A long non conducting plane of width b having surface charge density $σ = σ_{0} x$ is moving with velocity $v = v_{0} \hat{j}$ . The magnetic field (in Tesla) at point (a, a, 0) will be : [ given $σ_{0} = \frac{4 π}{μ _{0}} C / m^{3}$ , $a = 2 m$ and $b = 1 m$ , $v_{0} = 2 m / s$ ]

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Ans (1.52 to 1.60)

This strip can be assumed as current carrying wire.

$d B = \frac{μ _{0} d i}{2 π ( a - x )}$

$d i = σ_{0} v_{0} x d x$

$\int d B = \frac{μ _{0}}{2 π} σ_{0} v_{0} \int_{0}^{b} \frac{x}{a - x} d x$

$B_{n e t} = \frac{μ _{0}}{2 π} σ_{0} v_{0} [(a - x) - a ln (a - x)]_{0}^{b}$

$B_{n e t} = \frac{μ _{0}}{2 π} σ_{0} v_{0} [a ln (\frac{a}{a - b}) - b]$

$B_{n e t} = \frac{μ _{0}}{2 π} \times \frac{4 π}{μ _{0}} \times 2 [2 ln 2 - 1]$

$B_{n e t} = 4 [0.386] = 1.544$

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