Question

The magnetic field at one place on the earth's surface is $55\mu T$in strength and tilted ${60}^{\circ }$down from horizontal. A 200 turn coil having a diameter of 4.0 cm and a resistance of $2.0\mathrm{\Omega }$is connected to a $1.0\mu \mathrm{F}$capacitor rather than to a current meter. The coil is held in a horizontal plane and the capacitor is discharged. Then the coil is quickly rotated ${180}^{\circ }$so that the side that had been facing up is now facing down. Afterward, what is the voltage across the capacitor?

Short answer

The voltage across the capacitor,$\mathrm{\Delta }{V}_C=12\mathrm{V}$

Step-by-step solution

Magnetic Field

The flux density is a study of the magnet that runs through the area A coil. The magnetic field is carried out in a systematic way:

${\mathrm{\Phi }}_m=BA\cos \theta$

The diameter of the coil is d = 4 cm. So, the area of the coil is

$A=\pi {\left(\frac{d}{2}\right)}^2=\pi {\left(\frac{4\times {10}^{-2}\mathrm{m}}{2}\right)}^2=1.25\times {10}^{-3}{\mathrm{m}}^2$

A magnetic field is angled below the normal by, yielding in an angle at all between the magnetic force and the coil's normal line. At a certain length of time, the coil revolves by and so its angle gets. As an outcome, the magnetic field is varied from side to side, or the flow is modified.$d{\mathrm{\Phi }}_m=BA\left(\cos {30}^{\circ }-\cos {210}^{\circ }\right)$

Faraday's and Ohm's Law

From Faradays, the induced is the change of magnetic field in inside coil yet it is given by in the form

$\epsilon =N\frac{d{\mathrm{\Phi }}_{\mathrm{m}}}{dt}=\frac{NBA\left(\cos {30}^{\circ }-\cos {210}^{\circ }\right)}{dt}$

The induced current using Ohm's law

$I=\frac{\epsilon }{R}$

Capacitance of the Capacitor

the current is the change of capacitance in inside coil yet it is given by in the form

$\begin{array}{r}\frac{dq}{dt}=\frac{NBA\left(\cos {30}^{\circ }-\cos {210}^{\circ }\right)}{Rdt}\\ dq=\frac{NBA\left(\cos {30}^{\circ }-\cos {210}^{\circ }\right)}{R}\end{array}$

Related to the induced current in the capacitance to the capacitor

$\mathrm{\Delta }{V}_C=\frac{dq}{C}=\frac{NBA\left(\cos {30}^{\circ }-\cos {210}^{\circ }\right)}{RC}$

Those values provided by N, B, A, R and C$\mathrm{\Delta }{V}_C=\frac{NBA\left(\cos {30}^{\circ }-\cos {210}^{\circ }\right)}{RC}$

$=\frac{\left(200\right)\left(55\times {10}^{-6}\mathrm{T}\right)\left(1.25\times {10}^{-3}{\mathrm{m}}^2\right)\left(\cos {30}^{\circ }-\cos {210}^{\circ }\right)}{\left(2\mathrm{\Omega }\right)\left(1\times {10}^{-6}\mathrm{F}\right)}$

$=12\mathrm{V}$