NTA NEET 2020 Postponed due to Coronavirus Lockdown

Medical entrance exam, NTA NEET 2020 postponed due to COVID-19 outbreak. NEET 2020 Exam was scheduled to be held on May 3. The entrance exam is held annually for admission to undergraduate medical or MBBS, BDS and AYUSH programmes. Through this exam, admissions to veterinary courses are also done.

As per the orders of the HRD Minister, the NEET and JEE Main exams are postponed till last week of May. The admit cards will be issued after April 15. Here are the latest Update  by NTA through public notice issued on official website :

NTA NEET 2020 Postponed
.
With this NEET is the second major exam to have been postponed due to the nationwide lockdown due to COVID-19 outbreak.

Earlier this week, following the orders of the HRD Minister, NTA had postponed the engineering entrance exam, JEE Main. The exam was scheduled to begin on April 5.

 

As of now, the examination is proposed to be held in the last week of May 2020. The exact date will be announced later on after assessing the situation.

Accordingly, the Admit Cards for the Examination which were to be issued on 27th March 2020 will now be issued later on after assessing the situation after 15th April 2020 only.

We understand that the academic calendar and schedule is important but equally important is well being of every citizen including students.

Students and parents are advised to not get worried about the Examination. Moreover, parents are requested to help them in utilizing this time very efficiently for preparing for the Examination and focus on critical concepts in order to close learning gaps if any, NTA would keep students updates about the latest developments and would inform about changes with ample time.

All the best!

Stay home, Stay safe and study from home.

Click Here to Know How to Study Well in 21 Days Lockdown Period!!!

 

NEET UG 2020 Admit Card not Releasing Today: Exams Likely to be Postponed

In view of this pandemic situation due to Coronavirus (COVID-19), the admit cards for National Eligibility cum Entrance Test (NEET-UG) will not be released today (on March 27, 2020), as mentioned in the brochure issued by National Testing Agency (NTA). NEET UG 2020 admit card new date will be announced after the lockdown. Along with this, it is being highly expected that NEET 2020 exam might be postponed.

NEET 2020 Exam

NEET 2020 is scheduled to be conducted on May 3, however, the ongoing situation due to Coronavirus outbreak might affect the exam date as well. According to one of the senior official in the NTA “We are not releasing the admit cards tomorrow (March 27, 2020). The new dates for downloading the admit cards will be issued later following a review on April 14, 2020.”

There is, however, no decision yet on whether the date of the exam, scheduled to take place on 3 May, will be changed or not.

NEET 2020 Revised Dates

As per media reports, NTA is scheduled to hold a review meeting on 14th April 2020 in which the apex testing agency will decide the new exam dates for all the exams that have been postponed due to the COVID-19 outbreak. Until then, students are advised to take advantage of the postponement and prepare well for the NEET 2020 Exam which will be held after the lockdown period

The NEET exam is for admission to the undergraduate medical courses in all medical institutions.

NTA NEET 2020 application process was conducted from December 2nd, 2019 to January 1st, 2020. The exam is conducted in 11 languages and will test candidates on Physics, Chemistry, and Biology subjects with 180 MCQ questions for a total of 720 marks and for 3 hours’ duration. For more details, Click Here.

 

 

Bihar Board 12th Result 2020 Out || Check Here

Much awaited Bihar Board 12th result 2020 has been declared. BSEB has announced the results for Class 12th Board Exams at BSEB Official Website. BSEB releases results through online mode. The Students who have appeared in Bihar 12th Board Exam can check their result and download it.

The Examination of Bihar Board Class 12 has completed on 13th February 2020.

Students can also check the Bihar Board Inter Topper List, Merit List, Percentage of Result status on official site.

Steps to Check Bihar 12th Board Result 2020: 

Step 1:- Visit the Official Website or Click on direct link below.

Step 2:- Enter Roll Code and Roll Number.

Bihar board 12th result 2020Step 3:- Enter Code shown below on Result page.

Step 4:- Click on View Result Button.

Step 5:- BSEB 12th Result 2020 will display on screen.

Step 6:- Now you can check out Bihar 12th Result 2020

 

Click Here to Check BSEB 12th Board Result (Direct Link)

The students, after checking the Bihar Board Class 12 result 2020, will also be able to either download a PDF copy or take a printout of the scorecard. The students are advised to download / take a printout of the Bihar Board 12th Result 2020 mark sheet to use as a provisional result until official scorecards are issued by the BSEB. However, the provisional BSEB Result 2020 copy is only valid until the original mark sheets are issued by the board. Therefore, we advise all the students to collect their BSEB Board 12th Result 2020 mark sheets from their respective schools later.

