Answer:
The voltage across the capacitor = 0.8723 V
Explanation:
From the question, it is said that the coil is quickly pulled out of the magnetic field. Therefore , the final magnetic flux linked to the coil is zero.
The change in magnetic flux linked to this coil is:
[tex]\delta \phi = \phi _f - \phi _i[/tex]
= 0 - BA cos 0°
[tex]\delta \phi =-BA \\ \\ \delta \phi =-B( \pi r^2) \\ \\ \delta \phi =(1.0 \ mT)[ \pi ( \frac{d}{2} )^2][/tex]
[tex]\delta \phi =(1.0 \ mT)[ \pi ( \frac{0.01}{2} )^2][/tex]
[tex]\delta \phi = -7.85*10^8 \ Wb[/tex]
Using Faraday's Law; the induced emf on N turns of coil is;
[tex]\epsilon = N |\frac{ \delta \phi}{\delta t} |[/tex]
Also; the induced current I = [tex]\frac{ \epsilon}{R}[/tex]
[tex]\frac{ \delta q}{ \delta t} = \frac{1}{R} (N|\frac{\delta \phi}{\delta t } |)[/tex]
[tex]\delta q = \frac{1}{R} N|\delta \phi |[/tex]
[tex]\delta q = \frac{1}{0.30} (10)| -7.85*10^8 \ Wb |[/tex]
[tex]\delta q = 2.617*10^{-6} \ C[/tex]
The voltage across the capacitor can now be determined as:
[tex]\delta \ V =\frac{ \delta q}{C}[/tex]
= [tex]\frac{ 2.617*10^{-6} \ C}{3.0 \ *10^{-6} F}[/tex]
= 0.8723 V
