A rod of length $L$ rotates with a small uniform angular velocity $\omega$ about its perpendicular bisector. A

Question

A rod of length $L$ rotates with a small uniform angular velocity $\omega$ about its perpendicular bisector. A uniform magnetic field $B$ exists parallel to the axis of rotation. The potential difference between the centre of the rod and an end is:

Accepted Answer

The motional emf induced in a rod rotating in a magnetic field is given by the formula $\varepsilon = \frac{1}{2} B \omega r^2$, where $r$ is the length of the rod segment from the axis of rotation to the tip.

1.  **Identify the Axis and Length:** The rod rotates about its perpendicular bisector (center). Therefore, the distance from the axis of rotation (center) to one end of the rod is $r = \frac{L}{2}$.
2.  **Apply the Formula:** Substitute $r = L/2$ into the standard equation:
    $\varepsilon = \frac{1}{2} B \omega \left( \frac{L}{2} \right)^2$
3.  **Calculate:**
    $\varepsilon = \frac{1}{2} B \omega \left( \frac{L^2}{4} \right) = \frac{B \omega L^2}{8}$

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