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Q.

A charge of 1mC is placed at one end of a non-conducting rod of length 12 cm. The rod is rotated in a vertical plane about a horizontal axis passing through the other end of rod with angular frequency $100 π r a d / s .$ The magnetic field at a point in the axis of rotation at a distance of 16 cm from the centre of path is $π k^{2} \times 10^{- 8} T$ Where k is _______________.

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answer is 6.

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$I = \frac{q}{T} = \frac{q}{2 π / ω} = \frac{q ω}{2 π} = \frac{10^{- 3} \times 10 π}{2 π} = 0.05 A$
$B = \frac{μ_{0} I R^{2}}{2 {[R^{2} + x^{2}]}^{3 / 2}} = \frac{μ_{0} I {(0.12)}^{2}}{2 {[{(0.12)}^{2} + {(0.16)}^{2}]}^{3}}$
$= \frac{4 π \times 10^{- 7} \times 144 \times 10^{- 4}}{2 \times 8 \times 10^{- 3}} = 36 π \times 10^{- 8} T = π k^{2} \times 10^{- 8}$
$k = 6$

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