Paramagnetic Substances – Magnetism & Matters | Class 12 Physics Notes

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Paramagnetic substances

The substances which when placed in a magnetic field are feebly magnetised in the direction of magnetising field are called paramagnetic substances.

(1) The property of paramagnetism is found to exist in substances whose atoms or molecules have an excess of electron spinning in same direction.

(2) Atoms of paramagnetic substances possess a permanent magnetic dipole moment and thus behave like small bar magnets called atomic magnets.

(3) In absence of external magnetic field paramagnetic substances to not show any magnetism because atomic magnets are randomly oriented so net magnetic dipole moment is zero.

(4) In presence of external field each atomic magnet experiences a torque which tries to rotate and align them parallel to direction of magnetic field. The substance acquires net dipole moment and thus gets magnetised in direction of field.

(5) The property of paramagnetism is temperature dependent. The thermal agitation on increase of temperature spoils the alignment of atomic magnets which reduces net magnetic dipole moment.

             

  1. Some paramagnetic substances are $A \ell, N a,$ Sb, Pt, $\operatorname{CuC} \ell_{2}, M n,$ Cr, liquid oxygen etc.
HRD Minister Launches AI-powered Mock Test App for JEE & NEET Students

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Union HRD Minister Mr Ramesh Pokhriyal Nishank launched an AI-powered mock test App for JEE and NEET 2020 Students. Its an android app ‘National Test Abhyas’ for students who are going to appear in JEE Main 2020 and NEET 2020 examinations.

NTA plans to release one new mock test on the app everyday, which students can then download and attempt offline. Once the test is completed, students can go online again to submit the test and view their test report.

Every student is unique and requires specific guidance to detect and overcome gaps in knowledge and test-taking strategy.

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eSaral offers a broad categories of mock tests for students preparing for JEE 2020 and NEET 2020 Examinations. With the Part and full syllabus mock tests, eSaral App also provides topic wise tests that are very helpful in analyzing weak and strong areas and also helpful in revising important concepts from each and every chapter individually.

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Importance of Practicing Mock Tests

  1. Practicing Mock tests helps to re-examine the concepts that you have learned before.
  2. It also helps in Analyzing the weak and strong areas. Always try to strengthen strong points and work optimistically on weaker sections.
  3. Helps in improving Speed and presence of mind. As the students gets acclimatized to the types of questions asked in mock tests.
  4. Time-Management: Time allotted for the exam is limited. Since the Exam will held in Online Mode, so practicing online mock tests will help the students to get acquainted with Exam duration and attempting exam in online mode rather than pen – paper. More practice will help the students in managing time between easy and difficult questions.
  5. To Shed Exam Fear – To practice mock test works very efficiently in improving confidence level and eliminating exam fear. Practice 3 Hours at a stretch and solve questions for a sustained period of time without breaks.
  6. Helps in developing right strategy to attempt exam.

Practicing mock tests helps the students in knowing where they stand so they can prepare better next time.

Now we hope that you have understood the importance of JEE Main & NEET Mock Tests, make sure you sign up for mock test.

Start Early to take a lead. Time waits for none.


 

Mathematics Revision series for JEE, Class 11 and 12

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Revise complete Mathematics for Class 11, 12 & JEE on eSaral YouTube Channel. Get everything you are looking for JEE Preparation in this Mathematics Revision Series by N.K. Gupta Sir (M-tech from IIT Kanpur) and Vishnu Shrimal Sir (IIT-BHU).   Here you will get to know about important key concepts and all the important Formulae that are very important to solve conceptual problems that will come in main Exam.

 

CLASS XI

TOPIC Revision Videos Mind Maps
Logarithm Revise Mindmap
Trigonometric Ratios Revise Mindmap
Trigonometric Equations Revise Mindmap
Quadratic Equations Revise Mindmap
Limits Revise Mindmap
Continuity Revise Mindmap
Sequence and Series Revise Mindmap

CLASS XII

TOPIC Revision Videos Mind Maps
Vector Revise Mindmap
3D Geometry Revise Mindmap
Relations & Functions Revise Mindmap
Inverse trigonometric Functions Revise Mindmap
Differentiability Revise Mindmap
Methods of Differentiation Revise Mindmap

 


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CRACK JEE 2020 | How to Prepare for JEE Main in last 2 Months

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Less than 2 months remaining for JEE Main 2020 (July Attempt), but at the same time it is still possible to secure good percentile and an All India Rank to get qualified for JEE advanced and also to get admissions in some Top NIT’s. Read the complete article till the end to know How to Prepare for JEE Main in 2 Months.

After a long wait the dates for JEE Main 2020 has finally been announced. The exam will be held between 18th and 23rd July 2020. Although students have had an extended preparation time ever since we got into preventive lockdown due to COVID 19 pandemic.

Students were finding it bit difficult to concentrate in the absence of a specific target exam date. Now the students do have a specific date in front of them to look forward to. Whenever there is less time remaining, you have to put more effort into your studies to practice enough questions for good JEE preparation.

JEE Main 2020 Exam Preparation Tips for Last two remaining months: 

1. Practice mock tests and solve previous year question papers

Solving previous year question papers of JEE Main and taking mock tests is extremely important to get a good rank in the exam. Attempt the various free mock tests available on eSaral App and get the feel of the actual exam and improve your time management skills.

Solve at least ONE full syllabus online test every week. Solve questions by applying your reasoning and analytical skills. Develop test taking strategy, it will evolve with more tests that you attempt. Also solve previous years JEE Main question paper bank. While taking the tests students should look at building the right exam temperament, adjusting the body clock and getting comfortable with test environment.


CLICK HERE for JEE MAIN Previous Year Topicwise Questions. 


2. Practice more and more conceptual problems and Clear your Doubts.

No matter what how big or small your doubt is, get it cleared. Download eSaral App & we will assist you in clarifying your doubts about a particular topic related to JEE Main. For clearing doubts on important topics and concepts, you can join a crash course.

Crash Course will also help you to complete class 11 & 12 JEE Topics covering both Mains & Advanced syllabus with complete package of test series. To know more about JEE 2020 Crash Course, CLICK HERE

3. Effective Revision | Technique

Ensure you make a checklist of JEE Main syllabus with the list of topics/units against each subject – Physics, Chemistry and Mathematics. Tick off the important topics as and when you are done with revising the particular topic.

For Quick Revision with Mind- Maps Visit the given Links Below:

JEE Main is not difficult, rather it is predictable and tricky. Hence, clarity of concept and familiarity with the question type is essential to do well, so revision of concepts is the key. Prepare a list of concepts, topic-wise. Keep marking the concepts which are your weak areas. Keep referring to the notes&important formulae. 

It is extremely important to be in question solving mode. So, after revising every concept/topic, solve an unsolved question. One must NOT refer to solution without an attempt to solve the question. So, sufficient time must be devoted. Identify mistakes and/or weak area & fix them instantly.You have to be quick and accurate too. 

