Gauss law
Electric potential and capacitance
Kirchoff's current and voltage law
Wheatstone Networks
Biot-Savart's law
Ampere's law
Faraday's law and Lenz law
LC, RC and LCR circuits
LCR series and parallel connections
Important Concepts of Electricity and Magnetism
Name of Concept
Coulomb's law
This law of electrostatics is a fundamental statement that rules electrostatics.
This law states that the force of attraction or repulsion between two charges will be directly proportional to the product of two charges and inversely proportional to the square of the distance between them.
Electric dipole
An electric dipole is a couple of opposite charges q and –q separated by a finite distance d.
The gauss law introduces us to the concept of electric flux $\phi$.
Gauss law states that the total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity.
Ohm's law is a basic principle used in study of eleectrodynamics or current electricity.
The law gives a relationship between the voltage, current and resistance offered.
Mathematically stated as:
$V \propto I$ i.e., The voltage applied in the circuit will be directly proportional to the current flowing in the circuit.
Kirchhoff's Current Law and Voltage Law
Current law: The total current entering the node must be equal to the total current leaving the node.
I.e., Iin=Iout
Voltage law: In any closed circuit, the total voltage will be equal to the sum of the voltage drop within the circuit.
I.e., V=V1+V2+...
Fleming's Right Rule
If we arrange our thumb, forefinger and middle finger of the right-hand perpendicular to each other, then
The thumb points towards the direction of the motion of conductor relative to the magnetic field.
The middle finger points towards the direction of the induced current.
The forefinger points towards the direction of the magnetic field.
Fleming's Left Rule
If we arrange our thumb, forefinger and middle finger of the left-hand perpendicular to each other, then
The thumb points towards the direction of the force experienced by the conductor.
The middle finger points towards the direction of the electric current.
Biot Savart's Law
It states that:
"the magnetic field due to a tiny current element at any point is proportional to the length of the current element, the current, the sine of the angle between the current direction and the line joining the current element and the point, and inversely proportional to the square of the distance of that point. The direction of the magnetic field is in the direction of dl cross r."
"The magnetic field created by an electric current is proportional to the size of that electric current with a constant of proportionality equal to the permeability of free space."
Magnetic force
Magnetic force is given by $\vec F= q\left(\vec v\times \vec B\right)$
Hysteresis Loop
If we take a ferromagnetic material in completely demagnetized state and make it to undergo through a cycle of magnetization in which H is increased from zero to a maximum value Hmax, then decreases to zero, then reversed and again taken to –Hmax, and finally brought back to zero.
Magnetic substances
Pramagnetic: The substances which when placed in a magnetic field acquires a feeble magnetization in the same sense as the applied field
Diamagnetic: The magnetic substances which can be easily magnetised are known as diamagnetic substances.
Magnetic induction is the process by which an ordinary piece of iron briefly develops magnetic characteristics due to the presence of another magnet close by.
Electric field of electro magnetic waves:
E(x,t) = Emaxcos(kx - ωt + φ)
φ-The phase difference
Magnetic field of electro magnetic waves:
B(x,t) = Bmaxcos(kx - ωt + φ)
List of Important Formulae of Electricity and Magnetism
Electrostatic Force
$\vec F=\dfrac {k {q_1}{q_2}}{r^2}$
r- The distance between two charges
$\vec E=\dfrac {F}{q}$
F- The electrostatic force
$\phi=\dfrac {q}{\epsilon_o}$
Electric Potential
$V=\dfrac {kq}{r}$
$V=IR$
R- The resistance of the resistors
Combination of Resistors
Resistors in Series:
${R_s}={R_1}+{R_2}+....$
Resistors in Paralell:
${R_p}=\dfrac {{R_1}{R_2}}{{R_1}+{R_2}}$
Capacitance of the capacitor:
$C=\dfrac {q}{V}$
Combination of capacitors:
${C_s}=\dfrac {{C_1}{C_2}}{{C_1}+{C_2}}$
Parallel:
${C_p}={C_1}+{C_2}+....$
Electric field in parallel plate capacitor
$E=\dfrac {\sigma}{\epsilon_o}=\dfrac {Q}{{\sigma_o}A}$
Dielectric constant of a substance
$k=\dfrac {C}{C_o}$
Energy stored in capacitor
$W=\dfrac {1}{2} QV$
$dB=\dfrac {\mu_o}{4\pi} \dfrac {I\times dl\times \sin \theta}{r^2}$
$F=q\left [\vec E+ \left (\vec v\times \vec B\right)\right ]$
$\int B.dl={\mu_o}I$
Magnetic dipole
$\vec {P_m}=I \vec A$
Magnetic sucesptility
$\chi =\dfrac {I}{H}$
Tangent law
$B=H \tan \theta$
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.
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.
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 :
$T=\sqrt {\dfrac {2m}{IB}}$
$T=\sqrt {\dfrac {\pi m}{IB}}$
$T=\sqrt {\dfrac {\pi m}{2IB}}$
$T=\sqrt {\dfrac {2\pi m}{IB}}$
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:
$T=\dfrac {2\pi}{\omega}=\dfrac {2\pi}{\dfrac {2\pi IB}{m}}$
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)
$\dfrac {{\mu_o}IN}{2\left(b-a\right)} \log_e \left(\dfrac {b}{a}\right)$
$\dfrac {{\mu_o}I}{8}\dfrac {a+b}{a-b}$
$\dfrac {{\mu_o}I}{4\left(a-b\right)} \left[\dfrac {1}{a}-\dfrac {1}{b}\right]$
$\dfrac {{\mu_o}I}{8}\dfrac {a-b}{a+b}$
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)
(a), (c)
(a), (d)
(b), (c)
(b), (d)
Which of the following combination will be represented by a diamagnetic material?
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.
Trick: From graph eliminate the graphs that are having temperature dependency and porosity susceptibility.
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)
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)
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…
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FAQs on JEE Important Chapter - Electricity and Magnetism
1. Is electricity and magnetism important for JEE?
In terms of complexity and weightage, 'Electricity and Magnetism' is one of the most important topics in the JEE syllabus. You can be confident of a high overall score if you have a firm grasp on this subject.
2. What is the weightage of electricity and magnetism in JEE Mains?
Due to its vast syllabus electricity and magnetism bagged high weightage in exam. Nearly 5-6 questions have been asked repeatedly which carry about 15-20 marks.
3. Is electricity and magnetism chapters tough?
No, both the chapters are interrelated to each other. Hence if we prepare thoroughly for electricity part, we can easily understand magnetism concepts. So these are not tough chapters.
4. Who unified electricity and magnetism?
Unification of electricity and magnetism credit goes to physicist James Clerk Maxwell.
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