A parallel plate capacitor in the figure made of circular plates each of radius R=6.0 cm has a capacitance C=100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad `s^(-1)` .

Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

Explanation

We know, `B=(mu_(0))/(2pi)xx(pi)/(R^(2))xxI_(d)`
The formula is valid even if `I_(d)` is oscillating . As `I_(d)=I`, therefore
`B=(mu_(0)rI)/(2piR^(2))`
If `I=I_(0)` , the maximum value of current, then
Amplitude of B=max. value of B=`(mu_(0)rI_(0))/(2piR^(2))=(mu_(0)rsqrt(2)I_("rms"))/(2piR^(2))`
`=(4pixx10^(-7)xx0.03xxsqrt(2)xx6.9 xx 10^(-6))/(2xx3.14xx(0.06)^(2))=1.63xx10^(-11)T.`