 

Re-evaluation and Rechecking

The performance of the students in the BSEB 12th Result 2020 will lay the foundation for their academic future. Taking into account the importance of the Bihar Board 12th Result 2020, if the students are unhappy with the outcome of their hard work, they can apply for re-evaluation / rechecking of the answer sheets. The Bihar Board provides this facility to all the concerned students at the cost of a nominal fee. More information in this regard can be found at the respective schools.

 

Compartmental Exam

The BSEB also provides the students unable to pass in the Bihar Board Class 12 exam 2020 with an opportunity to prove themselves again. Students who failed in one or more exams in the board exam can opt for Bihar Board compartmental exam 2020. This exam will prevent an entire academic year from getting wasted. Students can appear in the compartmental exam of Bihar Board 2020 by filling in the online application form and paying the requisite application fee. For details, Visit BSEB Official Website.

Bihar 12th Board Exam Syllabus. 

Bihar 12th Board 2020 Topper’s of Science Stream:

RANK Reg. No. Candidates Name Total Marks
1 R-430700022-18 Neha Kumari 476
2 R-410440239-18 Vikky Kumar 474
2 R-360020136-18 Jahangeer Alam 474
3 R-230020391-18 Shivam Kumar Verma 473
3 R-710430106-18 Manish Kumar Jaiswal 473

 

 

JEE Main 2020 April Exam Postponed due to Corona-virus

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.

eSaral App comprises complete courses for Class 9 to 12 including Video Tutorials, Complete Study Material designed by Top IITians & Doctor Faculties.

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Share this Information among your dear and near ones and continue your learning with eSaral by being safe and keeping safe.

CRASH COURSE for JEE 2020

To know more about JEE Main 2020 CLICK HERE

jee advanced 2020 exam date
JEE Advanced 2020 Exam Date, Syllabus, Information Brochure (Released)

JEE Advanced 2020 will be organised by IIT Delhi this year. The Indian Institute of Technology, (IIT) Delhi has released the information brochure for JEE Advanced 2020 on its official website. JEE Advanced 2020 Exam Date has been announced. The Exam Will be held in 2 Sessions, Paper 1 and Paper 2.

JEE Advanced 2020 Exam Date: 17 May 2020 (Announced)

  • Session 1: Paper 1 – 9 am to 12 pm
  • Session 2: Paper 2 – 2 pm to 5 pm

As per the Released Information on Website, The Online Registration for JEE Advanced Exam will held from May 1 to May 6, 2020. 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.

IIT will release the admit card for the JEE Advanced 2020 on May 12, 2020, on its official website. It will be available to download until May 17, 2020.

 

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.

 

Online Crash Course for JEE Advanced 2020 Preparation : CLICK HERE FOR DETAILS

Information Bulletin (Brochure) for JEE Advanced 2020

JEE Advanced 2020 Information Bulletin available now : Download PDF [Online Registration, Examination and Admission Details].

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.

Previous Year Question Papers are available here.

 

JEE Advanced 2020 Syllabus

See theComplete & Detailed syllabus of JEE Advanced 2020 here : JEE Advanced Complete Syllabus

JEE Advanced Previous Year papers are available here: IIT JEE Papers

 

All The Best 🙂

True Dip and Apparent Dip – Magnetism and Matter Class 12

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

 

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

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

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

 

Click here for the Video tutorials of Magnetic Effect of Current Class 12

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

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

 

Click here for the Video tutorials of Magnetic Effect of Current Class 12

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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Magnetic Elements – Magnetism Class 12 || Physics Notes

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

 

Click here for the Video tutorials of Magnetic Effect of Current Class 12

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

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

 

Click here for the Video tutorials of Magnetic Effect of Current Class 12

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.

For free video lectures and complete study material, Download eSaral APP.

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

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

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

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

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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Saple papers & marking Scheme Class 12
Sample Papers Class 12 CBSE 2020 || All Subjects – Download PDF

CBSE Sample Papers for March 2020 Exams has been recently released by the Central Board of Secondary Education. Sample Papers Class 12 CBSE 2020 are given to download in PDF form. CBSE Class 12 Marking Scheme 2020 for these Sample Papers is also available in this article for download in PDF format.

These CBSE Sample Papers 2020 are based on updated format of CBSE Class 12 Question Papers and are very important for the preparation of upcoming CBSE Board Exam 2020.

Sample Papers & Marking Scheme

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When it is about preparing for the exam efficiently, CBSE Sample Papers for Class 12 are one of the best resources. It benefits students to get prior practice before they attempt for the final exam. Also, they will get to know if they are prepared for the exam completely or not. They can test their knowledge for all these subjects and get confident about the answers. CBSE sample papers for class 12 help students to identify topics that need to be focused more, types of tricky questions, and much more.

