A non-conducting disc of radius R has surface charge density which varies with distance from the centre as (r)= ₀ [1+ r R ] , where ₀ is a constant. The disc rotates about its axis with angular velocity . If B is the magnitude of magnetic induction at the centre, then B ₀ ₀ R will be

Step-by-step solution

Consider an elemental ring of radius r of thickness d r Charge on ring d q= Area of ring Charge density d q=(2 r d r) Current through ring, d i= d q T , as angular velocity of disc is . So, 2 =T d i= d q 2 = (2 r d r) 2 = r d r Now, magnetic induction due to elemental ring at the centre, d B= ₀ d i 2 r = ₀ or 2 r d r= ₀ 2 d r As = ₀ (1+ r R ) For magnetic induction due to entire disc, we integrate from r=0 to r=R B= ₀ ₀ 2 ₀^n (1+ r R ) d r IInd Part B= ₀ ₀ 2 [r+ 2 r^ 3 / 2 3 R ]₀^R aligned & B= ₀ ₀ 2 [R+ 2 3 R ] &