-->

Wave Optics

Question
CBSEENPH12039531

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 ω. 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 than the variation of the magnetic induction at the centre of the disc will be represented by the figure

Solution

A.


Consider ring like the element of the disc of radius r and thickness dr.
If σ is charge per unit area, then charge on the element
dq = σ(2πr dr)
current ‘i’ associated with rotating charge dq is
straight i space equals space fraction numerator left parenthesis dp right parenthesis straight w over denominator 2 straight pi end fraction space equals space straight sigma space straight w space straight r space dr
Magnetic field dB at center due to element

dB space equals space fraction numerator straight mu subscript 0 straight i over denominator 2 straight r end fraction space equals space fraction numerator straight mu subscript 0 σωdr over denominator 2 end fraction
straight B subscript net space equals space integral dB space equals space fraction numerator straight mu subscript straight o σω squared over denominator 2 end fraction space integral subscript 0 superscript straight R dr space equals space fraction numerator straight mu subscript straight o σωR over denominator 2 end fraction
rightwards double arrow space straight B subscript net space equals fraction numerator straight mu subscript straight o Qω over denominator 2 πR end fraction
So if Q and w are unchanged then
straight B subscript net space proportional to 1 over straight R

Some More Questions From Wave Optics Chapter