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JEE Important Chapter- Magnetic Effects of Current and Magnetism

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Magnetic Effects of Current and Magnetism for JEE

Magnetic Effects of Current and Magnetism for JEE

The chapter magnetic effects of electric current deals with the magnetic field produced by the current-carrying wire and moving charges and also magnetic force acting on a current-carrying wire. In magnetism, properties of a magnetic dipole, earth’s magnetism and different magnetic material are discussed. From JEE's point of view, students can score good marks in these chapters with proper practice and effort.


In magnetic effects of current, problems based on moving charges in a magnetic field and their applications are important. The magnetic field of various current-carrying conductors of various shapes and solenoids are derived using Biot Savart law and Ampere’s law. The force acting on a current-carrying conductor in a magnetic field is also important. Here, we will see many magnetic effects of electric current examples.


In the section on magnetism, earth’s magnetism and magnetic materials are very important and problems based on a magnetic bar placed in an external magnetic field are also discussed in detail. Also, new terms related to the magnetisation of material are also discussed along with the hysteresis curve.


Now, let us move on to the important concepts and formulae related to JEE and JEE Main exams from magnetic effects of current  and magnetism along with a few solved magnetic effects of electric current examples.


Important Topics of Magnetic Effects of Current and Magnetism

  • Magnetic force on a moving charge in a magnetic field

  • Trajectory of a charge moving in a perpendicular magnetic field

  • Cyclotron

  • Biot Savart Law

  • Ampere’s Law

  • Magnetic field due to a circular and straight wire

  • Magnetic field due to a solenoid

  • Force on a current-carrying conductor in a magnetic field.

  • The force between two parallel current-carrying conductors

  • Magnetic dipole moment of a magnet

  • Terms related to magnetism

  • Elements of Earth’s Magnetic Field

  • Magnetic Materials

  • Curie’s Law

  • Hysteresis curve


Important Concepts of Magnetic Effects of Current and Magnetism

Name of the Concept

Key Points of Concept

  1. Magnetic force on a moving charge in a magnetic field

  • A moving charge produces an electric field and magnetic field.

  • The charge (q) moving with a velocity (v) in a magnetic field (B) experiences a magnetic force(F) given by 

$\vec F=q(\vec v\times \vec B)$

$F=qvB\sin\theta$



  • The direction of magnetic force is given by right-hand rule as shown below



  1. The trajectory of a charge moving in a perpendicular magnetic field

  • When a charge (q) moves in a perpendicular magnetic field (B), then it follows a circular path and necessary centripetal force is given magnetic force.

  • The radius (R) of the circular path is given by

$R=\dfrac{mv}{qB}$

  • Time period of motion of charge in a circular path is

$T=\dfrac{2\pi m}{qB}$

  • Kinetic energy of circular path is given by

$K=\dfrac{q^2B^2R^2}{2m}$

  1. Cyclotron

  • Cyclotron is a device used to accelerate positively  charged particles.



  • The charged particles are accelerated by an alternating electric field created by two D shaped semicircular hollow chambers in a perpendicular magnetic field.

  • Since the charged particle is moving in a perpendicular magnetic field, it follows a circular path of increasing radius

  • The cyclotron frequency is given by

$f=\dfrac{qB}{2\pi m}$

  1. Biot Savart Law

  • According to Biot Savart law, the magnetic field (dB) at a point due to the current element (dl) is given by the expression



$\vec dB=\dfrac{\mu_0}{4\pi}\dfrac{\vec {Idl}\times\vec r}{r^3}$

$dB=\dfrac{\mu_0}{4\pi}\dfrac{Idl\sin\theta}{r^2}$

Where μ is the magnetic permeability of the medium and for free space, magnetic permeability μ0 = 4ℼ✕ 10-7 H/m

  • If you hold a current-carrying conductor in such a way that your thumb points towards the direction of current, then the remaining fingers give the direction of the magnetic field at that point.


(Image will be uploaded soon)


  • In the case of a circular current-carrying coil, if the four folding fingers of the right-hand point towards the direction of the current, then the thumb point towards the direction of magnetic field


(Image will be uploaded soon)


  1. Ampere’s Law

  • Ampere’s law gives a method to calculate the magnetic field due to a given current distribution.

