Question

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.30 Ω is in a 1.0 mT magnetic field, with the coil oriented for maximum flux. The coil is connected to an uncharged 3.0 μF capacitor rather than to a current meter. The coil is quickly pulled out of the magnetic field.
Afterward, what is the voltage across the capacitor?

Answer 1

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:

$\delta \phi = \phi _f - \phi _i$

    = 0 - BA cos 0°

$\delta \phi =-BA \\ \\ \delta \phi =-B( \pi r^2) \\ \\ \delta \phi =(1.0 \ mT)[ \pi ( \frac{d}{2} )^2]$

$\delta \phi =(1.0 \ mT)[ \pi ( \frac{0.01}{2} )^2]$

$\delta \phi = -7.85*10^8 \ Wb$

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

$\epsilon = N |\frac{ \delta \phi}{\delta t} |$

Also; the induced current I = $\frac{ \epsilon}{R}$

$\frac{ \delta q}{ \delta t} = \frac{1}{R} (N|\frac{\delta \phi}{\delta t } |)$

$\delta q = \frac{1}{R} N|\delta \phi |$

$\delta q = \frac{1}{0.30} (10)| -7.85*10^8 \ Wb |$

$\delta q = 2.617*10^{-6} \ C$

The voltage across the capacitor can now be determined as:

$\delta \ V =\frac{ \delta q}{C}$

= $\frac{ 2.617*10^{-6} \ C}{3.0 \ *10^{-6} F}$

= 0.8723 V