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If you solve the sample CBSE Sample question paper it will help students to test themselves. They will get to know their current preparation level. Accordingly, they can work on them.

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Solving CBSE Class 12 sample question papers will help students to identify silly mistakes. Note those points and try not to repeat them during the exam.

 

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

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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It is very crucial to know about the syllabus before starting your preparations for JEE. Knowing the syllabus beforehand will help in better understanding of the topics relevant to JEE Advanced Exam.

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

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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Magnetic Dipole Moment Definition, Formulas & Solved Examples | Class 12, JEE & NEET

A magnetic moment is a quantity that represents the magnetic strength and orientation of a magnet or any other object that produces a magnetic field. More precisely, a magnetic moment refers to a magnetic dipole moment, the component of magnetic moment that can be represented by a magnetic dipole.Know the Magnetic Dipole Moment Definition, Formulae and solved examples here.

Magnetic Dipole: An arrangement of two magnetic poles of equal and opposite strengths separated by a finite distance is called a magnetic dipole.

 

(1) Two poles of a magnetic dipole or a magnet are of equal strength and opposite nature.

(2) The line joining the poles of the magnet is called magnetic axis.

(3) The distance between the two poles of a bar magnet is called the magnetic length of magnet. It is denoted by 2$\ell$

(4) The distance between the ends of the magnet is called geometrical length of the magnet.

(5) The ratio of magnetic length and geometrical length is $\frac{5}{6}$ or 0.83

(6) A small bar magnet is treated like a magnetic dipole.

Magnetic Dipole Moment Definition:

The product of strength of either pole and the magnetic length of the magnet is called magnetic dipole moment. $\overrightarrow{ M }= m (\overrightarrow{2 \ell})$

 

Important Points to Remember

(1) It is a vector quantity whose direction is from south pole to north pole of magnet.

(2) The unit of magnetic dipole moment is ampere metre $^{2}\left( Am ^{2}\right)$ and Joule/Tesla (J/T). The dimensions are $M^{0} L^{2} T^{0} A^{1}$

(3) If a magnet is cut into two equal parts along the length then pole strength is reduced to half and length remains unchanged.

New magnetic dipole moment M’ = m’ $(2 \ell)=\frac{ m }{2} \times 2 \ell=\frac{ M }{2}$

The new magnetic dipole moment of each part becomes half of original value.

4. If a magnet is cut into two equal parts transverse to the length then pole strength remains unchanged and length is reduced to half. New magnetic dipole moment $M^{\prime}=m\left(\frac{2 \ell}{2}\right)=\frac{M}{2}$

The new magnetic dipole moment of each part becomes half of original value.

(5) In magnetism existence of magnetic monopole is not possible.

(6) The magnetic dipole moment of a magnet is equal to product of pole strength and distance between poles. M = m d

(7) As magnetic moment is a vector, in case of two magnets having magnetic moments $M_{1}$ and $M_{2}$ with angle $\theta$ between them, the resulting magnetic moment.

$M =\left[ M _{1}^{2}+ M _{2}^{2}+2 M _{1} M _{2} \cos \theta\right]^{1 / 2}$ with $\tan \phi=\left[\frac{M_{2} \sin \theta}{M_{1}+M_{2} \cos \theta}\right]$


Ex. The force between two magnetic poles in air is 9.604 mN. If one pole is 10 times stronger than the other, calculate the pole strength of each if distance between two poles is 0.1 m?

Sol. Force between poles $F =\frac{\mu_{0}}{4 \pi} \frac{ m _{1} m _{2}}{ r ^{2}}$ or $9.604 \times 10^{-3}=\frac{10^{-7} \times m \times 10 m}{0.1 \times 0.1}$

or $m ^{2}=96.04 N ^{2} T ^{-2}$ or m = 9.8 N/T

So strength of other pole is 9.8 × 10 = 98 N/T


Ex. A steel wire of length L has a magnetic moment M. It is then bent into a semicircular arc. What is the new magnetic moment?

Sol. If m is the pole strength then M = m . L or $m =\frac{ M }{ L }$

If it is bent into a semicircular arc then L = $\pi$ or $r=\frac{L}{\pi}$

So new magnetic moment $M^{\prime}=m \times 2 r=\frac{M}{L} \times 2 \times \frac{L}{\pi}=\frac{2 M}{\pi}$


Ex. Two identical bar magnets each of length L and pole strength m are placed at right angles to each other with the north pole of one touching the south pole of other. Evaluate the magnetic moment of the system.

Sol. $M _{1}= M _{2}= mL$


Also Read:

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