CBSE-XI-Physics
38: Electromagnetic Induction
- #2A 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
(a) zero
(b)
12Blω2
(c) Blω2
(d) 2Blω2.digAnsr: aAns : (a) 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=B\omega xdx``
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.
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