Question

A rod of length l rotates with a uniform angular velocity ω about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the two ends of the rod is ___________ .

Options

• zero

• `1/2Blomega^2`

• Blω²

• 2Blω²

Answer 1

zero

 

Let us consider a small element dx at a distance x from the centre of the rod rotating with angular velocity ω about its perpendicular bisector. The emf induced in the small element of the rod because of its motion is given by

`de = Bomegaxdx`

The emf induced between the centre of the rod and one of its end is given by

\[\int de = \int_0^l B\omega xdx\]

\[ \Rightarrow e = B\omega \left[ \frac{x^2}{2} \right]_0^{l/2} \]

\[ \Rightarrow e = \frac{1}{8}B\omega l^2\]

The emf at both ends is the same. So, the potential difference between the two ends is zero.