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Calculate the magnetic field at the center of a charged disc of radius ‘R’ having a uniform charge ‘Q’ surface charge density $' σ'$ rotating with angular velocity $ω$ . Also calculate current associated with disc

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$B = \frac{μ_{0} σ R ω}{2}$

$B = \frac{μ_{0} σ R ω}{4}$

$I = \frac{Q ω}{π}$

$I = \frac{Q ω}{2 π}$

answer is A, B.

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Detailed Solution

This rotating infinitesimal charge dq gives a current dI given by
$d I = \frac{d q}{π} = \frac{(\frac{Q}{π R^{2}}) (2 π r d r)}{(\frac{2 π}{ω})} \Rightarrow d I = (\frac{Q}{π R^{2}}) r d r$
Since, we know that $d B = \frac{μ_{0} (d I)}{2 r}$

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