JEE Main Study Notes for Magnetism: Check Quick Notes, Sample Questions
JEE Main Study Notes for Magnetism: An attractive and repulsive phenomenon that is produced by a moving electric charge is called magnetism. Electric field and magnetic field both are there in the affected region around a moving charge. A bar magnet is the most familiar example of magnetism, which is attracted to a magnetic field and can attract or repel other magnets.
Magnetic Effects of current and magnetism is one of the high weightage chapters of around 11%-12% weightage in JEE Main. In terms of preparation for JEE Main, candidates should make short notes for revision before the examination. These short notes will help them to make their preparation strategies stronger. Check JEE Main Physics Preparation Tips
JEE Main Study Notes for Magnetism cover some important topics such as Motion in a magnetic field, Torque on a current loop, Magnetic dipole, magnetic field due to the current element - Biot Savart Law, The Solenid & Toroid, and The moving Coil Galvanometer. Check JEE Main Physics Syllabus
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Magneto-static
It is the study of magnetic fields where the currents are steady i.e. not changing with time. It is the magnetic correspondent of electrostatics, where the charges are stationary.
Coulomb’s law in magnetism
The Coulomb force is inversely proportional to the squared distance between the magnetic poles Q1 and Q2 , repels between like poles (N–N, S–S) and attracts between unlike poles (N–S).
Magnitude of magnetic force between two poles
F = (µ0 /4π) (m1m2/r2)
where, µ0 is called the absolute magnetic permeability of free space.
µ0= 4π×10-7 Wb A -1 m-1
Magnetic field
Magnetic field, of any magnetic pole, is the region around it in which its magnetic influence can be realized.
Lines of Force
Line of force (Flux Lines) is the path along which a unit north pole would move if it were free to do so.
Properties of magnetic lines of force
Properties of magnetic lines of force
- Lines of force are directed away from a north pole and are directed towards a south pole. If they are isolated poles, a line of force starts from a north pole and ends at a south pole.
- Tangent, at any point, to the magnetic line of force gives the direction of magnetic intensity at that point.
- Two lines of force never cross each other.
Magnetic Dipole
Magnetic dipole is a combination of two isolated , equal and opposite magnetic poles separated by a small distance.
Magnetic Moment
Magnetic moment M of a magnetic dipole is the product of its pole strength and the magnetic length.
M = m×2l
Torque in Magnetic field
Work done in rotating a magnetic dipole in a magnetic field
W = MB (cosθ1 – cosθ2)
Potential Energy of a magnetic dipole in a magnetic field
Magnetic moment (M)
M = I×A
Other formulae of M
M = nIπr2
M= eVr/2 = er2ω/2 = er2 2πf/2 = er2 π/T
M = nµB
Resultant magnetic moment
When two bar magnets are lying mutually perpendicular to each other, then,
When two coils, each of radius r and carrying current i, are lying concentrically with their planes at right angles to each other, then
Must Read:
Atoms as a magnetic dipole
I = eω/2π
M = eωr2/2
M = n (eh/4πm)
eh/4πm is called Bohr’s magneton(smallest value of magnetic
moment which an electron can possesss).
