JEE Important Chapter - Electricity and Magnetism

When it comes to JEE Advanced exams, there is no chapter which can be ignored. Just like most of the chapters electricity and magnetism plays a vital role in scoring good grades and ranking. Because of its vast syllabus and concepts, electricity and magnetism are considered one of the most important chapters for JEE advanced exams.

The electricity and magnetism chapter consists of many sub-topics that are considered to be very basic and important. The first concept that we get to see in electricity and magnetism notes is electrostatics. The unit electrostatics and charges is a fundamental part of current electricity and magnetism. In this chapter, we study the link between electricity and magnetism. 

The key concepts of the electricity and magnetism chapter are electrostatics, current electricity and the laws related to it, capacitance, magnetostatics, electromagnetic induction, etc.

In this article, we will discuss the electricity and magnetism chapter and the important concepts of the chapter for JEE exam. Let’s start!

Important Topics of Electricity and Magnetism

• Coulomb's law

• Gauss law

• Electric potential and capacitance

• Ohm’s law

• Kirchoff’s current and voltage law

• Wheatstone Networks

• Biot-Savart’s law

• Ampere’s law

• Electromagnetic induction

• Faraday’s law and Lenz law

• LC, RC and LCR circuits

• LCR series and parallel connections

Important Concepts of Electricity and Magnetism

List of Important Formulae of Electricity and Magnetism

Solved Examples of Current Electricity and Magnetism

1. A particle of mass m and charge q moves with a constant velocity v along the positive x direction. It enters a region containing a uniform magnetic field B directed along the negative z direction, extending from x = 3 to x = 5. The minimum value of v required so that the particle can just enter the region.

Sol:

Given,

The mass of the particle=m

Charge of the particle=q

The particle is moving along positive x axis with constant velocity v directed along the negative z axis. Now, we are asked to determine the minimum value of v, for entering the field B.

In this problem, the magnetic field B is perpendicular to the motion of particle the particle in the given region.

Therefore, 

The magnetic force=The centripetal force

$qv B=\dfrac {m{v^2}}{r}$

$v=\dfrac {qBr}{m}$

Here, the region is extended from x=3 to x=5, thus the value of r will be:

r=5-3=2

Thus, we get:

$v=\dfrac {2qB}{m}$

Hence, the minimum value of v, for entering the field B is $\dfrac {2qB}{m}$.

Key Point: In this problem we can note one important formula i.e., the minimum value of v when B is perpendicular to the motion of the particle will be $\dfrac {\left({x_2}-{x_3}\right)qB}{m}$.

2. A small circular loop of conducting wire has radius a and carries current I. It is placed in a uniform magnetic field B perpendicular to its plane such that when rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period T. If the mass of the loop is m then :

1. $T=\sqrt {\dfrac {2m}{IB}}$

2. $T=\sqrt {\dfrac {\pi m}{IB}}$

3. $T=\sqrt {\dfrac {\pi m}{2IB}}$

4. $T=\sqrt {\dfrac {2\pi m}{IB}}$

Sol:

Given that, we have a loop of mass m and radius a, carrying a current I. The current loop is placed in field perpendicular to the magnetic field B.

Now, let us assume that the loop is slightly rotated by an angle $\theta$, then the torque acting on the loop will be:

$\tau = - MB \sin \theta$

$I\alpha=-MB \theta$ (Because, the angle $\theta$ is very small, thus $\sin \theta=\theta$)

$\dfrac {mr^2}{2}\alpha=- \left(\pi {r^2}I\right)B \theta$

$\alpha=\dfrac {2IB\pi}{m}\theta$.....(1)

We know that,

$\alpha = \omega^2 \theta$....(2)

From (1) and (2), we get:

$\omega^2 =\dfrac {2\pi IB}{m}$

$\omega = \sqrt{\dfrac {2\pi IB}{m}}$

Therefore, time period T of rotation will be:

$T=\dfrac {2\pi}{\omega}=\dfrac {2\pi}{\dfrac {2\pi IB}{m}}$

$T=\sqrt {\dfrac {2\pi m}{IB}}$

Therefore, option D is the right answer.

Key Point: To solve the questions like this we need to learn a few standard values of moment of inertia, to save our calculation time.

Previous Year Questions of Electricity and Magnetism

1. A coil having N turns is wound tightly in the form of a spiral with inner and outer radii 'a' and 'b' respectively. Find the magnetic field at centre, when a current I passes through coil: (JEE 2021)

1. $\dfrac {{\mu_o}IN}{2\left(b-a\right)} \log_e \left(\dfrac {b}{a}\right)$

2. $\dfrac {{\mu_o}I}{8}\dfrac {a+b}{a-b}$

3. $\dfrac {{\mu_o}I}{4\left(a-b\right)} \left[\dfrac {1}{a}-\dfrac {1}{b}\right]$

4. $\dfrac {{\mu_o}I}{8}\dfrac {a-b}{a+b}$

Sol:

Let us consider a small element dr at a distance r. Then the number of turns in the coil will be:

$dN=\dfrac {N}{b-a} dr$

Then the magnetic field due to the element dr at the center of the coil will be:

$dB=\dfrac {{\mu_o}I}{2r} dN$

$dB=\dfrac {{\mu_o}I N}{2 r\left (b-a)\right)}$

$\int_{r=a}^{r=b} dB=\int_{r=a}^{r=b}\dfrac {{\mu_o}I N}{2 r\left (b-a)\right)}$

$B=\dfrac {{\mu_o}I N}{2 \left (b-a)\right)} \log_e \dfrac {b}{a}$

Therefore, option A is the right answer.

Trick: The problem can be easily solved by looking at the the integration part, among all the option, only option A was having logarithmic value.

2. Following plots show magnetization (M) vs magnetizing field (H) and magnetic susceptibility () vs temperature (T) graph : (JEE 2021)

1. (a), (c)

2. (a), (d)

3. (b), (c)

4. (b), (d)

Which of the following combination will be represented by a diamagnetic material?

Sol:

We know that susceptibility of diamagnetic materials will be negative and independent of temperature. From above graphs, graph (a) and graph (c) represents the diamagnetic material.

Therefore, option a is the right answer.

Trick: From graph eliminate the graphs that are having temperature dependency and porosity susceptibility.

Practice Questions

1. Electric field of plane electromagnetic wave propagating through a non-magnetic medium is given by E = 20cos(2  1010 t-200x) V/m. The dielectric constant of the medium is equal to : (Take $\mu_r$ = 1) (Ans: 9)

2. A current of 1.5 A is flowing through a triangle, of side 9 cm each. The magnetic field at the centroid of the triangle is (Assume that the current is flowing in the clockwise direction.) (Ans: $3\times {10^{-5}}$ T, inside the plane of triangle)

Conclusion

In this article we had an overview of the unit electricity and magnetism which is having many subtopics. In the article we discussed a few important concepts of electricity and magnetism in exam point of view. We have also discussed about the list of important formulae along with solved examples, previous year question, etc…