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A non0conducting disk of radius R has surface charge density which varies with distance from the centre as sigma (r ) = sigma _(0)[1+ sqrt((r )/(R ))], where sigma _(0) is a constant. The disc rotates about its axis with angular velocity omega. If B is the magnitude of magnetic induction at the centre, then (B)/(mu_(0) sigma _(o)omega R) will be |
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Answer» `3/4`  Charge on ring `dq = `Area of ring `xx` Charge density `implies dq =(2pi r. dr) sigma` `implies` CURRENT through ring, di `= (dq)/(T),` as angular VELOCITY of disc is `omega.` So, `(2pi)/(omega) =T` `implies di = (omega dq)/(2pi) = (omega (2pi r. dr)sigma)/(2pi) = omega r sigma dr` Now, manetic INDUCTION due to elemental ring at thte centre, `DB= (mu_(0)di)/(2r) =(mu_(0) omega r sigma)/(2r).dr = (mu_(0) omega sigma)/(2).dr` As` sigma = sigma _(0) (1+ sqrt((r)/(R)))` For magnetic induction due to entire dusc, we integrate from `r =0` to `r =R` `implies B = (mu_(0) omega sigma _(0))/(2) int _(0) ^(pi) (1+ sqrt((r)/(R))).dr` IInd part `implies B= (mu_(0) omega sigma _(0))/(2) [r+ (2r ^(3//2))/(3 sqrtR)]_(0) ^(R)` `implies B = (mu_(0) omega sigma _(0))/(2) [ R = (2)/(3)R]` `implies B= 5/6 mu _(0)omega sigma _(0) R implies (B)/(mu_(0) omega sigma _(0)R) =5/6` |
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