Magnetic flux density at a distance from a magnetic dipole in free space
B= (µ0 /4π) (m1/r2)
Force
F = (µ0 /4π) (m1 m2/r2)
Magnetic intensity at any point due to a magnetic pole in free space
F = (µ0 /4π) (m/r2)
Magnetic intensity due to a bar magnet in free space
Point situated on the axial line (End-on position)
F = (µ0 /4π) [2Mr/(r2-l2)2]
In case of a magnetic-dipole, F = (µ0 /4π) [2M/r3]
Point situated on equatorial line (Broad side-on position)
F = (µ0 /4π) [M/(r2+l2) 3/2]
In case of a magnetic-dipole, F = (µ0 /4π) [M/r3]
Point situated anywhere
F= (µ0 /4π) [M/r3] √1+3 cos2θ
Direction, tan β = ½ tan θ
Combined magnetic field due to bar magnet and earth – “Neutral Points”
Bar magnet placed in a magnetic meridian
North pole facing north of earth
B = (µ0 /4π) [M/(r2+l2) 3/2]
At neutral points, B=H
So, H = (µ0 /4π) [M/(r2+l2) 3/2]
North pole facing south of earth
B = (µ0 /4π) [2Mr/(r2-l2)2]
At neutral points, B=H
So, H = (µ0 /4π) [2Mr/(r2-l2)2]
Intensity of magnetization (I)
Intensity of magnetization (I)
Intensity of magnetization (I), is the magnetic moment (M) developed per unit volume (V) of the specimen, when subjected to a uniform magnetic field.
I = M/V = m/a
where,
m is the pole strength
a is the area of the specimen.
Relation between magnetic field (B) and field intensity (H)
B = H+4πI
Permeability (µ)
It is the ratio between magnetic induction to the strength of magnetic field.
µ= B/H
For paramagnetic and ferromagnetic substances, B > H. So, µ > 1
For diamagnetic substances, B < H. So, µ < 1
Susceptibility (k)
Susceptibility of a magnetic substance is the ratio between intensity of magnetization (I) to the strength of magnetic field (H).
k = I/H
Relation between µ and k
µ= 1+4πk
For substances which get magnetized in the direction of magnetic field,
- I is positive
- k is positive
- B > H
- µ > 1
For substances which get magnetized in the direction opposite to that of magnetic field,
- I is negative
- k is negative
- B < H
- µ < 1
Check: JEE Main Study Notes for Current Electricity
Magnetic Substance
Magnetic Substance
A substance that is affected by a magnetic field is called a magnetic substance.
- Diamagnetic substances:- substances that are repelled by magnets.
Example- antimony, bismuth, lead, tin, zinc, mercury, gold, phosphorus
- Paramagnetic substances:- substances that are weakly attracted by the magnets.
- Example-aluminium, platinum, oxygen, manganese, chromium
- Ferromagnetic substances:- substances that are strongly attracted by magnets.
Example- iron, cobalt, nickel
Curie-Weiss Law
Magnetic flux
Magnetic flux
Magnetic flux lined with the surface is the product of area and component of B perpendicular that area.
When θ = 90º, cosθ = 0.
This signifies, no magnetic flux is linked with a surface when the field is
parallel to the surface.
When θ = 0º, cosθ = 1.
This signifies magnetic flux linked with a surface is maximum when the area is held perpendicular to the direction of field.
Biot-Savart Law or Ampere’s Theorem
Or, dB = (µ0 /4π) (I dl sin θ/r2)
Field due to straight current – carrying conductor of finite length at a point P, perpendicular distance a from the linear conductor XY
B =(µ0I/4πa)× (sin θ1 + sin θ2)
Direction:-
For current in the conductor
- from X to Y, the direction of B is normal to the plane of conductor downwards.
- from Y to X, the direction of B is normal to the plane of conductor downwards.
Field due to straight carrying conductor of infinite length at a point P, perpendicular distance R from the linear conductor XY
B = (µ0I/2πa)
Field due to two concentric coils of radii r1 and r2 having turns N1 and N2 in which same current I is flowing in anticlockwise direction at their common center O
B = µ0 I/2 [N1/r1+ N2/r2]
If the number of turns in them is same,
B = µ0 NI/2 [1/r1+ 1/r2]
Direction:- Direction of B will be normal to the plane of paper
upwards.
Field due to two concentric coils of radii r1 and r2 having turns N1 and N2 in which same current I is flowing in mutually opposite direction at their common center O
B = µ0 I/2 [N1/r1 - N2/r2]
If the number of turns in them is same, B = µ0NI/2 [1/r1 - 1/r2]
Direction:- Direction of B will be normal to the plane of paper
upwards.