4. Effective Revision with the help of Mind Maps & Flashcards

Have a large stack of flashcards handy. It should contain all the important JEE Main formulae, concepts and diagrams. These could also be used in the last minutes before JEE Main.Here are the list of Complete Mind Maps for Organic Chemistry & Physics

5. Analyse & Strengthen your weak areas

Every time you take a mock test, Analyze your mistakes and weak areas. Work on the subjects, topics and units which are you are weak at. Resolve and strengthen them well before the actual exam takes place. 

6. Performance Analysis during practice test

Analyze your performance after every test. Subject wise weak areas needs to be identified and revisit those concepts all over again to eradicate them. Attempt Mock Tests available on eSaral App to get an instant test analysis with performance report.

Don’t panic about less months remaining for JEE preparation. You just need to focus on more study hours and practice to crack the JEE exam in remaining time.

Your preparation in these final months before the exam will have a major impact on your performance and JEE Main percentile, ranks and scores. Furthermore, your time management skills clubbed with a positive attitude will make your chances of getting into the top tech institutes of the country better. Remember, it’s important to utilize each day efficiently to secure success in JEE Main 2020.

Stay away from social networking sites for a few month. Also maintain a good health, as fitness goal is paramount during this time. Light exercise, healthy diet, good sleep are the key aspects to remain fit. Take all necessary precaution as prescribed by our government & health department. 

Whatever is the score in JEE Main 2020 January exam, there will be an option to increase score & percentile in the upcoming attempt of JEE Main 2020.

Always have a positive mindset and Believe it that whatever you have prepared is more than enough for exams. A positive mindset helps you to retain more and you will be able to score well in exams.

Kindly share your reviews and let us know the queries regarding JEE Main 2020 Exam. Drop your messages in the comment box below & let us help you with the correct guidance.

All the best 🙂

 

JEE Advanced 2020 Admit Card (Postponed) | Exam on 23rd August ’20

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IIT Delhi has postponed the JEE Advanced 2020 Admit Card release along with other events. Candidates who will complete the registration process before the last date, will be able to access the JEE Advanced admit card 2020 in online mode by providing their registration number, date of birth, mobile number and email ID. Information regarding the examination such as the venue address, timings, date and more will be made known through JEE Advanced admit card 2020. Without the admit card of JEE Advanced 2020, the candidates will not be allowed to attempt the examination. Since it is an important document, the candidates are advised to keep the JEE Advanced 2020 admit card safely till the end of the admission procedure.

How to Download JEE Advanced 2020 Admit Card?

To download JEE Advanced hall ticket 2020, candidates need to follow the steps given below:

  • Visit the official website of JEE Advanced 2020
  • Navigate to the ‘Candidate Login’ link
  • Enter your JEE Advanced 2020 registration number, Date of Birth, Mobile number and email id to login
  • JEE Advanced 2020 hall ticket will be displayed on the screen
  • Check all the details printed on admit card thoroughly
  • Download and take a printout for future reference

In case of any discrepancy in downloading the admit card of JEE Advanced 2020, aspirants have to immediately contact the officials of their respective zonal IIT.  

JEE Advanced Admit Card Important Dates:

Events

Dates

Release of JEE Advanced 2020 admit card/hall ticket To be announced
Last date to download JEE Advanced 2020 admit card To be announced
JEE Advanced 2020 exam  23-Aug-2020

 

Issues you might face while downloading JEE Advanced admit card 2020

  • Server issues: Once JEE Advanced hall tickets will be released, successful applicants can download their respective admit card. The server may be down due to a sudden traffic surge on the website.
  • Incorrect credentials: Candidates who forgot their login credentials will not be able to download JEE Advanced admit card 2020. However, they can retrieve their password by clicking the ‘Forgot Password’ tab on the official website.
  • Slow internet: This is one of the most common issues faced by JEE Advanced aspirants while downloading the admit card.

Errors in JEE Advanced admit card and ways to resolve them

  • Discrepancy/ies in JEE Advanced admit card 2020: For discrepancy in the particulars (photograph/signature) mentioned on the JEE Advanced hall ticket or on the confirmation page, candidates should immediately get in touch with the JEE Advanced authorities.
  • Admit card not received: Please note that JEE Advanced authorities will not send the admit cards by post to candidates. Candidates can download the admit card only from the official website of JEE Advanced 2020.

Documents to Carry to JEE Advanced Test Centre

Along with JEE Advanced 2020 admit card, candidates need to carry one of the following original photo identity proofs:

  • Aadhaar Card
  • Driving License
  • School/ College ID
  • Pan Card
  • Passport

JEE Advanced Previous Year Question Papers are available here.


JEE Advanced 2020 Syllabus

Click Here for JEE Advanced Complete Syllabus


 

NTA JEE Main 2020, NEET 2020 Exam dates Announced | Check here for details

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The exam dates for NTA JEE Main 2020, NEET 2020 announced by the Union Minister of Human Resource Development Ramesh Pokhriyal Nishank on tuesday (5th May’20).

The JEE (Main) 2020 and NEET 2020 exams were earlier re-scheduled by the ministry, due to the nationwide lockdown in the wake of coronavirus outbreak. The new date for the JEE (Main) 2020 exams was since then awaited.

  1. Public Notice by NTA for JEE 2020 Students
  2. Public Notice by NTA for NEET 2020 Students

The JEE Main exam will be conducted from July 18 to July 23, 2020, Ramesh Pokhriyal Nishank said.

The NEET 2020 exam will be conducted on July 26, he said, while the JEE Advanced on 23rd August ’20.

JEE Main 2020: Public Notice by NTA

NEET 2020: Public Notice by NTA

According to the announcement, Union Minister of Human Resource Development Ramesh Pokhriyal Nishank said the examination process for the JEE (Main) 2020 will begin from July this year, while the session will begin by August.

The JEE Main exam will be conducted from July 18 to July 23, 2020, Ramesh Pokhriyal Nishank said.

The NEET 2020 exam will be conducted on July 26, he said, while the JEE Advanced on 23rd August ’20.

CRASH COURSE to CRACK JEE MAIN 2020 | Register NOW 

The date for the release of admit cards for JEE Main, NEET 2020 is yet not available.

This was the second interaction between Ramesh Pokhriyal Nishank and students. This session was earlier scheduled to be held on May 2, 2020. However, the same was postponed for May 5, for some reason.

Earlier he interacted on 27th April 2020 and answered some critical questions of students and their parents.

According to the National Testing Agency (NTA), for JEE Main over nine lakh candidates have registered, on the other hand over 15.93 lakh students are likely to take up the NEET 2020 exams.


JEE Advanced Previous Year Question Papers are available here.


