Verified Solution • 2025–26 Session moderate

QJEEM12Ph0f05: JEE Main 2012 Physics - Class 12 JEEM12Ph0f05 Solution: A charge is uniformly distributed over the surface of…

A charge QQ is uniformly distributed over the surface of non-conducting disc of radius R.R. The disc rotates about an axis perpendicular to its plane and passing through its center with an angular velocity ω.\omega . As a result of this rotation a magnetic field of induction BB is obtained at the center of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and very the radius of the disc then the variation of the magnetic induction at the center of the disc will be represented by the figure :

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/c0c54863cd47ed8448bba71cf1038806d86b98eae07ad027a13174651c9c3c35.webp
  • B https://ik.imagekit.io/novatra/photos/7c5f2bdaa91426df7db92afc60fc357309a58fdc1e3ea9656357f55d102b2583.webp
  • C https://ik.imagekit.io/novatra/photos/6a6fb74a819bc3c0e85b372a912141824b07c5c260d47f0bfd55f95ea3decf6e.webp
  • D https://ik.imagekit.io/novatra/photos/b015e19fc2f8a45b7259e4aa3ee28e1cbea7c54546eca3548459c0776f03a230.webp

Step-by-Step Verified Solution

The magnetic field due a disc is given as

B=h0ωQ2πRB = {{{h_0}\omega Q} \over {2\pi R}} i.e., B∝1RB \propto {1 \over R}

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