What Does A Changing Electric Field Induce

Learning Objectives

By the finish of this section, you will exist able to:

• Connect the human relationship between an induced emf from Faraday's police force to an electric field, thereby showing that a changing magnetic flux creates an electric field
• Solve for the electric field based on a changing magnetic flux in time

The fact that emfs are induced in circuits implies that work is being washed on the conduction electrons in the wires. What can possibly exist the source of this work? We know that it'southward neither a bombardment nor a magnetic field, for a battery does not take to be present in a excursion where current is induced, and magnetic fields never practice work on moving charges. The answer is that the source of the work is an electric field $\overrightarrow{E}$ that is induced in the wires. The piece of work done by $\overrightarrow{\mathrm{Due\; east}}$ in moving a unit charge completely around a circuit is the induced emf ε; that is,

where ${\displaystyle \oint }$ represents the line integral effectually the circuit. Faraday'due south law can be written in terms of the induced electric field every bit

$${\displaystyle \oint \overrightarrow{E}·d\overrightarrow{\mathrm{50}}=-}\frac{d{\text{Φ}}_{\text{m}}}{dt}.$$

13.10

In that location is an important distinction between the electric field induced by a changing magnetic field and the electrostatic field produced past a fixed charge distribution. Specifically, the induced electric field is nonconservative considering it does net piece of work in moving a accuse over a airtight path, whereas the electrostatic field is conservative and does no internet work over a closed path. Hence, electrical potential tin can exist associated with the electrostatic field, but not with the induced field. The post-obit equations stand for the stardom betwixt the two types of electric field:

$$\begin{array}{ccc}\hfill {\displaystyle \oint \overrightarrow{\mathrm{Due\; east}}·d\overrightarrow{l}}& \ne \hfill & 0(\text{induced});\hfill \\ \hfill {\displaystyle \oint \overrightarrow{E}·d\overrightarrow{l}}& =\hfill & 0(\text{electrostatic}).\hfill \end{array}$$

thirteen.xi

Our results can be summarized past combining these equations:

$$\epsilon ={\displaystyle \oint \overrightarrow{\mathrm{East}}·d\overrightarrow{\mathrm{50}}=\text{−}}\frac{d{\text{Φ}}_{\text{m}}}{dt}.$$

13.12

Case 13.seven

Induced Electric Field in a Round Coil

What is the induced electric field in the circular coil of Example 13.2 (and Figure thirteen.nine) at the three times indicated?

Strategy

Using cylindrical symmetry, the electrical field integral simplifies into the electric field times the circumference of a circle. Since nosotros already know the induced emf, nosotros can connect these two expressions by Faraday's law to solve for the induced electric field.

Solution

The induced electric field in the gyre is constant in magnitude over the cylindrical surface, like to how Ampere's law bug with cylinders are solved. Since $\overrightarrow{E}$ is tangent to the roll,

$${\displaystyle \oint \overrightarrow{E}·d\overrightarrow{l}=}{\displaystyle \oint Ed\mathrm{50}}=2\pi r\mathrm{Due\; east}.$$

When combined with Equation xiii.12, this gives

$$E=\frac{\epsilon }{\mathrm{ii}\pi r}.$$

The direction of $\epsilon$ is counterclockwise, and $\overrightarrow{E}$ circulates in the same direction effectually the coil. The values of E are

$$\begin{array}{}\\ \\ E(t_{\mathrm{ane}})=\frac{\mathrm{six.0}\text{Five}}{\mathrm{two}\pi (0.50\text{m})}=\mathrm{ane.9}\text{5/k;}\hfill \\ E(t_2)=\frac{\mathrm{4.seven}\text{V}}{\mathrm{ii}\pi (\mathrm{0.l}\text{m})}=1.5\text{Five/thou;}\hfill \\ E(t_3)=\frac{0.040\text{V}}{2\pi (0.50\text{one thousand})}=0.013\text{Five/thousand}\text{.}\hfill \end{array}$$

Significance

When the magnetic flux through a circuit changes, a nonconservative electrical field is induced, which drives current through the circuit. Just what happens if $dB\text{/}dt\ne 0$ in free space where there isn't a conducting path? The respond is that this case tin can be treated as if a conducting path were present; that is, nonconservative electrical fields are induced wherever $dB\text{/}dt\ne 0,$ whether or not in that location is a conducting path present.

These nonconservative electrical fields e'er satisfy Equation xiii.12. For case, if the circular coil of Figure 13.9 were removed, an electric field in free infinite at $r=0.50\text{thousand}$ would yet be directed counterclockwise, and its magnitude would still exist 1.ix V/grand at $t=0$ , i.5 5/m at $t=5.0\times \mathrm{one}{0}^{\text{−}\mathrm{ii}}\text{due south},$ etc. The beingness of induced electric fields is certainly not restricted to wires in circuits.

