An AC rms voltage of 2 V having a frequency of 50 kHz is applied to a condenser of capacity of 10 μF. The maximum value of the magnetic field between the plates of the condenser, if the radius of plate is 10 cm is

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0.4 $π$ μT

4 $π$ μT

$4 \sqrt{2}$ $π$ μT

40 $π$ μT

answer is C.

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Detailed Solution

$\begin{array}{rcl} i_{0} & = & \frac{V_{0}}{X_{C}} \\ = & \frac{\sqrt{2} V_{rms}}{X_{C}} \\ = & \frac{\sqrt{2} V_{rms}}{\frac{1}{ωc}} \\ = & \sqrt{2} V_{rms} ω c \\ = & \sqrt{2} \times 2 \times 2 π \times 50 \times 10^{3} \times 10 \times 10^{- 6} \\ = & 2 π \sqrt{2} Amp \\ \oint \bar{B} . \bar{d l} & = & μ_{0} i_{0} \\ \Rightarrow & B . 2 πR = μ_{0} i_{0}; \\ \Rightarrow & B . 2 π 10^{- 1} = 4 π \times 10^{- 7} \times 2 π \sqrt{2} \\ \Rightarrow & B = 4 \sqrt{2} π μT \end{array}$