JEE Advanced 2020 Syllabus

Click Here for JEE Advanced Complete Syllabus

JEE Main 2020 April Exam Postponed due to Corona-virus

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JEE Main 2020 April Exam has been postponed by NTA due to Corona-virus Outbreak. The exam was scheduled to be held on April 5, 7, 9 and 11. In recent notice at official website, National Testing Agency on Wednesday (18th march) took the decision to postpone the JEE Main April 2020 Examination in pursuance of MHRC letter. New dates of JEE Main likely to be decided on March 31, 2020.

The new date will be decided in accordance with the board exams schedule and other competitive exams to ensure there is no clash.

The Ministry of Human Resource Development (MHRD) on Wednesday directed the Central Board of Secondary Education (CBSE) and all educational institutions in the country to postpone all exams including JEE Main till March 31 in view of coronavirus outbreak.

MHRD also stated that “While maintenance of academic calendars and exam schedule is important, equally important is the safety and security of the students who are appearing in various exams as also that of their teachers and parents”.

The Public Notice by NTA:

 

To ensure, your Learning never gets affected, eSaral has taken a lead in this regard and has made eSaral App FREE till 15th April’20.  

This step is taken by eSaral keeping in mind Safety and Future of students amidst ongoing health crisis. Now students can study from home safely and continue their learning.

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JEE Advanced 2020 Exam Date Announced, Check Syllabus – Important Information

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Joint Entrance Examination, JEE Advanced or IIT JEE Exam dates 2020 have been announced. Union HRD Minister Ramesh Pokhriyal today shared that the JEE Advanced 2020 would be conducted on August 23, 2020. The dates for JEE Main 2020 were announced on May 5 by the HRD Minister during a webinar with the students.

JEE Main 2020 would be conducted from July 18 to July 23. Given the date of the JEE Advanced 2020, it can be suggested that the results of JEE Main 2020 are expected to be announced by August 10. The online application process for JEE Advanced 2020 would begin only after the final results of JEE Main 2020 are announced along with the JEE Main 2020 Ranks

JEE Advanced 2020 Exam Date: 23rd August ’20

JEE Main 2020 (Revised dates by MHRD): July 18th to July 23rd ’20

JEE Advanced is organised by one of seven Zonal Coordinating IIT’s guided by the Joint Admission Board (JAB). It is a National Level Engineering Entrance Exam for admission in the Bachelors, Integrated Masters and Dual Degree Programs. The exam will be conducted in online mode. Candidate needs to clear the JEE Main Exam to be eligible for the JEE Advanced Examination.

 

JEE Advanced Result 2020

Results for JEE Advanced 2020 will tentatively be released by the IIT Delhi in the second week of August. The result will comprise the exam scores, All India Rank (AIR) of candidates, subject-wise marks, etc.

 

Students may please note that the timeline provided above is only based on the usual time frame followed by the IITs in regards the examination. The actual schedule would be announced by IIT Delhi in a few days time. IIT JEE or JEE Advanced 2020 is conducted by the IITs while JEE Main, the qualifier is conducted by National Testing Agency. Both JEE Main and JEE Advanced 2020 would be online or computer based tests.

Eligibility Criteria

Candidates must fulfill one of the following two criteria in order to be eligible for admission at IIT:

  1. Candidates must have secured at least 75% aggregate marks in the Class XII (or an equivalent) Board examination. The aggregate marks for SC, ST and PwD candidates should be at least 65%. Physics, Chemistry, and Mathematics are required as compulsory subjects in Class XII (or equivalent) Board examination in 2019 or 2020.
  2. Candidates must be within the category-wise top 20 percentile of successful candidates in their respective Class XII (or equivalent) board examination in 2019 or 2020 with Physics, Chemistry, and Mathematics as compulsory subjects.

 

JEE Advanced Exam Pattern

The details regarding the JEE Advanced Exam pattern is given below:

  • Online Mode: From 2020, the exam is conducted via online mode only.
  • Number of Papers: JEE Advanced consists of two compulsory papers (Paper 1 and Paper 2).
  • Duration of Exam: 3 hours/each paper. Some extra time will be given to the PwD candidates.
  • Type of Questions: Objective type (MCQs).
  • Language of Question Paper: English and Hindi.
  • Subjects: Physics, Chemistry, and Mathematics subjects will be asked in the exam.
  • Negative Marking: Yes, there is a provision of negative marking and different for paper 1 & 2.

JEE Advanced Previous Year Question Papers are available here.


JEE Advanced 2020 Syllabus

Click Here for JEE Advanced Complete Syllabus


All The Best 🙂

True Dip and Apparent Dip – Magnetism and Matter Class 12

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Here we will study about True Dip and Apparent Dip.

Apparent Dip

The dip at a place is determined by a dip circle. It consists of magnetized needle capable of rotation in vertical plane about a horizontal axis. The needle moves over a vertical scale graduated in degrees.

If the plane of the scale of the dip circle is not in the magnetic meridian then the needle will not indicate correct direction of earth’s magnetic field. The angle made by the needle with the horizontal is called Apparent Dip.

True Dip

When the plane of scale of dip circle is in the magnetic meridian the needle comes to rest in direction of earth’s magnetic field. The angle made by the needle with the horizontal is called True Dip.

Suppose dip circle is set at angle $\alpha$ to magnetic meridian.

Horizontal component $${B_H}’ = {B_H}\cos \alpha $$

Vertical component $${B_V}’ = {B_V}$$ (remains unchanged)

Apparent dip is $${\theta ^\prime }\tan {\theta ^\prime } = {{{B_V}’} \over {{B_H}’}} = {{{B_V}} \over {{B_H}\cos \alpha }} = {{\tan \theta } \over {\cos \alpha }}\left( {\tan \theta = {{{B_V}} \over {{B_H}}} = {\rm{ true dip }}} \right)$$

                                 

  1. For a vertical plane other than magnetic meridian $\alpha>0$ or $\cos \alpha<1$ so $\theta^{\prime}>\theta$ In a vertical plane other than magnetic meridian angle of dip is more than in magnetic meridian.
  2. For a plane perpendicular to magnetic meridian $\alpha=\frac{\pi}{2}$ $\therefore \tan \theta^{\prime}=\infty \quad$ so $\quad \theta^{\prime}=\frac{\pi}{2}$ So in a plane perpendicular to magnetic meridian dip needle will become vertical.

At magnetic equator :

  1. Angle of dip is zero.
  2. Vertical component of earths magnetic field becomes zero $B_{V}=B \sin \theta=B \sin 0=0$
  3. A freely suspended magnet will become horizontal at magnetic equator.
  4. At equator earth’s magnetic field is parallel to earth’s surface i.e., horizontal.

At magnetic poles :

  1. Angle of dip is $90^{\circ}$
  2. Horizontal component of earth’s magnetic field becomes zero. $B_{H}=B \cos \theta=B \cos 90=0$
  3. A freely suspended magnet will become vertical at magnetic poles.
  4. At poles earth’s magnetic field is perpendicular to the surface of earth i.e. vertical.