  • According to ampere’s law, the line integral of the magnetic field around any closed curve is equal to μ0 times the net current enclosed by the closed curve. 



  • $\oint\vec B\cdot\vec{dl}=\mu_0i_{enclosed}$

  • $\oint\vec B\cdot\vec{dl}=\mu_0(i_1+i_3-i_2)$

  1. Magnetic field due to a circular and straight wire

  • Magnetic field due to a circular coil having current I and radius R at a distance x along its axis is given by

$B=\dfrac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}$




  • Magnetic field due to an infinitely long current-carrying wire having current I at a distance d  is

$B=\dfrac{\mu_0I}{2\pi d}$


  1. Magnetic field due to a solenoid

  • A solenoid is a tightly wounded cylindrical coil of insulated wire having a smaller diameter than its length

  • Magnetic field due to a solenoid having n number of turns per unit length along its axis outside the solenoid

(Image will be uploaded soon)

$B=\mu_0nI$

  1. Force on a current-carrying conductor in a magnetic field.

  • When a current-carrying conductor having length L is placed in a magnetic field B, it will experience a force given by the formula,



$F=ILB\sin\theta$

The direction of force is given by fleming’s left-hand rule as given below


(Image will be uploaded soon)


  1. Force between two parallel current-carrying conductors

  • Force  acting between two parallel current-carrying conductors having length l is given by

$F_1=\dfrac{\mu_0}{4\pi}\dfrac{2I_1I_2}{d}\times l$

$F_2=\dfrac{\mu_0}{4\pi}\dfrac{2I_1I_2}{d}\times l$



  • If the current in the conductors is in the same direction, there will be the force of attraction and if the currents are in the opposite direction, then there will be the force of repulsion

  1. The magnetic dipole moment of a magnet

  • Magnetic dipole moment of a magnet gives the strength of a magnet. 

  • It is a vector quantity directed from south to north and its magnitude is given by

$\vec M=m(2\vec l)$

Where m is the pole strength and l is the magnetic length.



  1. Terms related to magnetism

  • Intensity of magnetising field (H):

Intensity of magnetising field or magnetising field is the extent to which a magnetic field can magnetise a substance and its SI unit is A/m

  • Intensity of Magnetisation(I):

It is the degree to which a substance is magnetised when it is placed in a magnetic field given by the formula,

$I=\dfrac{M}{V}$

Where M is the induced dipole moment and V is the volume

  • Magnetic susceptibility (ꭕm)

Magnetic susceptibility of a material is defined as the ratio of the intensity of magnetisation (I) to the magnetic intensity(H) applied to the substance.

$\chi_m=\dfrac{I}{H}$

  1. Elements of Earth’s Magnetic Field

  • There are three elements used to completely determine the earth’s magnetic field at a place. They are given below

  • Magnetic declination(θ): It is the angle between geographic meridian and magnetic meridian planes at a place

  • Angle of inclination or dip(ɸ): It is the angle that the intensity of the magnetic field of earth makes with horizontal line.

  • The horizontal component of earth’s magnetic field (BH): The earth’s magnetic field (B) at a place can be resolved into horizontal(BH) and vertical (Bv)  components.

$B_V=B\sin\phi$

$B_H=B\cos\phi$

$\tan\phi=\dfrac{B_V}{B_H}$

Where ɸ is the angle of inclination.

  1. Magnetic Materials

  • Paramagnetic Materials

These materials experience a weak force of  attraction in a magnetic field. They get feebly magnetised in the direction of the applied magnetic field.


Eg:- Al, Na, O2(at S.T.P)

  • Diamagnetic materials

Diamagnetic materials experience a weak force of repulsion in an external magnetic field. They also get weakly magnetised in the opposite direction of the applied magnetic field.

Eg: Cu, Bi, N2(at STP), Pd

Ferromagnetic materials

These materials experience a strong force of attraction in a magnetic field. They get strongly magnetised in the direction of the applied magnetic field.

Eg: Fe, Co, Ni,


  1. Curie’s Law

  • According to curie’s law, the magnetic susceptibility of a paramagnetic substance is inversely proportional to its absolute temperature.