Example 13.8

Electric Field Induced past the Changing Magnetic Field of a Solenoid

Part (a) of Figure xiii.18 shows a long solenoid with radius R and due north turns per unit of measurement length; its current decreases with fourth dimension according to $I=I_0{e}^{\text{−}\alpha t}.$ What is the magnitude of the induced electric field at a point a altitude r from the primal centrality of the solenoid (a) when $r>R$ and (b) when $r<R$ [encounter role (b) of Figure 13.18]. (c) What is the direction of the induced field at both locations? Assume that the infinite-solenoid approximation is valid throughout the regions of involvement.

Figure 13.xviii (a) The current in a long solenoid is decreasing exponentially. (b) A cantankerous-sectional view of the solenoid from its left stop. The cantankerous-section shown is about the middle of the solenoid. An electric field is induced both within and outside the solenoid.

Strategy

Using the formula for the magnetic field inside an infinite solenoid and Faraday's law, nosotros calculate the induced emf. Since we have cylindrical symmetry, the electric field integral reduces to the electric field times the circumference of the integration path. Then we solve for the electric field.

Solution

1. The magnetic field is confined to the interior of the solenoid where

   $$B={\mu }_{0}nI={\mu }_0\mathrm{due\; north}{I}_0{e}^{\text{−}\alpha t}.$$

   Thus, the magnetic flux through a circular path whose radius r is greater than R, the solenoid radius, is

   $${\text{Φ}}_{\text{thou}}=BA={\mu }_0\mathrm{due\; north}{I}_0\pi R^{\mathrm{ii}}{e}^{\text{−}\alpha t}.$$

   The induced field $\overrightarrow{E}$ is tangent to this path, and considering of the cylindrical symmetry of the system, its magnitude is constant on the path. Hence, we take

   $$\begin{array}{ccc}\hfill \left|{\displaystyle \oint \overrightarrow{\mathrm{East}}·d\overrightarrow{l}}\right|& =\hfill & \left|\frac{d{\text{Φ}}_{\text{m}}}{dt}\right|,\hfill \\ \\ \\ \hfill E(2\pi r)& =\hfill & \left|\frac{d}{dt}({\mu }_0\mathrm{due\; north}{I}_0\pi R^{\mathrm{two}}{e}^{\text{−}\alpha t})\right|=\alpha {\mu }_0\mathrm{north}{I}_0\pi R^2{\mathrm{eastward}}^{\text{−}\alpha t},\hfill \\ \\ \\ \hfill E& =\hfill & \frac{\alpha {\mu }_0\mathrm{north}{I}_0{R}^2}{2r}{\mathrm{east}}^{\text{−}\alpha t}(r>R).\hfill \end{array}$$

2. For a path of radius r inside the solenoid, ${\text{Φ}}_{\text{grand}}=B\pi r^2,$ and then

   $$E(2\pi r)=\left|\frac{d}{dt}({\mu }_0\mathrm{northward}{I}_0\pi r^{\mathrm{ii}}{\mathrm{eastward}}^{\text{−}\alpha t})\right|=\alpha {\mu }_0\mathrm{northward}{I}_0\pi r^2{\mathrm{due\; east}}^{\text{−}\alpha t},$$

   and the induced field is

   $$E=\frac{\alpha {\mu }_0\mathrm{north}{I}_{0}r}{2}{e}^{\text{−}\alpha t}(r<R).$$

3. The magnetic field points into the page as shown in role (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 exist so directed also.

Significance

In part (b), notation that $\left|\overrightarrow{E}\right|$ increases with r inside and decreases as i/r exterior the solenoid, as shown in Effigy 13.xix.

Figure thirteen.nineteen The electrical field vs. distance r. When $r<R,$ the electric field rises linearly, whereas when $r>R,$ the electrical field falls of proportional to one/r.

Check Your Understanding thirteen.vi

Check Your Understanding Suppose that the coil of Example 13.two is a square rather than circular. Can Equation 13.12 be used to calculate (a) the induced emf and (b) the induced electric field?

Check Your Understanding 13.7

Check Your Agreement What is the magnitude of the induced electrical field in Example thirteen.8 at $t=0$ if $r=6.0\text{cm},$ $R=\mathrm{two.0}\text{cm},n=2000$ turns per meter, $I_0=\mathrm{two.0}\text{A},$ and $\alpha =200{\text{due south}}^{\text{−}\mathrm{one}}?$

Check Your Understanding thirteen.8

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

Check Your Understanding 13.9

Check Your Understanding A long solenoid of cross-exclusive area $\mathrm{five.0}{\text{cm}}^2$ is wound with 25 turns of wire per centimeter. Information technology is placed in the middle of a closely wrapped curlicue 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 $dI\text{/}dt=\mathrm{-0.20}\text{A}\text{/}\text{south}?$ (b) What is the electric field induced in the coil?

Source: https://openstax.org/books/university-physics-volume-2/pages/13-4-induced-electric-fields

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