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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Waves
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A parallel plate capacitor as shown in the figure is made of circular plates each of radius $R = 6\: cm$, and it has a capacitance $C = 100\: pF$. The capacitor is connected to a $230\: V$ ac supply with a frequency of 300 rad/s. The amplitude $B$ at a point $3\: cm$ from the axis between the plates is

 

$\begin {array} {1 1} (a)\;16.3 \times 10^{-11}T & \quad (b)\;16.3 \times 10^{-13}T \\ (c)\;1.63 \times 10^{-11}T & \quad (d)\;1.63 \times 10^{-13}T \end {array}$

 

Can you answer this question?
 
 

1 Answer

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$R = 6 \: cm = 0.06 \: m, r= 3\: cm =0.03 \: m$
$C = 100 \: pF = 100 \times 10^{-12}F$
$V = 230\: V$
$\omega = 300 \: rad/s$
rms value of conduction current,$ I = \large\frac{V}{X_c}$
$X_c =$ Capacitive resistance = $ \large\frac{1}{\omega C} $
So, I = V × ωC.
$= 230 \times 300 \times 100 \times 10^{-12}$
$= 6.9 \times 10^{-6} A = 6.9\: \mu A$
Magnetic field is $B = \large\frac{\mu_or I_o}{2\pi R^2}$
where,$ I_o$ = maximum value of current = $ \sqrt{2} I$
$=\large\frac{ 4\pi \times 10^{-7} \times 0.03 \times \sqrt2 \times 6.9 \times 10^{-6}}{ 2\pi \times (0.06)^2 }$
$= 1.63 \times 10^{-11}T$
Ans : (c)

 

answered Mar 4, 2014 by thanvigandhi_1
edited Oct 8, 2014 by thagee.vedartham
 

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