Ex. If $\theta_{1}$ and $\theta_{2}$ are angles of dip in two vertical planes at right angle to each other and $\theta$ is true dip then prove $\cot ^{2} \theta=\cot ^{2} \theta_{1}+\cot ^{2} \theta_{2}$.

Sol. If the vertical plane in which dip is $\theta_{1}$ subtends an angle $\alpha$ with meridian than other vertical plane in which dip is $\theta_{2}$ and is perpendicular to first will make an angle of $90-\alpha$ with magnetic meridian. If $\theta_{1}$ and $\theta_{2}$ are apparent dips than

$\tan \theta_{1}=\frac{B_{V}}{B_{H} \cos \alpha}$

$\tan \theta_{2}=\frac{B_{V}}{B_{H} \cos (90-\alpha)}=\frac{B_{V}}{B_{H} \sin \alpha}$

$\cot ^{2} \theta_{1}+\cot ^{2} \theta_{2}=\frac{1}{\left(\tan \theta_{1}\right)^{2}}+\frac{1}{\left(\tan \theta_{2}\right)^{2}}=\frac{B_{H}^{2} \cos ^{2} \alpha+B_{H}^{2} \sin ^{2} \alpha}{B_{V}^{2}}=\frac{B_{H}^{2}}{B_{V}^{2}}=\left(\frac{B \cos \theta}{B \sin \theta}\right)^{2}=\cot ^{2} \theta$

So $\quad \cot ^{2} \theta_{1}+\cot ^{2} \theta_{2}=\cot ^{2} \theta$


Also Read:

Properties of Paramagnetic & Diamagnetic Materials

 

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Neutral Point of Magnet – Magnetism and Matter Class 12 Physics Notes

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A neutral point of Magnet is a point at which the resultant magnetic field is zero. In general the neutral point is obtained when horizontal component of earth’s field is balanced by the produced by the magnet. When the N pole of the magnet points South and the magnet in the magnetic meridian.

When we plot magnetic field of a bar magnet the curves obtained represent the superposition of magnetic fields due to bar magnet and earth.

DEFINITION
A neutral point in the magnetic field of a bar magnet is that point where the field due to the magnet is completely neutralized by the horizontal component of earth’s magnetic field. At neutral point field due to bar magnet (B) is equal and opposite to horizontal component of earth’s magnetic field $\left( B _{ H }\right) or \quad B = B _{ H }$

 

  • Neutral point when north pole of magnet is towards geographical north of earth.

                                   

The neutral points $N_{1}$ and $N_{2}$ lie on the equatorial line. The magnetic field due to magnet at neutral point is $B =\frac{\mu_{0}}{4 \pi} \frac{ M }{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}$ Where M is magnetic dipole moment of magnet, $2 l$is its length and r is distance of neutral point.

At neutral point $B = B _{ H } \cdot \quad$ so $\quad \frac{\mu_{0}}{4 \pi} \frac{ M }{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}= B _{ H }$

For a small bar magnet $\left(\ell^{2}<< r ^{2}\right)$ then $\frac{\mu_{0}}{4 \pi} \frac{ M }{ r ^{3}}= B _{ H }$

  • Neutral point when south pole of magnet is towards geographical north of earth.

                       

The neutral points $N_{1}$ and $N_{2}$ lie on the axial line of magnet. The magnetic field due to magnet

at neutral point is $B =\frac{\mu_{0}}{4 \pi} \frac{2 Mr }{\left( r ^{2}-\ell^{2}\right)^{2}}$

At neutral point $B =\frac{\mu_{0}}{4 \pi} \frac{2 Mr }{\left( r ^{2}-\ell^{2}\right)^{2}}$ so $\frac{\mu_{0}}{4 \pi} \frac{2 Mr }{\left( r ^{2}-\ell^{2}\right)^{2}}= B _{ H }$

For a small bar magnet $\left(\ell^{2}<< r ^{2}\right)$ then $\frac{\mu_{0}}{4 \pi} \frac{2 M }{ r ^{3}}= B _{ H }$

 

Special Point

When a magnet is placed with its S pole towards north of earth neutral points lie on its axial line. If magnet is placed with its N pole towards north of earth neutral points lie on its equatorial line. So neutral points are displaced by $90^{\circ}$ on rotating magnet through $180^{\circ}$ In general if magnet is rotated by angle $\theta$ neutral point turn through an angle $\frac{\theta}{2}$

 

Neutral Point in Special Cases

                                 

(a) If two bar magnets are placed with their axis parallel to each other and their opposite poles face each other then there is only one neutral point (x) on the perpendicular bisector of the axis equidistant from the two magnets.

(b) If two bar magnets are placed with their axis parallel to each other and their like poles face each other then there are two neutral points on a line equidistant from the axis of the magnets.


Also Read:

Properties of Paramagnetic & Diamagnetic Materials

 

About eSaral
At eSaral we are offering a complete platform for IIT-JEE & NEET preparation. The main mission behind eSaral is to provide education to each and every student in India by eliminating the Geographic and Economic factors, as a nation’s progress and development depends on the availability of quality education to each and every one. With the blend of education & technology, eSaral team made the learning personalized & adaptive for everyone.

 

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Bar Magnet as an Equivalent Solenoid – Magnetism || Class 12 Physics Notes

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Bar Magnet as an Equivalent Solenoid

In solenoid each turn behaves as a small magnetic dipole having dipole moment $\mathrm{I} \mathrm{A}$. A solenoid is treated as arrangement of small magnetic dipoles placed in line with each other. The number of dipoles is equal to number of turns in a solenoid. The south and north poles of each turn cancel each other except the ends. So solenoid can be replaced by single south and north pole separated by distance equal to length of solenoid. The magnetic field produced by a bar magnet is identical to that produced by a current carrying solenoid.

Derivation of Bar Magnet as an Equivalent Solenoid

To demonstrate the similarity of a current carrying solenoid to a bar magnet, let us calculate axial field of a finite solenoid carrying current.

