Question

State Lenz’s law. A rod MN of length \( L \) is rotated about an axis passing through its end M perpendicular to its length, with a constant angular velocity \( \omega \) in a uniform magnetic field \( \vec{B} \) parallel to the axis. Obtain an expression for emf induced between its ends.

Hint:

In rotating rods, each element contributes to the total emf due to varying linear velocity \( v = \omega x \). Integrating gives the total emf.

Answer 1

Lenz’s Law: Lenz’s law states that the direction of induced emf is such that it opposes the cause producing it. Mathematically, this is expressed as: \[ \mathcal{E} = -\frac{d\Phi}{dt} \] EMF Induced in a Rotating Rod:
Let a rod of length \( L \) rotate in a uniform magnetic field \( \vec{B} \), with angular velocity \( \omega \), about one of its ends (M). The magnetic field is parallel to the axis, i.e., perpendicular to the plane of rotation. Consider a small element at a distance \( x \) from the axis. Its linear velocity: \[ v = \omega x \] Small emf induced in this element: \[ d\mathcal{E} = B \cdot v \cdot dx = B \cdot \omega x \cdot dx \] Total emf across the rod: \[ \mathcal{E} = \int_0^L B \omega x \, dx = B \omega \int_0^L x \, dx = B \omega \left[ \frac{x^2}{2} \right]_0^L = \frac{1}{2} B \omega L^2 \] Final Expression: \[ \boxed{ \mathcal{E} = \frac{1}{2} B \omega L^2 } \]