Field due to circular coil at the center O
B = µ0 I/2R
Field due to two parallel very long linear conductors carrying current in same direction
- At point P i.e. at a distance r/2 from both conductors,
B=0
- At a point Q i.e. at a distance x from first and r+x from second conductor,
B =µ0 2I/4π [(1/x) + (1/r+x)]
Direction:-B is normal to the plane of paper downwards.
At a point P i.e. at a distance x from first and r-x from second
conductor,
B =µ0 2I/4π [(1/x) - (1/r-x)]
If B is positive, then its direction will be normal to the plane of paper
upwards.
If B is negative, then its direction will be normal to the plane of paper
downwards.
Field due to two parallel very long linear conductors carrying current in opposite direction (same figure as above)
At point P distance x from first conductor,
B = µ0 2I/4π [(1/x) + (1/r-x)]
Direction:- of B will be normal to the plane of paper downwards.
At point Q distance x from first conductor,
B = µ0 2I/4π [(1/x) - (1/r-x)]
Direction:- of B will be normal to the plane of paper upwards.
Field due to semicircular arc of wire at the center O of the arc
B = (µ0 /4π) (πI/r)
Direction:- Direction of B will be at right angle to the plane of circular arc downwards. If the direction current is in anticlockwise, then the direction of field B will be a right angle to the plane of circular arc upwards.
Field due to straight wire and loop at the center O of the loop
- If the current in the loop in anticlockwise direction
B = (µ0 /4π) [2πI/r + 2I/r]
Direction:- Normal to the plane of paper upwards.
- If the current in the loop in clockwise direction
B = (µ0 /4π) [2πI/r - 2I/r]
Direction:- Normal to the plane of paper downwards
Field due to two semicircular arc of wire
B = µ0I/4 [1/a-1/b]
B = µ0I/4 [1/R+1/r]
Direction:- Normal to the plane of paper downward.
Field due to two concentric circular arcs at O
B = (µ0 /4π) Iθ [1/r1 -1/r2]
where,
r 1 is the radius of inner circle
r 2 is the radius of outer circle.
Direction:- Normal to the plane of paper upwards
Field due to semicircular arc and straight conductor at point P
B = (µ0I/4πr) [π+2]
Direction:- Normal to the plane upward.
Field due to semicircular arc and straight conductor at point O
B = (µ0I/4πr) [π+1]
Direction: Normal to the plane upward.
Field due to square loop having length of side a at center C
B = 2√2(µ0I/ πa)
Direction:- Normal to the plane of paper downwards.
Field at the center of a circular coil carrying current I
B = (µ0 /2) (NI/r)
Magnetic field at any point on the axis of a circular coil carrying current I
B = (µ0 /2) [NIa2/(a2 +x2) 3/2]
- Magnetic field at the center of the coil
B = (µ0 /2) [NI/a]
- Magnetic field at a point situated large distance away on the axis
B = (µ0 /2π) [NIA/x3]
- Current loop as a magnetic dipole
B = (µ0 /4π) [2M/x3]
where, M (=IA) is the magnetic moment of the magnetic dipole.
Magnetic field at any point on the axis of a solenoid carrying current:-
B = (µ0 NI/2l) [cosθ1 - cos θ2]
For an infinitely long solenoid, θ1 = 0 and θ2 = π. So, B=µ0 NI
At one end, B = µ0 NI/2
Field due to a current in cylindrical rod
- Outside:- B = µ0I/2πR
- Surface:- B = µ0I/2πR
- Inside:- B = µ0 IR/2πR2
Field due to a toroid
- Inside:- B = µ0 NI - µ0 NI/2πR
- Outside:- B =0
Force on electric current
Force on a moving charge in a magnetic field
Lorentz Force
Motion of a charged particle at right angles to a magnetic field Radius,
r = mv/qB
Force on a conductor carrying current and placed in a magnetic field
Fleming’s left-hand rule
Check: JEE Main Study Notes for Work, Power, and Energy
Fleming’s left-hand rule
Stretch the first finger, central finger, and the thumb of your left hand in mutually perpendicular directions. If the first finger points towards the magnetic field, central finger points towards electric current then the thumb gives the direction of force acting on the conductor.
