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ExamsJEE AdvancedPhysics

A composite rod of total length l rotates freely with angular velocity w about one end in a region free of gravity and of any external electric or magnetic field. The portion adjacent to the rotation axis, of length l/4, is an insulator, while the remaining outer portion (from l/4 to l) is a conductor. Due to the rotation, free electrons in the conducting part redistribute until equilibrium. Find the potential difference developed across the conducting region. (mₑ = electron mass, e = electronic charge magnitude.)

  1. 3*mₑ*w²*l²/(4e)
  2. (1/4)*mₑ*w²*l²/e
  3. (1/16)*mₑ*w²*l²/e
  4. 15*mₑ*w²*l²/(32e)

Correct answer: 15*mₑ*w²*l²/(32e)

Solution

When the conducting rod spins, the free electrons require a centripetal force to follow circular paths. This force is provided by an internal electric field that arises from a slight charge separation. At equilibrium, eE = mₑ*w²*r, so E(r) = mₑ*w²*r/e directed inward. The potential difference between the two ends of the conducting segment is the integral of E dr from r = l/4 to r = l. Integrating r dr gives (r²)/2, evaluated between l and l/4, yielding (l² - l²/16)/2 = (15/32) l². Hence V = (15/32) mₑ*w²*l²/e.

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