Consider:
$a=$ radius of solenoid

$2 l=$ length of solenoid with centre O

$n=$ number of turns per unit length
$I=$ current passing through solenoid

$O P=r$

Consider a small element of thickness $d x$ of solenoid at distance $x$ from
O. and number of turns in element $=n d x$

We know magnetic field due to n turns coil at axis of solenoid is given
by

$d B=\frac{\mu_{0} n d x I a^{2}}{2\left[(r-x)^{2}+a^{2}\right]^{\frac{3}{2}}}$

The magnitude of the total field is obtained by summing over all the elements $-$ in other words by integrating from $x=-1$ to $x=+1 .$ Thus,

$B=\frac{\mu_{0} n I a^{2}}{2} \int_{-1}^{l} \frac{d x}{\left[(r-x)^{2}+a^{2}\right]^{3 / 2}}$

This integration can be done by trigonometric substitutions. This exercise, however, is not
necessary for our purpose. Note that the range of $x$ is from $-1$ to $+1 .$ Consider the far axial field of the solenoid, i.e., $r>>$ a and $r>>1 .$ Then the
denominator is approximated by

$\begin{aligned}\left[(r-x)^{2}+a^{2}\right]^{3 / 2} &=r^{3} \\ \text { and } B &=\frac{\mu_{0} n I a^{2}}{2 r^{3}} \int_{-1}^{1} d x \\ &=\frac{\mu_{0} n I}{2} \frac{2 l a^{2}}{r^{3}} \end{aligned}$

Note that the magnitude of the magnetic moment of the solenoid is, (total number of turns $\times$ current $\times$ cross-sectional area). Thus,

$B=\frac{\mu_{0}}{4 \pi} \frac{2 m}{r^{3}}$

It is clear from the above expression that magnetic moment of a bar magnet is equal to the magnetic moment of an equivalent solenoid that produces the same magnetic field.

 

 

 

Also Read:

Biot Savart’s Law

 

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Terrestrial Magnetism – Earth’s Magnetism || Class 12 Physics Notes

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Do you know that the Earth is also a magnet? Yes!!! How do you think then that the suspended bar magnet always points in the north-south direction? Adware about the concept of the Terrestrial Magnetism that we are going to discuss in this chapter. It is really interesting to study and analyze this concept of earth’s magnetism.

The branch of Physics which deals with the study of earth’s magnetic field is called terrestrial magnetism.

  1. William Gilbert suggested that earth itself behaves like a huge magnet. This magnet is so oriented that its S pole is towards geographic north and N pole is towards the geographic south.
  2. The earth behaves as a magnetic dipole inclined at small angle $11.5^{\circ}$ to the earth’s axis of rotation with its south pole pointing geographic north.
  3. The idea of earth having magnetism is supported by following facts.
  4. A freely suspended magnet always comes to rest in N-S direction.
  5. A piece of soft iron buried in N-S direction inside the earth acquires magnetism.
  6. Existence of neutral points. When we draw field lines of bar magnet we get neutral points where magnetic field due to magnet is neutralized by earth’s magnetic field.
  7. The magnetic field at the surface of earth ranges from nearly 30 $\mu T$ near equator to about 60$\mu T$ near the poles. The magnetic field on the axis is nearly twice the magnetic field on the equatorial line.

 

Also Read:

Biot Savart’s Law

 

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Magnetic Elements – Magnetism Class 12 || Physics Notes

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The physical quantities which determine the intensity of earth’s total magnetic field completely both in magnitude and direction are called magnetic elements.

Angle of Declination $(\phi)$:

The angle between the magnetic meridian and geographic meridian at a place is called angle of declination.

(a) Isogonic Lines : Lines drawn on a map through places that have same declination are called isogonic lines.

(b) Agonic Line : The line drawn on a map through places that have zero declination is known as an agonic line.


Angle of Dip or Inclination

The angle through which the N pole dips down with reference to horizontal is called the angle of dip. At magnetic north and south pole angle of dip is $90^{\circ}$. At magnetic equator the angle of dip is zero.

OR The angle which the direction of resultant field of earth makes with the horizontal line of magnetic meridian is called angle of dip.

(a) Isoclinic Lines : Lines drawn up on a map through the places that have same dip are called isoclinic lines.

(b) Aclinic Line : The line drawn through places that have zero dip is known as aclinic line. This is the magnetic equator.

Horizontal component of earth’s magnetic field

The total intensity of the earth’s magnetic field makes an angle  with horizontal. It has

(i) Component in horizontal plane called horizontal component $B_{H}$.

(ii) Component in vertical plane called vertical component $B_{V}$

$B _{ V }= B \sin \theta \quad B _{ H }= B \cos \theta$

So $\frac{ B _{ V }}{ B _{ H }}=\tan \theta \quad$ and $\quad B =\sqrt{ B _{ H ^{2}}+ B _{ V ^{2}}}$

 

IMPORTANT POINTS

  1. If $\theta$ and $\phi$ are known we can find direction of B.
  2. If $\theta$ and $B_{H}$ are known we can find magnitude of B.
  3. So if $\theta$, $\phi$ and $B_{H}$ are known we can find total field at a place. So these are called as Elements of earth’s magnetic field.
  4. The declination gives the plane, dip gives the direction and horizontal component gives magnitude of earth’s magnetic field.
  5. If declination is ignored, then the horizontal component of earth’s magnetic field is from geogrophic south to geographic north.
  6. Angle of dip is measured by instrument called dip circle.

 

Also Read:

Biot Savart’s Law

 

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Bohr Magneton: Unit of Bohr Magneton – Magnetism || Class 12 Physics Notes

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Bohr Magneton is defined as the magnetic dipole moment associated with an atom due to orbital motion of an electron in the first orbit of hydrogen atom. This is the smallest value of magnetic moment. Unit of Bohr Magneton is-

  • In CGS units it is defined by the equation:

$\mu \mathrm{B}=(e \cdot h) / 2 m_{\mathrm{e}} \mathrm{c}$

  • In SI units it is defined by the equation:

$\mu_{\mathrm{B}}=e \cdot h / 2 m_{e}$

 

  1. The electron possesses magnetic moment due to its spin motion also $\overrightarrow{ M }_{ s }=\frac{ e }{ m _{ e }} \overrightarrow{ s }$ .where $\overrightarrow{ s }$ is spin angular momentum of electron and $S=\pm \frac{1}{2}\left(\frac{h}{2 \pi}\right)$
  2. The total magnetic moment of electron is the vector sum of its magnetic moments due to orbital and spin motion.
  3. The resultant angular momentum of the atom is given by vector sum of orbital and spin angular momentum due to all electrons. Total angular momentum $\vec{J}=\vec{L}+\vec{S}$
  4. The resultant magnetic moment $\overrightarrow{ M _{j}}=- g \left(\frac{ e }{2 m }\right) \overrightarrow{ J }$

where g is Lande’s splitting factor which depends on state of an atom.

For pure orbital motion g = 1 and pure spin motion g = 2.

 

Also Read:

Biot Savart’s Law

 

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Current loop as a Magnetic Dipole – Magnetism | Class 12th Physics Notes

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Current loop as a Magnetic Dipole

Ampere found that the distribution of magnetic lines of force around a finite current carrying solenoid is similar to that produced by a bar magnet. This is evident from the fact that a compass needle when moved around these two bodies show similar deflections. After noting the close resemblance between these two, Ampere demonstrated that a simple current loop behaves like a bar magnet and put forward that all the magnetic phenomena is due to circulating electric current. This is Ampere’s hypothesis.

We consider a circular coil carrying current I. When seen from above current flows in anti clockwise direction.

