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A disc of radius R rotates at an angular velocity omega about the axis perpendicular to its surface and passing through its centre. If the disc has a uniform surface charge density sigma, find the magnetic induction on the axis of rotation at a distance x from the centre. (Given R=3m, x=4m and mu_(0)sigmaomega=20Tesla//m.) |
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Answer» Solution :Consider a ring of RADIUS `r` and width `dr`. `dq=(2pirdr)sigma` Current due to the ring is `DI=(omegadq)/(2pi)=sigmaomegardr` Magnetic field due to this ring at point `P` is, `dB=(mu_(0)dIr^(2))/(2(r^(2)+X^(2))^(3//2))` or `intdB=(mu_(0)sigmaomega)/(2)int_(0)^(R )(r^(3)dr)/((r^(2)+x^(2))^(3//2))` Substituting `r^(2)+x^(2)=t^(2)`, and `2rdr=2tdt` and integrating we have `B-(mu_(0)sigmaomega)/(2)[(R^(2)+2X^(2))/(sqrt(T^(2)+x^(2)))-2x]`. 
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