$\chi\varpropto\dfrac{1}{T}$

$\chi=\dfrac{C}{T}$

Where C is the curie’s constant and T is the absolute temperature. 

  • The temperature above which a ferromagnetic material becomes paramagnetic material is called curie temperature.

  1. Hysteresis Curve

  • It is a B-H curve of a ferromagnetic material by noting down the variation in magnetic field density (B) with respect to changing magnetising field (I).



  • The lagging of magnetic flux density behind the magnetising field is called hysteresis and the B-H curve is called the hysteresis curve.

  •  The remainder value of magnetic flux density (ob) even after magnetising field (H) is reduced to zero is called retentivity.

  • The value of magnetising in the opposite direction to reduce the magnetisation of the material to zero is called coercivity

  • The area of the B-H curve gives the energy loss per one cycle of magnetisation and demagnetisation.


List of Important Formulae

Sl. No

Name of the Concept

Formula

1.

Lorentz force formula

The force acting on a charged particle (q) in a combined electric field (E) and magnetic field (B)

$\vec F=q\vec E+q(\vec v\times \vec B)$

2.

Pitch and radius of a helical path of a charged particle (q) moving at an angle θ to the magnetic field(B) with a velocity (v) 

$\text{Pitch}=\dfrac{2\pi mv\cos\theta}{qB}$

$r=\dfrac{mv\sin\theta}{qB}$

3.

Magnetic field inside a torroid 



$B=\mu_0nI$

4.

The magnetic moment of a current-carrying coil having an area of cross-section (A) and N number of turns

$ M=NAI$


5.

The torque acting on a current-carrying coil placed in a magnetic field B

$ \tau = NAIB\sin\theta$

Where θ is the angle between area vector and magnetic field and N is the number of turns 

7.

The torque acting on a magnetic dipole placed in a magnetic field B

$\vec \tau = \vec M\times \vec B$

$ \tau = MB\sin\theta$

Where θ is the angle between magnetic dipole moment M and magnetic field B.

7.

The potential energy of a magnetic dipole in an external magnetic field B.

$ P.E = -MB\cos\theta$

Where θ is the angle between magnetic dipole moment M and magnetic field B.


8.

Relation between relative permeability (μr) and magnetic susceptibility(𝜒m)

$\mu_r=1+\chi_m$





Solved Examples 

  1. A current of 10 A is flowing in a wire of a length of 1.5 m. A force of 15 N acts on it when it is placed in a uniform magnetic field of 2 T. The angle between the direction of the current and magnetic field is…

Sol:

Given, 

Uniform magnetic field, B = 2T

Current passing through the wire, I = 10 A

Length of wire, L = 1.5 m 

The force acting on the wire placed in the magnetic field, F = 15 N

The formula to calculate the magnetic force acting on the wire is given by,

$F=ILB\sin\theta$

Substitute values for each term in the formula to calculate the angle between the magnetic field and direction of the current.

$15=10\times 1.5\times 2\sin\theta$

$\sin\theta=\dfrac{1}{2}$

$\theta=\sin^{-1}(0.5)$

$\theta=30^{0}$

Therefore, the angle between the direction of the current and magnetic field is 30o.


Key Point: A current-carrying placed in a magnetic field experience a force.


  1. If the angle of dip at two places are 300 and 450 respectively, then the ratio of horizontal components of earth’s magnetic field at two places will be  

Sol:

Given,

The angle of dip of two places are θ1= 300 and  θ2=450

The horizontal component of earth’s magnetic field where θ1= 300 is given by,

$B_{H1}=B\cos\theta_1$

$B_{H1}=B\cos(30^0)$

$B_{H1}=\dfrac{\sqrt{3}B}{2}$...(1)

Similarly, the horizontal component of earth’s magnetic field where θ1= 450 is given by,

$B_{H2}=B\cos\theta_2$

$B_{H2}=B\cos(45^0)$

$B_{H2}=\dfrac{B}{\sqrt{2}}$....(2)

Divide equation (1) by equation (2) to obtain the ratio of horizontal component of earth’s magnetic field.

