Answer:
The voltage across the capacitor = 0.8723 V
Step-by-step 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:

= 0 - BA cos 0°
![\delta \phi =-BA \\ \\ \delta \phi =-B( \pi r^2) \\ \\ \delta \phi =(1.0 \ mT)[ \pi ( (d)/(2) )^2]](https://img.qammunity.org/2021/formulas/physics/college/gayske07nsvtf31cvcm1k24xbld1t1dbaw.png)
![\delta \phi =(1.0 \ mT)[ \pi ( (0.01)/(2) )^2]](https://img.qammunity.org/2021/formulas/physics/college/9ixk5zl8jf8x0x7xihvbwljmuuvbi1hcn8.png)

Using Faraday's Law; the induced emf on N turns of coil is;

Also; the induced current I =





The voltage across the capacitor can now be determined as:

=

= 0.8723 V