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 $$\omega .$$ 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 :
A.
B.
C.
D.
Answer :
Solution :
The magnetic field due to a disc is given as
$$\eqalign{ & B = \frac{{{\mu _0}\omega Q}}{{2\pi R}} \cr & {\text{i}}{\text{.e}}{\text{.,}}\,B \propto \frac{1}{R} \cr} $$
The magnetic field due to a disc is given as
$$\eqalign{ & B = \frac{{{\mu _0}\omega Q}}{{2\pi R}} \cr & {\text{i}}{\text{.e}}{\text{.,}}\,B \propto \frac{1}{R} \cr} $$