$\dfrac{B_{H1}}{B_{H2}}=\dfrac{\left(\dfrac{\sqrt{3}B}{2}\right)}{\left(\dfrac{B}{\sqrt{2}}\right)}$

$\dfrac{B_{H1}}{B_{H2}}=\dfrac{\sqrt{3}}{\sqrt{2}}$

Therefore, the ratio of horizontal components of earth’s magnetic field at two places is $\dfrac{\sqrt{3}}{\sqrt{2}}$.


Key point: The horizontal component of earth’s magnetic field and dip angle is related by the formula. 


Previous Year Questions from JEE paper

  1. A proton, a deuteron and an α particle are moving with the same momentum in a uniform magnetic field. The ratio of magnetic forces acting on them is _______ and their speeds are in the ratio______.(JEE 2021)

a. 2 : 1 : 1 and 4 : 2 : 1

b. 1 : 2 : 4 and 2 : 1 :1

c. 1 : 2 : 4 and 1 : 1 : 2

d. 4 : 2 : 1 and 2 : 1 : 1

Sol: 

We know velocity(v) of a body is related to its momentum by the formula

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

The magnetic force acting on a charged particle is given by the formula,

Let the mass of a proton, duetron and  α particle be m, 2m and 4m respectively.

Let the charge of a proton, duetron and  α particle be q, 2q and 4q respectively.

Force acting on proton in a magnetic field B is given by,

$F_1=qvB$

$F_1=q\left(\dfrac{P}{m}\right)B$

Force acting on duetron in a magnetic field B is given by,

$F_2=qvB$

$F_2=q\left(\dfrac{P}{2m}\right)B$

Force acting on α particle in a magnetic field B is given by,

$F_3=2qvB$

$F_3=2q\left(\dfrac{P}{4m}\right)B$

$F_3=q\left(\dfrac{P}{2m}\right)B$

The ratio of magnetic forces on the charged particles are,

$F_1:F_2:F_3=q\left(\dfrac{P}{m}\right)B:q\left(\dfrac{P}{2m}\right)B:q\left(\dfrac{P}{2m}\right)B$

$F_1:F_2:F_3=2:1:1$

The ratio of velocities of charge particles are,

$v_1:v_2:v_3=\left(\dfrac{P}{m}\right):\left(\dfrac{P}{2m}\right):\left(\dfrac{P}{4m}\right)$

$v_1:v_2:v_3=4:2:1$

Therefore, correct option is option (a)


Key Point: Knowing the charge and mass of proton, duetron and alpha particle is important in order to solve the above problem. 


  1. The figure gives experimentally measured B vs H variation in a ferromagnetic material. The retentivity, coercivity and saturation, respectively, of the material, are(JEE Main 2021)



a. 1.5 T, 50 A/m, 1 T

b. 1 T, 50 A/m, 1.5 T

c. 1.5 T, 50 A/m, 1 T

d. 150 A/m, 1 T, 1.5 T

Sol:


Retentivity is the ability of an object to retain magnetism even after the magnetising field is removed. It is denoted by the value of X = 1T

Coercivity is the opposing magnetising field required to demagnetise the material and it is denoted by the value of Y = 50 A/m

Saturation is the value of magnetic field density in the B-H curve denoted by the value of Z = 1.5 T


Practice Questions

  1. A magnet is parallel to a uniform magnetic field. If it is rotated by 60o, the work done is 0.8 J. How much work is done in moving 30o further? (Ans: 0.8 J)

  2. A charged particle (charge q) is moving in a circle of radius R with a uniform speed v. The associated magnetic moment is given by μ (Ans: qvR/2)


Conclusion

In this article we have provided important information regarding the chapter magnetic effects of electric current and magnetism such as important concepts, formulae, etc.. Students should work on more solved examples along with  previous year question papers for scoring good grades in the JEE exams.

