Engineering
Physics
Physical Quantities
Question

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

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Solution
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 dB=µ0dI2r

 Integrating we get B1R

 dp=σ(2πrdr)=2QrdrR2

 dI=(dq)ω2π=QωrdrπR2

 

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