Question

A conducting rod $AB$ of mass $M$ and length $L$ is hinged at its end $A$. It can rotate freely in the vertical plane (in the plane of figure). A long straight wire is vertical and carrying current $I$. The wire passes very close to $A$. If the rod is released from its vertical position of unstable equilibrium, choose the correct statements at the instant shown in the figure.

• A. The emf between the ends of the rod when it has rotated through an angle $\theta$. (see figure).
  $E=\frac{{\mu }_{0}I}{2\pi \sin \theta }\sqrt{3gL(1-\cos \theta )}$
  
• B. The emf between the ends of the rod when it has rotated through an angle $\theta$. (see figure)
  $E=\frac{{\mu }_{0}I}{\pi \sin \theta }\sqrt{gL(1-\cos \theta )}$
• C. The angular velocity at that instant is,
  $\omega =\sqrt{\frac{3g(1-\cos \theta )}{L}}$
• D. The angular velocity at that instant is, $\omega =\sqrt{\frac{g(1-\cos \theta )}{L}}$

Answer 1

Answer: (A), (C)

The correct options are
A The emf between the ends of the rod when it has rotated through an angle $\theta$. (see figure).
$E=\frac{{\mu }_{0}I}{2\pi \sin \theta }\sqrt{3gL(1-\cos \theta )}$

C The angular velocity at that instant is,

$\omega =\sqrt{\frac{3g(1-\cos \theta )}{L}}$

If we apply the energy conservation principle to find the angular speed $\omega$ of the rod,
$\frac{1}{2}{I}_A{\omega }^2=$ loss in gravitational potential energy
$\frac{1}{2}\left(\frac{M{L}^2}{3}\right){\omega }^2=Mgh$
here $h$ is change in position of COM of rod
$\therefore h=\frac{L}{2}[1-\cos \theta ]$

$\Rightarrow \frac{1}{2}\left(\frac{M{L}^2}{3}\right){\omega }^2=Mg\frac{L}{2}[1-\cos \theta ]$
$\Rightarrow \omega =\sqrt{\frac{3g(1-\cos \theta )}{L}}$

Now, consider an element of length $dx$ on the rod.
Speed of the element is, $v=\omega x$
Magnetic field at the location of the element due to current carrying wire is,
$B=\frac{{\mu }_{0}I}{2\pi d}=\frac{{\mu }_{0}I}{2\pi x\sin \theta }$
$\therefore$ EMF induced in the element is,
$dE=Bvdx=\frac{{\mu }_{0}I\omega }{2\pi \sin \theta }dx$
Emf in the rod is, $E=\int dE=\frac{{\mu }_{0}I\omega }{2\pi \sin \theta }{\int }_0^{L}dx$

$\Rightarrow E=\frac{{\mu }_{0}I\omega L}{2\pi \sin \theta }$
$E=\frac{{\mu }_{0}I}{2\pi \sin \theta }\sqrt{3gL(1-\cos \theta )}$