Induced Electric Field

Induced Electric Field

Edited By Vishal kumar | Updated on Sep 03, 2024 02:42 PM IST

The concept of the induced electric field is fundamental in understanding electromagnetic phenomena. It arises when a changing magnetic field generates an electric field, a principle elegantly captured by Faraday's Law of Induction. This phenomenon is pivotal in the workings of many everyday technologies. For instance, in power generators, the rotation of a coil within a magnetic field induces an electric field, which in turn generates electric current, powering homes and industries. Similarly, electric induction is at play in transformers that modify voltage levels for efficient power transmission. In this article, we will discuss the induced electric field. It is not just a theoretical concept; it is a driving force behind the technological advancements that shape our daily lives, from the electric motors in household appliances to the wireless charging of devices.

This Story also Contains
  1. Induced Electric Field
  2. Solved Examples Based on Induced Electric Field
  3. Summary

Induced Electric Field

The concept of the induced electric field is a cornerstone of electromagnetism, describing how a changing magnetic field generates an electric field. This relationship is encapsulated in Faraday's Law of Induction, which states that a time-varying magnetic field creates a circulating electric field.

Whenever a magnetic field varies with time, an induced electric field $E_{i n}$ is produced in any closed path, whether in the matter
or in empty space. This Induced electric field is directly proportional to induced emf as $\varepsilon=\oint \overrightarrow{E_{i n}} \cdot \overrightarrow{d l}$.

This is given as

$\varepsilon=\oint \overrightarrow{E_{i n}} \cdot \overrightarrow{d l}=\frac{-d \phi}{d t}$

This is known as an integral form of Faraday’s laws of EMI.

Properties of Induced Electric Field

  • The induced electric field is different from the electrostatic field. As it is non-conservative and non-electrostatic in nature.
  • Its field lines are concentric circular closed curves.
  • This field is not created by source charges.
  • Its direction is along the tangent to its field lines.

We can understand better through video.

Solved Examples Based on Induced Electric Field

Example 1: A coil having n turns and resistance R is connected with a galvanometer of resistance 4R. This combination is moved in time t seconds from a magnetic field W1 Weber to W2 Weber. The induced current in the circuit is

1) $-\frac{W_2-W_1}{5 R n t}$
2) $-\frac{n\left(W_2-W_1\right)}{5 R t}$
3) $-\frac{\left(W_2-W_1\right)}{R n t}$
4) $-\frac{n\left(W_2-W_1\right)}{R t}$

Solution:

Induced current $I=\frac{-n}{R^{\prime}} \frac{d \phi}{d t}=\frac{-n}{R^{\prime}} \frac{d W}{d t}$ where,

$\phi=W=$ flux $\times$ per unit turn of the coil

Change in flux $=W_2-W_1$

Total current per coil

$\begin{aligned} & \therefore I=\frac{\xi}{R_{e q}}=\frac{n}{R_{e q}} \frac{\Delta \phi}{\Delta t} \\ & I=\frac{n\left(W_2-W_1\right)}{(R+4 R) t}=\frac{n\left(W_2-W_1\right)}{5 R t}\end{aligned}$

The induced current is opposite to its cause of production

$I=\frac{-n\left(W_2-W_1\right)}{5 R t}$

Hence, the answer is the option (2).

Example 2: A small circular loop of wire of radius a is located at the centre of a much larger circular wire loop of radius b. The two loops are in the same plane. The outer loop of radius b carries an alternating current I=Io cos (ωt). The emf induced in the smaller inner loop is nearly :

1) $\frac{\pi \mu_0 I_0}{2} * \frac{a^2}{b} \omega \sin \omega t$
2) $\frac{\pi \mu_0 I_0}{2} * \frac{a^2}{b} \omega \cos \omega t$
3) $\pi \mu_0 I_0 * \frac{a^2}{b} \omega \sin \omega t$
4) $\pi \mu_0 I_0 * \frac{b^2}{a} \omega \cos \omega t$

Solution:

The magnetic field produced by the outer loop $=\frac{\mu_o I}{2 R}=\frac{\mu_o I_o \cos \omega t}{2 b}$

$\begin{aligned} & \phi=B \cdot A=\left(\frac{\mu_o I_o \cos \omega t}{2 b}\right) \pi a^2 \\ & \xi=\left|\frac{-d \phi}{d t}\right|=\frac{\mu_o I_o \pi}{2 b} a^2 \cdot \omega \sin \omega t\end{aligned}$

Hence, the answer is the option (2).