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23rd June 2022 - Maths, Physics and Chemistry
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24th June 2022 - Maths, Physics and Chemistry
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July
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JEE Main 2022 Book Solutions and PDF Download

JEE Main 2022 Book Solutions and PDF Download

View all JEE Main Important Books
In order to prepare for JEE Main 2022, candidates should know the list of important books i.e. RD Sharma Solutions, NCERT Solutions, RS Aggarwal Solutions, HC Verma books and RS Aggarwal Solutions. They will find the high quality readymade solutions of these books on Vedantu. These books will help them in order to prepare well for the JEE Main 2022 exam so that they can grab the top rank in the all India entrance exam.
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books
Maths
NCERT Book for Class 12 Maths
books
Physics
NCERT Book for Class 12 Physics
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JEE Main Mock Tests

JEE Main Mock Tests

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JEE Main 2022 free online mock test series for exam preparation are available on the Vedantu website for free download. Practising these mock test papers of Physics, Chemistry and Maths prepared by expert teachers at Vedantu will help you to boost your confidence to face the JEE Main 2022 examination without any worries. The JEE Main test series for Physics, Chemistry and Maths that is based on the latest syllabus of JEE Main and also the Previous Year Question Papers.
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JEE MAIN MOCK TEST - 1
3 hr  • 75 questions • OBJECTIVE
JEE MAIN MOCK TEST - 3
3 hr  • 75 questions • OBJECTIVE
JEE MAIN MOCK TEST - 2
3 hr  • 75 questions • OBJECTIVE
Toppers

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Shreyas
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Shreyas

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NIT Nagpur

Nidhi Sharma
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Nidhi Sharma

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Luv Mehan
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Luv Mehan

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JEE Main 2022 Cut-Off

JEE Main 2022 Cut-Off

JEE Main Cut Off
NTA is responsible for the release of the JEE Main 2022 June and July Session cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Main 2022 June and July Session is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Main qualifying marks for 2021 ranged from 87.8992241 for general-category, while for OBC/SC/ST categories, they ranged from 68.0234447 for OBC, 46.8825338 for SC and 34.6728999 for ST category.
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JEE Main 2022 Results

JEE Main 2022 Results

JEE Main 2022 June and July Session Result - NTA has announced JEE Main result on their website. To download the Scorecard for JEE Main 2022 June and July Session, visit the official website of JEE Main NTA.
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Rank List
Counselling
Cutoff
JEE Main 2022 state rank lists will be released by the state counselling committees for admissions to the 85% state quota and to all seats in NITs and CFTIs colleges. JEE Main 2022 state rank lists are based on the marks obtained in entrance exams. Candidates can check the JEE Main 2022 state rank list on the official website or on our site.
The NTA will conduct JEE Main 2022 counselling at https://josaa.nic.in/. There will be two rounds of counselling for admission under All India Quota (AIQ), deemed and central universities, NITs and CFTIs. A mop-up round of JEE Main counselling will be conducted excluding 15% AIQ seats, while the dates of JEE Main 2022 June and July session counselling for 85% state quota seats will be announced by the respective state authorities.
NTA is responsible for the release of the JEE Main 2022 June and July Session cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Main 2022 June and July Session is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Main qualifying marks for 2021 ranged from 87.8992241 for general category, while for OBC/SC/ST categories, they ranged from 68.0234447 for OBC, 46.8825338 for SC and 34.6728999 for ST category.
Want to know which Engineering colleges in India accept the JEE Main 2022 scores for admission to Engineering? Find the list of Engineering colleges accepting JEE Main scores in India, compiled by Vedantu. There are 1622 Colleges that are accepting JEE Main. Also find more details on Fees, Ranking, Admission, and Placement.
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Counselling

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FAQs on JEE Important Chapter- Magnetic Effects of Current and Magnetism

FAQ

1. What is the weightage of the laws of motion in JEE?

Nearly 1-2 questions are asked in the exam from this chapter covering about 10 marks.

2. What is the difficulty level of the chapter on magnetic effects of current and magnetism?

Students can easily answer the questions coming from the magnetism section if they have gone through the concepts thoroughly. However, the questions from magnetic effects of current requires knowledge and problem solving skills to score good marks.

3. Are previous year's questions Important for JEE Main?

Yes, previous year's JEE Main questions are important. Once the student is finished with learning concepts and solving problems, they must go through the previous year's question papers of the last 20 years to understand the test pattern and frequent questions asked in the exam.