Time Varying Magnetic Field

A time-varying magnetic field is a magnetic field whose strength and direction change over time. This phenomenon is fundamental in the study of electromagnetism and has a wide range of practical applications in everyday life. When a magnetic field changes, it induces an electric field, a principle known as electromagnetic induction. This principle is the backbone of many modern technologies. For example, it is used in the generation of electricity in power plants, where rotating turbines create time-varying magnetic fields that induce electric currents. Another common application is in transformers, which adjust the voltage of electric power for safe and efficient transmission and distribution. The concept also finds use in wireless communication devices, where varying magnetic fields are used to transmit data over distances. Understanding time-varying magnetic fields is crucial for developing and improving the technology that powers and connects our world.

This Story also Contains

1. Time-Varying Magnetic Field
2. Solved Examples Based on Time-Varying Magnetic Field
3. Summary

Time-Varying Magnetic Field

A time-varying magnetic field is a magnetic field whose magnitude and direction change over time. This dynamic behaviour is a fundamental aspect of electromagnetism and plays a critical role in various technological applications. In essence, when a magnetic field varies with time, it induces an electric field, a phenomenon described by Faraday's Law of Electromagnetic Induction. This principle is at the heart of many devices we use daily.

As we learn Induced electric field is given by

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

But by using $\phi=B. A$ so we can also write

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

Where

A $\rightarrow$ constant Area

B $\rightarrow$ Varying Magnetic field

Example

A uniform but time-varying magnetic field B(t) exists in a circular region of radius ‘a’ and is directed into the plane of the paper as shown in the below figure, the magnitude of the induced electric field $\left(E_{i n}\right)$ at point P lies at a distance r from the centre of the circular region is calculated as follows.

Due to the time-varying magnetic field induced electric field will be produced whose electric field lines are concentric circular closed curves of radius r.

$
\begin{aligned}
\text { if } r & \leq a \\
\text { then } E_{\text {in }}(2 \pi r) & =\pi r^2\left|\frac{d B}{d t}\right| \\
\Rightarrow & E_{\text {in }}=\frac{r}{2}\left|\frac{d B}{d t}\right|
\end{aligned}
$

For $r>R$,
$
\begin{aligned}
& E_{\text {in }} * 2 \pi r=\pi a^2\left|\frac{d B}{d t}\right| \\
& \Rightarrow E_{\text {in }}=\frac{a^2}{2 r}\left|\frac{d B}{d t}\right|
\end{aligned}
$

The graph of E vs r

where E=induced electric field

Recommended Topic Video

Solved Examples Based on Time-Varying Magnetic Field

Example 1: The flux linked with a coil at any instant t is given by $\phi=10 t^2-50 t+250$. The induced emf (in Volts) at t = 3s is

1) -10

2) -190

3) 190

4) 10

Solution:

Rate of change of magnetic Flux

$\varepsilon=\frac{-d \phi}{d t}$

wherein

$\begin{aligned} & d \phi \rightarrow \phi_2-\phi_1 \\ & \phi=10 \mathrm{t}^2-50 t+250 \\ & \therefore \quad \frac{d \phi}{d t}=20 t-50 \\ & \text { Induced em } f, \varepsilon=\frac{-d \phi}{d t} \\ & \text { or } \varepsilon=-(20 t-50)=-[(20 \times 3)-50]=-10 \text { volt } \\ & \text { or } \varepsilon=-10 \text { volt }\end{aligned}$

Hence, the answer is the option (1).

Example 2: Figure shows three regions of the magnetic field, each of area A, and in each region magnitude of the magnetic field decreases at a constant rate a. If $\vec{E}$ is an induced electric field then the value of line integral $\oint \vec{E}.{d \vec{r}}$. along the given loop is equal to

1) $\alpha A$
2) $-\alpha A$
3) $3 \alpha A$
4) $-3 \alpha A$

Solution:

Rate of change of magnetic Flux

$\varepsilon=\frac{-d \phi}{d t}$
wherein

$
d \phi \rightarrow \phi_2-\phi_1
$
$\phi_2-\phi_1-$ change in flux
Potential
$
\int \vec{E} \cdot d \vec{r}=-\frac{d \phi}{d t}
$

and take the sign of flux according to the right-hand curl rule.

$\int \vec{E} \cdot d \vec{r}=-((\alpha A)+(\alpha A)+(-\alpha A))=-\alpha A$

Hence, the answer is the option (2).

Example 3: 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 4: Faraday's law of electromagnetic induction states that the induced emf is

1) Proportional to the change in magnetic flux linkage

2) Equal to the change in magnetic flux linkage

3) Equal to the change of magnetic flux

4) Proportional to the rate of change of magnetic flux

Solution:

Flux may change with time in several ways

$
\varepsilon=N \frac{-d}{d t}(B A \cos \Theta)
$

From Faraday's law
$
\varepsilon=-N \frac{d \phi}{d t}
$

Where $\phi=B A \cos \theta$

Hence, the answer is the option (4).

Example 5: 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 $I=I_0 \cos (\omega t)$. 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 (1).

Summary

A time-varying magnetic field induces an electric field, a principle described by Faraday's Law of Electromagnetic Induction. This phenomenon is crucial in various applications, such as electricity generation, transformers, and wireless communication. Understanding the induced electric fields, as demonstrated through practical examples and mathematical equations, highlights the importance of this concept in both theoretical and applied electromagnetism.