Question

A rod of mass $m$ and length $\ell$ is released from the position shown with its upper end hinged in a uniform horizontal magnetic field $B$. Find the maximum induced emf in the rod: 

• A. $B\ell\sqrt{\dfrac{3g\ell}{8}}$
• B. $B\ell\sqrt{\dfrac{g\ell}{8}}$
• C. $B\ell\sqrt{\dfrac{7g\ell}{8}}$
• D. $B\ell\sqrt{\dfrac{5g\ell}{8}}$

Hint:

For rotating conductors in a magnetic field, first find the maximum angular speed using energy conservation, then apply $\mathcal{E}_{\max}=\dfrac{1}{2}B\ell^2\omega$.

Answer 1

The Correct Option is A

Concept: Key ideas involved:
Conservation of mechanical energy for a rotating rod
Moment of inertia of a rod about one end: $I=\dfrac{1}{3}m\ell^2$
Motional emf induced in a rotating rod in a magnetic field The maximum emf is induced when the angular speed of the rod is maximum, i.e., when the rod passes through the vertical position.
Step 1: Loss of gravitational potential energy The centre of mass of the rod is at a distance $\dfrac{\ell}{2}$ from the hinge. Initial angle with vertical is $60^\circ$. Vertical fall of centre of mass: \[ h=\frac{\ell}{2}(1-\cos60^\circ) = \frac{\ell}{2}\left(1-\frac{1}{2}\right) = \frac{\ell}{4} \] Loss in potential energy: \[ \Delta U = mg\frac{\ell}{4} \]
Step 2: Rotational kinetic energy at lowest position \[ \frac{1}{2}I\omega^2 = mg\frac{\ell}{4} \] Using $I=\dfrac{1}{3}m\ell^2$: \[ \frac{1}{2}\cdot\frac{1}{3}m\ell^2\omega^2 = mg\frac{\ell}{4} \] \[ \omega^2 = \frac{3g}{2\ell} \]
Step 3: Expression for induced emf For a rod rotating in a uniform magnetic field perpendicular to the plane of motion: \[ \mathcal{E}_{\max}=\frac{1}{2}B\ell^2\omega \] Substitute $\omega$: \[ \mathcal{E}_{\max}=\frac{1}{2}B\ell^2\sqrt{\frac{3g}{2\ell}} = B\ell\sqrt{\frac{3g\ell}{8}} \] Step 4: Thus, the maximum induced emf is: \[ \boxed{B\ell\sqrt{\dfrac{3g\ell}{8}}} \]