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Q.

A disc of radius R rotates with constant angular velocity about its own axis. Surface charge density of this disc varies as $σ = {αr}^{2}$ where, r is the distance from the centre of disc. Determine the magnetic field intensity at the centre of disc

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a

$μ_{0} {αωR}^{3}$

b

$\frac{μ_{0} {αωR}^{3}}{6}$

c

$\frac{μ_{0} {αωR}^{3}}{8}$

d

$\frac{μ_{0} {αωR}^{3}}{3}$

answer is B.

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Detailed Solution

$\begin{array}{l} dA = (2 πr) dr \\ dq = σ \cdot dA = (2 {παr}^{3}) \cdot dr \\ i = (dq) f = (dq) \frac{ω}{2 π} = ({αωr}^{3}) dr \\ dB = \frac{μ_{0} i}{2 r} = \frac{{αμ}_{0} {ωr}^{2} dr}{2} \\ B = \int_{0}^{R} dB = \frac{μ_{0} {αωR}^{3}}{6} \end{array}$

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