Chapter 29 Problem 78

Problem

A circular-shaped circuit of radius $r$, containing a resistance $R$ and capacitance $C$, is situated with its plane perpendicular to a spatially uniform magnetic field ${\vec {B}}$ directed into the page. Starting at time $t=0$, the voltage difference $V_{ba}=V_{b}-V_{a}$ across the capacitor plates is observed to increase with time $t$ according to $V_{ba}=V_{0}(1-e^{-t/\tau })$, where $V_{0}$ and $\tau$ are positive constants. Determine $dB/dt$ , the rate at which the magnetic field magnitude changes with time. Is $B$ becoming larger or smaller as time increases?

Solution

The emf around the loop is equal to the time derivative of the flux. Since the area of the coil is constant, the time derivative of the flux is equal to the derivative of the magnetic field multiplied by the area of the loop.

$I={\frac {dQ}{dt}}={\frac {dCV}{dt}}={\frac {d}{dt}}\left[CV_{0}\left(1-e^{-t/\tau }\right)\right]={\frac {CV_{0}}{\tau }}e^{-t/\tau }$

$I={\frac {V_{0}}{R}}e^{-t/\tau }$

${\mathcal {E}}=IR+V_{c}=\left({\frac {V_{0}}{R}}e^{-t/\tau }\right)R+V_{0}\left(1-e^{-t/\tau }\right)=V_{0}$

$V_{0}={\frac {d\Phi _{B}}{dt}}={\frac {AdB}{dt}}=\pi r^{2}{\frac {dB}{dt}}$

${\frac {dB}{dt}}={\frac {V_{0}}{\pi r^{2}}}$

The top plate of the capacitor is going to higher potential. It means it is under a charging emf in the opposite direction. Thus the induced ccurrent flows in clockwise direction. Clockwise current means the induced magnetic field is aligned with the magnetic field shown. So the magnetic field must be decreasing.