A thin non-conducting disc of radius 10 cm has a charge 2 mC is distributed uniformly over its surface. The disc is rotated with a constant angular velocity 100 rad/s about its axis. The magnetic field at the centre of the disc is $k \times 10^{- 7} T,$ where k is ___________________.

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answer is 2.

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Detailed Solution

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Surface charge density $σ = \frac{Q}{π R^{2}}$
$d q = \frac{Q}{π R^{2}} \cdot 2 π r d r = \frac{2 Q r d r}{R^{2}}$
$I = \frac{d q}{2 π / ω} = \frac{Q ω r}{π R^{2}} d r$
$d B = \frac{μ_{0} I}{2 r} = \frac{μ_{0} Q ω}{2 π R^{2}} d r$
$B = \frac{μ_{0} Q ω}{2 π R^{2}} \int_{0}^{R} d r = \frac{μ_{0} Q ω}{2 π R^{}} = \frac{(2 \times 10^{- 7}) (10^{- 3}) (100)}{(0.1)} = 2 \times 10^{- 7}$
$k \times 10^{- 7} T$
$k = 2$