13.4 Induced electric fields  (Page 2/4)

Electric field induced by the changing magnetic field of a solenoid

Strategy

Solution

1. The magnetic field is confined to the interior of the solenoid where
   
   $B={\mu }_{0}nI={\mu }_{0}n{I}_0{e}^{\text{−}\alpha t}.$
   
   Thus, the magnetic flux through a circular path whose radius r is greater than R , the solenoid radius, is
   
   ${\text{Φ}}_{\text{m}}=BA={\mu }_{0}n{I}_0\pi R^2{e}^{\text{−}\alpha t}.$
   
   The induced field $\overrightarrow{E}$ is tangent to this path, and because of the cylindrical symmetry of the system, its magnitude is constant on the path. Hence, we have
   
   $\begin{array}{ccc}\hfill \left|{\displaystyle \oint \overrightarrow{E}·d\overrightarrow{l}}\right|& =\hfill & \left|\frac{d{\text{Φ}}_{\text{m}}}{dt}\right|,\hfill \\ \\ \\ \hfill E(2\pi r)& =\hfill & \left|\frac{d}{dt}({\mu }_{0}n{I}_0\pi R^2{e}^{\text{−}\alpha t})\right|=\alpha {\mu }_{0}n{I}_0\pi R^2{e}^{\text{−}\alpha t},\hfill \\ \\ \\ \hfill E& =\hfill & \frac{\alpha {\mu }_{0}n{I}_0{R}^2}{2r}{e}^{\text{−}\alpha t}(r>R).\hfill \end{array}$
2. For a path of radius r inside the solenoid, ${\text{Φ}}_{\text{m}}=B\pi r^2,$ so
   
   $E(2\pi r)=\left|\frac{d}{dt}({\mu }_{0}n{I}_0\pi r^2{e}^{\text{−}\alpha t})\right|=\alpha {\mu }_{0}n{I}_0\pi r^2{e}^{\text{−}\alpha t},$
   
   and the induced field is
   
   $E=\frac{\alpha {\mu }_{0}n{I}_{0}r}{2}{e}^{\text{−}\alpha t}(r<R).$
3. The magnetic field points into the page as shown in part (b) and is decreasing. If either of the circular paths were occupied by conducting rings, the currents induced in them would circulate as shown, in conformity with Lenz’s law. The induced electric field must be so directed as well.

Significance

Check Your Understanding The magnetic field shown below is confined to the cylindrical region shown and is changing with time. Identify those paths for which $\epsilon ={\displaystyle \oint \overrightarrow{E}·d\overrightarrow{l}\ne 0.}$

$P_1,P_2,P_4$

Got questions? Get instant answers now!

Check Your Understanding A long solenoid of cross-sectional area $5.0{\text{cm}}^2$ is wound with 25 turns of wire per centimeter. It is placed in the middle of a closely wrapped coil of 10 turns and radius 25 cm, as shown below. (a) What is the emf induced in the coil when the current through the solenoid is decreasing at a rate $dI\text{/}dt=\mathrm{-0.20}\text{A}\text{/}\text{s}?$ (b) What is the electric field induced in the coil?

a. $3.1\times 10^{\text{−}6}\text{V};$ b. $2.0\times {10}^{\text{−}7}\text{V}\text{/}\text{m}$

Got questions? Get instant answers now!