Question

A disc of radius $R$ is rolling without sliding on a horizontal surface with a velocity of center of mass $v$ and angular velocity $\omega$ in a uniform magnetic field $B$ which is perpendicular to the plane of the disc as shown in figure. $O$ is the center of the disc and $P,Q,R$ and $S$ are the four points on the disc. Which of the following statements is true?

• A. Due to translation, induced emf across $PS=Bvr$
• B. Due to rotation, induced emf across $QS=0$
• C. Due to translation, induced emf across $RO=0$
• D. Due to rotation, induced emf across $OQ=Bvr$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Due to translation, induced emf across $PS=Bvr$
B Due to rotation, induced emf across $QS=0$
C Due to translation, induced emf across $RO=0$

Consider an infinitesimally small element on the surface of disc between point P and S of length $rd\theta$, where $\theta$ is the angle line connecting that point with center forms with horizontal.

Thus emf induced across the element=$Bvrd\theta \cos \theta$

Hence net emf between points P and S is ${\int }_0^{\pi /2}BvR\cos \theta d\theta =Bvr$

Since points R and O lie along same line of along velocity of translation, there is no emf induced across RO. And, for the same reason, there is no emf induced across QS and OQ.