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Last updated date: 01st Feb 2023

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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!

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

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 :

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

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

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

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

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.

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)

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

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

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

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

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)

(a), (c)

(a), (d)

(b), (c)

(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.

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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JEE Advanced 2023 exam date have been announced by the NTA. JEE Advanced 2023 will now be conducted on 4-June-2023, and the exam registration closes on 4-May-2023. You can check the complete schedule on our site. Furthermore, you can check JEE Advanced 2023 dates for application, admit card, exam, answer key, result, counselling, etc along with other relevant information.

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Last updated date: 01st Feb 2023

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IIT Guwahati has announced the JEE Advanced 2023 application form release date on the official website https://jeeadv.ac.in/. JEE Advanced 2023 Application Form is available on the official website for online registration. Besides JEE Advanced 2023 application form release date, learn about the application process, steps to fill the form, how to submit, exam date sheet etc online. Check our website for more details.

Last updated date: 01st Feb 2023

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It is crucial for the the engineering aspirants to know and download the JEE Advanced 2023 syllabus PDF for Maths, Physics and Chemistry. Check JEE Advanced 2023 syllabus here along with the best books and strategies to prepare for the entrance exam. Download the JEE Advanced 2023 syllabus consolidated as per the latest NTA guidelines from Vedantu for free.

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JEE Advanced 2023 Study Materials: Strengthen your fundamentals with exhaustive JEE Advanced Study Materials. It covers the entire JEE Advanced syllabus, DPP, PYP with ample objective and subjective solved problems. Free download of JEE Advanced study material for Physics, Chemistry and Maths are available on our website so that students can gear up their preparation for JEE Advanced exam 2023 with Vedantu right on time.

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Download JEE Advanced Question Papers & Answer Keys of 2022, 2021, 2020, 2019, 2018 and 2017 PDFs. JEE Advanced Question Paper are provided language-wise along with their answer keys. We also offer JEE Advanced Sample Question Papers with Answer Keys for Physics, Chemistry and Maths solved by our expert teachers on Vedantu. Downloading the JEE Advanced Sample Question Papers with solutions will help the engineering aspirants to score high marks in the JEE Advanced examinations.

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Last updated date: 01st Feb 2023

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In order to prepare for JEE Advanced 2023, candidates should know the list of important books i.e. RD Sharma Solutions, NCERT Solutions, HC Verma books and RS Aggarwal Solutions. They will find the high quality readymade solutions of these books on Vedantu. These books will help them in order to prepare well for the JEE Advanced 2023 exam so that they can grab the top rank in the all India entrance exam.

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JEE Advanced 2023 free online mock test series for exam preparation are available on the Vedantu website for free download. Practising these mock test papers of Physics, Chemistry and Maths prepared by expert teachers at Vedantu will help you to boost your confidence to face the JEE Advanced 2023 examination without any worries. The JEE Advanced test series for Physics, Chemistry and Maths that is based on the latest syllabus of JEE Advanced and also the Previous Year Question Papers.

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Last updated date: 01st Feb 2023

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IIT Bombay is responsible for the release of the JEE Advanced 2023 cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Advanced 2023 is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Advanced qualifying marks for 2021 ranged from 17.50%, while for OBC/SC/ST categories, they ranged from 15.75% for OBC, 8.75% for SC and 8.75% for ST category.

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JEE Advanced 2023 Result - IIT Bombay has announced JEE Advanced result on their website. To download the Scorecard for JEE Advanced 2023, visit the official website of JEE Advanced i.e https://jeeadv.ac.in/.

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Want to know which Engineering colleges in India accept the JEE Advanced 2023 scores for admission to Engineering? Find the list of Engineering colleges accepting JEE Advanced scores in India, compiled by Vedantu. There are 1622 Colleges that are accepting JEE Advanced. Also find more details on Fees, Ranking, Admission, and Placement.

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FAQ

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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