega }}{{2R}}^2}\]
Therefore, the average power loss in the loop due to Joule heating is \[{\dfrac{{{\pi ^2}{a^2}{b^2}{B^2}\omega }}{{2R}}^2}\].

Hence, option C is the correct answer.

Note: In order to resolve this problem, it is important to remember the formula for flux in a coil, induced emf, average power loss and area of the ellipse. Moreover, a measurement of the total magnetic field that traverses a specific region is called magnetic flux. It is a helpful tool for explaining how the magnetic force affects things inhabiting a certain region. The precise area selected will determine how magnetic flux is measured.