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 posits that the direction of an induced electromotive force (emf) opposes the very cause that produces it. This relationship is mathematically represented as: \[ \mathcal{E} = -\frac{d\Phi}{dt} \] EMF Induced in a Rotating Rod:
Consider a rod of length \( L \) rotating with angular velocity \( \omega \) in a uniform magnetic field \( \vec{B} \). The rotation is about one end (M), and the magnetic field is aligned with the axis of rotation, thus perpendicular to the plane of rotation. For a small segment of the rod at a distance \( x \) from the axis, its linear velocity is given by: \[ v = \omega x \] The small emf induced in this element is: \[ d\mathcal{E} = B \cdot v \cdot dx = B \cdot \omega x \cdot dx \] The total emf induced across the entire rod is obtained by integrating this expression from one end to the other: \[ \mathcal{E} = \int_0^L B \omega x \, dx = B \omega \int_0^L x \, dx = B \omega \left[ \frac{x^2}{2} \right]_0^L = \frac{1}{2} B \omega L^2 \] Final Expression: \[ \boxed{ \mathcal{E} = \frac{1}{2} B \omega L^2 } \]