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NEET PHYSICSELECTROMAGNETIC INDUCTIONEasy

Question

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}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTROMAGNETIC INDUCTION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSELECTROMAGNETIC INDUCTIONlengthrotatesuniformangularvelocity

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