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NEET PHYSICSEasy

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

A

BωL28\frac{B\omega L^2}{8}

B

BωL22\frac{B\omega L^2}{2}

C

BωL24\frac{B\omega L^2}{4}

D

zero

Step-by-Step Solution

The motional emf induced in a rod rotating in a magnetic field is given by the formula ε=12Bωr2\varepsilon = \frac{1}{2} B \omega r^2, where rr 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=L2r = \frac{L}{2}.
  2. Apply the Formula: Substitute r=L/2r = L/2 into the standard equation: ε=12Bω(L2)2\varepsilon = \frac{1}{2} B \omega \left( \frac{L}{2} \right)^2
  3. Calculate: ε=12Bω(L24)=BωL28\varepsilon = \frac{1}{2} B \omega \left( \frac{L^2}{4} \right) = \frac{B \omega L^2}{8}
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