Example 3: A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity ω, the maximum e.m.f. induced in the coil will be :

1) $3 n B A \omega$
2) $(3 / 2) \mathrm{nBA} \omega$
3) $n B A w$
4) $(1 / 2) n B A \omega$

Solution:

If Angle $\theta$ - change
$$
\begin{aligned}
& \varepsilon=-N A B d \frac{(\cos \Theta)}{d \Theta} \times \frac{d \Theta}{d t} \\
& \varepsilon=+N B A \omega \sin \Theta
\end{aligned}
$$

Emf induced is $\mathrm{e}=($ NBA $\omega)$ since
The maximum value of $e$ is NBA $\omega$

Hence, the answer is the option (3).

Example 4: A small conducting loop of radius a and resistance r is pulled with velocity V perpendicular to a long straight conductor carrying current i as shown in Fig the current I in terms of V is

1) $\frac{\mu_0 i a^2}{2 x^2} V$
2) $\frac{\mu_0 i a^2 V^2}{2 x^2 r}$
3) $\frac{\mu_0 i a^2}{2 x^2 r^2} V$
4) $\frac{\mu_0 i a^2 V}{2 x^2 r} V$

Solution:

Induced Current

$$
I=\frac{\varepsilon}{R}=\frac{-N}{R} \frac{d \phi}{d t}
$$
wherein
$R \rightarrow$ Resistance
$\frac{d \phi}{d t} \rightarrow_{\text {Rate of change of flux }}$
Magnetic field at the position of the loop
$$
B=\frac{\mu_0 i}{2 x r}
$$

Since the loop is small, the magnetic field can be assumed to be constant.

So, flux across the loop $\phi=\int \vec{B} \cdot d \vec{A}$
$$
\begin{aligned}
& =\vec{B} \int d \vec{A} \\
& =B A \\
& =\frac{\mu_0 i \pi a^2}{2 \pi x} \\
& \frac{\mu_0 i a^2}{2 x}
\end{aligned}
$$

Now induced emf,
$$
\begin{aligned}
& \varepsilon=\frac{-d \phi}{d t} \\
& =\frac{-\mu_0 i a^2}{2}\left(\frac{d(1 / x)}{d t}\right) \\
& =\frac{\mu_0 i a^2}{2 x^2}\left(\frac{d(x)}{d t}\right) \\
& =\frac{\mu_0 i a^2}{2 x^2} V
\end{aligned}
$$

Now induced emf,

$$
\begin{aligned}
& \varepsilon=\frac{-d \phi}{d t} \\
& =\frac{-\mu_0 i a^2}{2}\left(\frac{d(1 / x)}{d t}\right) \\
& =\frac{\mu_0 i a^2}{2 x^2}\left(\frac{d(x)}{d t}\right) \\
& =\frac{\mu_0 i a^2}{2 x^2} V
\end{aligned}
$$

Now induced current
$$
\begin{aligned}
& I=\frac{\varepsilon}{r} \\
& I=\frac{\mu_0 i a^2}{2 x^2 r} V
\end{aligned}
$$

Hence, the answer is the option (2).

Example 5: For induced electric field, the electric field lines are

1) Its field lines are concentric circular closed curves.

2) Its field lines are parallel straight lines.

3) Its field lines are diverging parabola

4) Its field lines are converging hyperbola curves.

Solution:

Properties of Induced electric field

  • The induced electric field is different from the electrostatic field. As it is non-electrostatic.
  • Its field lines are concentric circular closed curves.
  • This field is not created by source charges.
  • Its direction is along the tangent to its field lines.

Hence, the answer is the option (2).

Summary

The induced electric field, governed by Faraday's Law of Induction, emerges from time-varying magnetic fields, creating circulating electric fields. Unlike electrostatic fields, induced electric fields are non-conservative and characterized by concentric circular field lines. This phenomenon is integral to technologies such as power generators, transformers, and wireless charging devices, demonstrating its crucial role in modern electrical applications.

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