  1. The magnetic field lines due to each elementary portion of the circular coil are circular near the element and almost straight near center of circular coil.
  2. The magnetic lines of force seem to enter at lower face of coil and leave at upper face.
  3. The lower face through which lines of force enter behaves as south pole and upper face through which field lines leave behaves as north pole.
  4. A planar loop of any shape behaves as a magnetic dipole.
  5. The dipole moment of current loop $(\mathrm{M})=$ ampere turns (nI) $\times$ area of coil (A) or $\mathrm{M}=\mathrm{nIA}$.
  6. The unit of dipole moment is ampere meter $^{2}\left( A – m ^{2}\right)$
  7. Magnetic dipole moment is a vector with direction from S pole to N pole or along direction of normal to planar area.

 

Atoms as a Magnetic Dipole

In an atom electrons revolve around the nucleus. These moving electrons behave as small current loops. So atom possesses magnetic dipole moment and hence behaves as a magnetic dipole.

The angular momentum of electron due to orbital motion $L = m _{ e } vr$

The equivalent current due to orbital motion $I =-\frac{ e }{ T }=-\frac{ ev }{2 \pi r }$

–ve sign shows direction of current is opposite to direction of motion of electron.

Magnetic dipole moment $M=I A=-\frac{e v}{2 \pi r} \cdot \pi r^{2}=-\frac{e v r}{2}$

Using $L=m_{e}$ vr we have $\quad M=-\frac{e}{2 m_{e}} L$

In vector form $\overrightarrow{ M }=-\frac{ e }{2 m _{ e }} \overrightarrow{ L }$

The direction of magnetic dipole moment vector is opposite to angular momentum vector.

According to Bohr’s theory $L=\frac{n h}{2 \pi} \quad n=0, \quad 1, \quad 2 \ldots \ldots$

So $M=\left(\frac{e}{2 m_{e}}\right) \frac{n h}{2 \pi}=n\left(\frac{e h}{4 \pi m_{e}}\right)=n \mu_{B}$

Where $\mu_{ B }=\frac{ eh }{4 \pi m _{ e }}=\frac{\left(1.6 \times 10^{-19} C \right)\left(6.62 \times 10^{-34} Js \right)}{4 \times 3.14 \times\left(9.1 \times 10^{-31} kg \right)}=9.27 \times 10^{-24} Am ^{2}$ is called Bohr Magneton. This is natural unit of magnetic moment.

 

 

Also Read:

Biot Savart’s Law

 

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Class 12 Magnetism – Gauss Law Definition || Solved Examples

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As you know that the science is filled with fun facts. The deeper one dives into the concepts of science and its related fields, the greater amount of knowledge and information there is to learn in there. One such topic of study is the Gauss Law, which studies electric Charge along with a surface and the topic of Electric Flux. Let us study about the Gauss Law definition , Formula, Solved Examples in this Article,.

Gauss’s law states that the net flux of an Electric Field in a closed surface is directly proportional to the enclosed electric charge. It is one of the four equations of Maxwell’s laws of electromagnetism. It was initially formulated by Carl Friedrich Gauss in the year 1835 and relates the electric fields at the points on a closed surface and the net charge enclosed by that surface.

The surface integral of magnetic field $\overrightarrow{ B }$ over a closed surface S is always zero Mathematically $\oint_{S} \vec{B} \cdot \overrightarrow{d a}=0$

  1. Isolated magnetic poles do not exist is a direct consequence of gauss law in magnetism.
  2. The total magnetic flux linked with a closed surface is always zero.
  3. If a number of magnetic field lines are leaving a closed surface, an equal number of field lines must also be entering the surface.

Ex. A bar magnet of length 0.1 m has a pole strength of 50 Am. Calculate the magnetic field at a distance of 0.2 m from its centre on its equatorial line.

Sol. $B _{ equi }=\frac{\mu_{0}}{4 \pi} \frac{ M }{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}=\frac{10^{-7} \times 50 \times 0.1}{\left(0.2^{2}+0.05^{2}\right)^{\frac{3}{2}}}=\frac{5 \times 10^{-7}}{(0.04+0.0025)^{\frac{3}{2}}}$ or $B _{\text {equi }}=5.7 \times 10^{-5}$ Tesla


Ex. What is the magnitude of the equatorial and axial fields due to a bar magnet of length 5 cm at a distance of 50 cm from its mid-point. The magnetic moment of the bar magnet is 0.40 $Am ^{2}$

Sol. Here r >> $\ell$. So equatorial field $B _{\text {equi }}=\frac{\mu_{0}}{4 \pi} \frac{ M }{ r ^{3}}=\frac{10^{-7} \times 0.4}{(0.5)^{3}}=3.2 \times 10^{-7} T$

Axial field $B _{\text {axial }}=\frac{\mu_{0}}{4 \pi} \frac{2 M }{ r ^{3}}=2 \times 3.2 \times 10^{-7}=6.4 \times 10^{-7} T$


Ex. Find the magnetic field due to a dipole of magnetic moment 1.2 $Am ^{2}$ at a point 1 m away from it in a direction making an angle of 60° with the dipole axis?

Sol. $B =\frac{\mu_{0}}{4 \pi} \frac{ M }{ r ^{3}} \sqrt{1+3 \cos ^{2} \theta}=\frac{10^{-7} \times 1.2 \sqrt{1+3 \cos ^{2} 60}}{1}=\frac{10^{-7} \times 1.2 \times \sqrt{7}}{2}=1.59 \times 10^{-7} T$

$\tan \theta^{\prime}=\frac{1}{2} \tan \theta=\frac{1}{2} \tan 60^{\circ}=\frac{\sqrt{3}}{2}=0.866$

So $\theta^{\prime}=\tan ^{-1} 0.866=40.89^{\circ}$


Ex. A circular coil of 100 turns and having a radius of 0.05 m carries a current of 0.1 A. Calculate the work required to turn the coil in an external field of 1.5 T through $180^{\circ}$ about an axis perpendicular to the magnetic field. The plane of coil is initially at right angles to magnetic field?

Sol. Work done W = MB

$\left(\cos \theta_{1}-\cos \theta_{2}\right)=N I A B\left(\cos \theta_{1}-\cos \theta_{2}\right)$

or $W = NI _{ B }^{2} B \left(\cos \theta_{1}-\cos \theta_{2}\right)=100 \times 0.1 \times 3.14 \times(0.05)^{2} \times 1.5\left(\cos 0^{\circ}-\cos \pi\right)=0.2355 J$


Ex. A bar magnet of magnetic moment $1.5 HT ^{-1}$ lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required to turn the magnet so as to align its magnetic moment. (i) Normal to the field direction? (ii) Opposite to the field direction? (b) What is the torque on the magnet in case (i) and (ii)?

