Question:

A plane circular coil is rotated about its vertical diameter with a constant angular speed \(\omega\) in a uniform horizontal magnetic field. Initially the plane of the coil is parallel to the magnetic field. Draw the plot showing the variation of induced emf \(e\) in the coil as a function of \(\omega t\), where \(t\) represents the time elapsed.

Show Hint

Whenever \[ \phi=\phi_0\sin(\omega t), \] Faraday's law gives \[ e=-\frac{d\phi}{dt} =-e_0\cos(\omega t). \] Thus, the induced emf leads or lags the flux by \(90^\circ\) depending upon the chosen sign convention.
Show Solution
collegedunia
Verified By Collegedunia

Solution and Explanation

Concept: According to Faraday's law of electromagnetic induction, \[ e=-\frac{d\phi}{dt}. \] The induced emf is equal to the negative rate of change of magnetic flux. Since the magnetic flux varies sinusoidally, the induced emf will also vary sinusoidally but will be phase shifted by \(90^\circ\).

Step 1:
Write the expression for magnetic flux. From part (a), \[ \phi=\phi_0\sin(\omega t). \]

Step 2:
Apply Faraday's law. Using \[ e=-\frac{d\phi}{dt}, \] we obtain \[ e = -\frac{d}{dt} \left[ \phi_0\sin(\omega t) \right]. \] Differentiating, \[ e = -\phi_0\omega\cos(\omega t). \] Let \[ e_0=\phi_0\omega. \] Hence, \[ \boxed{e=-e_0\cos(\omega t)}. \]

Step 3:
Determine important points of the graph. At \[ \omega t=0, \] \[ e=-e_0. \] At \[ \omega t=\frac{\pi}{2}, \] \[ e=0. \] At \[ \omega t=\pi, \] \[ e=+e_0. \] At \[ \omega t=\frac{3\pi}{2}, \] \[ e=0. \] At \[ \omega t=2\pi, \] \[ e=-e_0. \] Thus the graph is a negative cosine curve. Required Plot: \[ e=-e_0\cos(\omega t) \] \[ \begin{array}{c} \text{Induced emf }(e) e_0 \quad\quad\quad\quad\bullet \quad\quad\quad / \backslash \quad\quad\quad/ \quad \backslash 0 \quad\bullet\quad\quad\quad\quad\bullet\quad\quad\quad\quad\bullet \quad\quad\quad\backslash \quad / \quad\quad\quad \backslash / -e_0 \bullet\quad\quad\quad\quad\quad\quad\quad\quad\bullet \end{array} \] \[ 0 \qquad \frac{\pi}{2} \qquad \pi \qquad \frac{3\pi}{2} \qquad 2\pi \] along the \(\omega t\)-axis. The graph starts from \(-e_0\), reaches zero at \(\frac{\pi}{2}\), becomes \(+e_0\) at \(\pi\), and then repeats periodically.
Was this answer helpful?
0
0

Top CBSE CLASS XII Faraday’s Law of Induction Questions

View More Questions