Sol. Here, M = $1.5 JT ^{-1}, B =0.22 T$

(a) P.E. with magnetic moment aligned to field = – MB

P.E. with magnetic moment normal to field = 0

P.E. with magnetic moment antiparallel to field = + MB

(i) Work done = increase in P.E. = 0 – (–MB) = MB = 1.5 × 0.22 = 0.33 J.

(ii) Work done = increase in P.E. = MB – (–MB) = 2MB = 2 × 1.5 × 0.22 = 0.66 J.

(b) We have $\tau$ = MB sin $\theta$

(i) $\theta=90^{\circ}, \sin \theta=1, \tau= MB \sin \theta=1.5 \times 0.22 \times 1=0.33 J$

This torque will tend to align M with B.

(ii) $\theta=180^{\circ}, \sin \theta=0, \tau= MB \sin \theta=1.5 \times 0.22 \times 0=0$


Ex. A short bar magnet of magnetic moment 0.32 J/T is placed in uniform field of 0.15 T. If the bar is free to rotate in plane of field then which orientation would correspond to its (i) stable and (ii) unstable equilibrium? What is potential energy of magnet in each case?

Sol. (i) If M is parallel to B then $\theta=0^{\circ}$ So potential energy $U=U_{\min }=-M B$

$U_{\min }=-M B=-0.32 \times 0.15 J=-4.8 \times 10^{-2} J$

This is case of stable equilibrium

(ii) If M is antiparallel to B then $\theta=\pi^{\circ}$ so potential energy

$U=U_{\max }=+M B=+0.32 \times 0.15=4.8 \times 10^{-2} J$

This is case of unstable equilibrium.


Also Read:

Biot Savart’s Law

 

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Potential Energy of Magnetic Dipole in Magnetic Field || Magnetism Class 12 Physics Notes

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Potential Energy of magnetic dipole in magnetic field is defined as the amount of work done in rotating the dipole from zero potential energy position to any desired position.

A current loop does not experience a net force in a magnetic field. It however, experiences a torque. This is very similar to the behavior of an electric dipole in an electric field. A current loop, therefore, behaves like a magnetic dipole.

Potential Energy of a Bar Magnet in Uniform Magnetic Field

When a bar magnet of dipole moment M is kept in a uniform magnetic field B it experiences a torque $\tau=M B \sin \theta$ which tries to align it parallel to direction of field.

If magnet is to be rotated against this torque work has to be done.

The work done in rotating dipole by small angle d$$\theta $$ is $d W =\tau d \theta$

Total work done in rotating it from angle $\theta_{1}$ to $\theta_{2}$ is

$W =\int d W =\int_{\theta_{1}}^{\theta_{2}} \tau d \theta= MB \int_{\theta_{1}}^{\theta_{2}} \sin \theta d \theta= MB \left(\cos \theta_{1}-\cos \theta_{2}\right)$

This work done in rotating the magnet is stored inside the magnet as its potential energy.

So U = MB $\left(\cos \theta_{1}-\cos \theta_{2}\right)$

The potential energy of a bar magnet in a magnetic field is defined as work done in rotating it from a direction perpendicular to field to any given direction.

$U = W _{ \theta }- W _{\frac{\pi}{2}}=- MB \cos \theta=-\overrightarrow{ M } \cdot \overrightarrow{ B }$

 

Also Read:

Biot Savart’s Law

 

 

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Magnetic Field due to a Short Bar Magnet | Magnetic Dipole – Class 12 physics Notes

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The magnetic field due to a short bar magnet at any point on the axial line is twice the magnetic field at a point on the equatorial line of that magnet at the same distance. S.l. unit of torque acting on the bar magnet is Nm.

Magnetic field due to a short bar magnet (magnetic dipole) :

On Axial Point or End on Position

The magnetic field $\overrightarrow{ B }_{\text {axial }}$ at a point P due to bar magnet will be the resultant of the magnetic fields $\overrightarrow{ B _{1}}$ due to N-pole of magnet and $\overrightarrow{ B _{2}}$ due to S-pole of magnet.

$\overrightarrow{ B }_{\text {axial }}=\overrightarrow{ B _{1}}+\overrightarrow{ B _{2}}$

$\overrightarrow{ B }_{1}=\frac{\mu_{0}}{4 \pi} \frac{ m }{( r -\ell)^{2}}(\hat{ r })$ and $\overrightarrow{ B }_{2}=\frac{\mu_{0}}{4 \pi} \frac{ m }{( r +\ell)^{2}}(-\hat{ r })$

$\therefore \quad \overrightarrow{ B }_{ axial }=\left[\frac{\mu_{0}}{4 \pi} \frac{ m }{( r -\ell)^{2}}-\frac{\mu_{0}}{4 \pi} \frac{ m }{( r +\ell)^{2}}\right] \hat{ r }=\frac{\mu_{0} m }{4 \pi}\left[\frac{4 r \ell}{( r -\ell)^{2}( r +\ell)^{2}}\right] \hat{ r }$

$\overrightarrow{ B }_{ axial }=\frac{\mu_{0}}{4 \pi} \frac{2 \overrightarrow{ Mr }}{\left( r ^{2}-\ell^{2}\right)^{2}}$

Magnetic field due to a bar magnet at an axial point has same direction as that of its magnetic dipole moment vector.

For a bar magnet of very small length $\ell<< r \overrightarrow{ B }_{ axial }=\frac{\mu_{0}}{4 \pi} \frac{2 \overrightarrow{ M }}{r^{3}}$

 

Browse More Topics Related Magnetism:

On Equatorial Point or Broadside Position 

The magnetic field $\overrightarrow{ B }_{\text {equi }}$ at a point P due to bar magnet will be the resultant of the magnetic fields $\overrightarrow{ B _{1}}$ due to N-pole of magnet and $\overrightarrow{ B _{2}}$ due to S-pole of magnet $\overrightarrow{ B }_{ equi }=\overrightarrow{ B _{1}}+\overrightarrow{ B _{2}}$ due to S-pole of magnet $\overrightarrow{ B }_{ equ i }=\overrightarrow{ B _{1}}+\overrightarrow{ B _{2}}$

$\left|\overrightarrow{ B }_{1}\right|=\frac{\mu_{0}}{4 \pi} \frac{ m }{ NP ^{2}}=\frac{\mu_{0}}{4 \pi} \frac{ m }{ r ^{2}+\ell^{2}}$ along $NP$

$\left|\overrightarrow{ B }_{2}\right|=\frac{\mu_{0}}{4 \pi} \frac{ m }{ SP ^{2}}=\frac{\mu_{0}}{4 \pi} \frac{ m }{\left( r ^{2}+\ell^{2}\right)}$ along $PS$

So $\left|\vec{B}_{1}\right|=\left|\vec{B}_{2}\right|$ On resolving $\vec{B}_{1}$ and $\overrightarrow{ B _{2}}$ along PX’ and PY we find $\left|\vec{B}_{1}\right|$ $\sin \theta$ and $$|\overrightarrow {{B_2}} |\,\sin \theta $$

are equal and opposite so they cancel each other. So resultant field

$\overrightarrow{ B }_{ equi }=\overrightarrow{ B _{1}} \cos \theta(-\hat{ r })+\overrightarrow{ B _{2}} \cos \theta(-\hat{ r })=\left[\frac{\mu_{0}}{4 \pi} \frac{ m }{\left( r ^{2}+\ell^{2}\right)} \cos \theta+\frac{\mu_{0}}{4 \pi} \frac{ m }{\left( r ^{2}+\ell^{2}\right)} \cos \theta\right](-\hat{ r })$

$=2 \cdot \frac{\mu_{0}}{4 \pi} \frac{ m }{\left( r ^{2}+\ell^{2}\right)} \cdot \frac{\ell}{\sqrt{ r ^{2}+\ell^{2}}}(-\hat{ r })$

$\overrightarrow{ B }_{ equi }=-\frac{\mu_{0}}{4 \pi} \frac{\overrightarrow{ M }}{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}$

$\overrightarrow{ B }_{ equi }=-\frac{\mu_{0}}{4 \pi} \frac{\overrightarrow{ M }}{\left( r ^{2}+\ell^{2}\right)^{\frac{3}{2}}}$

The direction of magnetic field at a point on equitorial line is opposite to magnetic dipole moment vector.

For a bar magnet of very small length $\overrightarrow{ B }_{ equi }=-\frac{\mu_{0}}{4 \pi} \frac{\overrightarrow{ M }}{r^{3}}$

At An Arbitrary Point

The point P is on axial line of magnet S’N’ with magnetic moment Mcos$$\theta $$ Magnetic flux density $B _{1}=\frac{\mu_{0}}{4 \pi} \frac{2 M \cos \theta}{ r ^{3}}$

The point P is simultaneously on the equatorial line of other magnet N”S” with magnetic moment Msin $$\theta $$ Magnetic flux density $B _{2}=\frac{\mu_{0}}{4 \pi} \frac{ M \sin \theta}{ r ^{3}}$

Total magnetic flux density at P.

$B =\sqrt{ B _{1}^{2}+ B _{2}^{2}}=\frac{\mu_{0}}{4 \pi} \frac{ M }{ r ^{3}} \sqrt{4 \cos ^{2} \theta+\sin ^{2} \theta}$ or

$\tan \theta^{\prime}=\frac{B_{2}}{B_{1}}=\frac{\frac{\mu_{0}}{4 \pi} \frac{M \sin \theta}{r^{3}}}{\frac{\mu_{0}}{4 \pi} \frac{2 M \cos \theta}{r^{3}}}=\frac{1}{2} \tan \theta$ or $\theta^{\prime}=\tan ^{-1}\left(\frac{1}{2} \tan \theta\right)$

 

Also Read:

 

 

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JEE Advanced Previous Year Questions
JEE Advanced Previous Year Question Papers – Download PDF

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Aspirants who are willing to clear Joint Entrance Examination JEE 2020 have to solve Previous Year question Papers of Both JEE Main & Advanced. Here you will get JEE Advanced Previous Year Question Papers PDF’s. This is probably the final and important step of completing the overall JEE Exam preparation. Considering the numerous benefits of it owns, eSaral offers a wide range of topic wise Previous Year questions for JEE Main and Advanced.

The exam conducting authority NTA will conduct JEE main twice a year in computer-based mode. If you are aiming to successfully crack JEE Advanced Exams, eSaral offers JEE Advanced previous year question papers from Year (2007-2019).

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Magnetic Field | Properties of Magnetic Lines of Force Class 12, JEE

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Magnetic Field is defined as the space around a magnet (or a current carrying conductor) in which its magnetic effect can be experienced. Magnetic Lines of Force can be defined as curved lines used to represent a magnetic field, drawn such that the number of lines relates to the magnetic field’s strength at a given point and the tangent of any curve at a particular point is along the direction of magnetic force at that point. The Properties of Magnetic Lines of Force are also discussed.

Magnetic Field:

Magnetic Field is defined as the space around a magnet (or a current carrying conductor) in which its magnetic effect can be experienced.

(1) The magnetic field in a region is said to be uniform if the magnitude of its strength and direction is same at all points in that region.

(2) A magnetic field in a region is said to be uniform if the magnitude of its strength and direction is same at all the points in that region.

(3) The strength of magnetic field is also known as magnetic induction or magnetic flux density.

(4) The $\mathrm{SI}$ unit of strength of magnetic field is Tesla (T)

1 Tesla = 1 newton ampere $^{-1}$ metre $^{-1}\left( NA ^{-1} m ^{-1}\right)=1$ Weber metre $^{-2}\left( Wb m ^{-2}\right)$

  1. The CGS unit is Gauss (G)

1 Gauss (G) $=10^{-4}$ Tesla (T)

Properties of Magnetic Lines of Force:

The magnetic field lines is the graphical method of representation of magnetic field. This was introduced by Michael Faraday.

(1) A line of force is an imaginary curve the tangent to which at a point gives the direction of magnetic field at that point.

(2) The magnetic field line is the imaginary path along which an isolated north pole will tend to move if it is free to do so.

(3) The magnetic lines of force are closed curves. They appear to converge or diverge at poles. outside the magnet they run from north to south pole and inside from south to north.

(4) The number of lines originating or terminating on a pole is proportional to its pole strength.

Magnetic flux = number of magnetic lines of force = $\mu_{0} \times m$

Where $${\mu _0}$$ is number of lines associated with unit pole.

(5) Magnetic lines of force do not intersect each other because if they do there will be two directions of magnetic field which is not possible.

(6) The magnetic lines of force may enter or come out of surface at any angle.

(7) The number of lines of force per unit area at a point gives magnitude of field at that point. The crowded lines show a strong field while distant lines represent a weak field.

(8) The magnetic lines of force have a tendency to contract longitudnally like a stretched elastic string producing attraction between opposite pole.

(9) The magnetic lines of force have a tendency to repel each other laterally resulting in repulsion between similar poles.

(10) The region of space with no magnetic field has no lines of force. At neutral point where resultant field is zero there cannot be any line of force.

(11) Magnetic lines of force exist inside every magnetised material.                                           

Important points :

(i) Magnetic lines of force always form closed and continuous curves whereas the electric lines of force are discontinuous.

(ii) Each electric line of force starts from a positive charge and ends at a negative charge. Electric lines of force are discontinuous because no such lines exist inside a charged body.

(iii) In magnetism, as there are no monopoles, therefore, the magnetic field lines will be along closed loops with no starting or ending. The magnetic lines of force would pass through the body of the magnet.

(iv) At very far off points, the lines due to an electric dipole and a magnetic dipole appear identical.

 

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