The force between two parallel conductors carrying currents
F = µ0I1I2 l /2πd
Torque on a current loop= NIBA cos θ = NMB cos θ (Since, M = IA)
Moving Coil Galvanometer
I = (C/nBA)θ = Kθ
where K = C/nBA is known as the reduction factor of the moving coil galvanometer.
Sensitivity of a Galvanometer
Current Sensitivity:- Si = C/nAH
Smaller the value of Si, more sensitive is the galvanometer.
Voltage Sensitivity:- Sv =V/G = CG/nAH
Smaller the value of Sv, more sensitive is the galvanometer
Conversion of a galvanometer into an ammeter
- Is /Ig = G/S
- S=GIg/Ig = GIg/I-Ig
Conversion of a galvanometer into a voltmeter
R =(V/Ig ) – G
Ampere’s current law
Cyclotron
- T = 2πm/qB
- v = qB/2πm
- ω = θB/m
- radius of particle acquiring energy E, r = (√2mE)/qB
- velocity of particle at radius r, v = qBr/m
- the maximum kinetic energy (with upper limit of radius = R)
Kmax = ½ [q2B2R2/m]
Magnetic field produced by a moving charge
Tips to prepare for Magnetism
Tips to prepare for Magnetism
- Candidates should have good command over the concepts of Magnetic Effects of Current and Magnetism and should know how to apply them well at the time of solving questions in the exam.
- Solve all the questions at home with proper concentration and try to do all calculations on your own. Understand the derivation of each formula clearly as if you don’t remember the formulas, you can easily derive them but also try to remember all the formulas because, in some questions, you will be required to directly apply these formulas.
- Make a plan to prepare for the chapter and follow the time table regularly.
- Understand all the laws from this chapter along with their applications. Aspirants can easily solve the questions from this chapter if they have studied properly.
- Try to solve previous year’s question papers and give mock tests.
Check JEE Main Question Paper
Solved Sample Questions
Solved Sample Questions
Question. The horizontal component of the earth's magnetic field at any place is 0.36×10–4 Weber/m2. If the angle of dip at that place is 60o then the value of a vertical component of earth's magnetic field will be-(in Wb/m2)-
(A) 0.12 × 10–4
(B) 0.24 × 10–4
(C) 0.40 × 10–4
(D) 0.62 × 10–4
Sol. As we have done in the last problem,
BH tan Φ = 0.36 × 10–4 tan 60o
= 0.36 × 10–4 √3 = 0.62 × 10–4 Wb / m2
Question. The value of angle of dip at a place on earth is 45o. If the horizontal component of earth's magnetic field is 5×10–5 Tesla then the total magnetic field of earth will be-
(A)5/√2 × 10–5 Tesla
(B) 10/√2 × 10–5 Tesla
(C) 15/√2 × 10–5 Tesla
(D) zero
Sol. Using the same formula again, we know
Total Magnetic Field B = √B2
V + B2
H = √B2
H tan2 45o + B2
H = BH √2Hence, the answer is (A) 5√2 × 10–2 Wb / m2.
Question. The ratio of intensities of magnetic field, at distances x and 2x from the center of a magnet of length 2cm on its axis, will be-
(A) 4 : 1
(B) 4 : 1 approx
(C) 8 : 1
(D) 8 : 1 approx
Sol. Given that the length of the magnet is 2 cm.
We know that for finding the ratios of intensities of magnetic field
B = m0 / 4π 2M / (l3 + x3) >> m0 / 4π 2M / x3
Hence, the correct option is (A).
Question. The length of a bar magnet is 10 cm and its pole strength is 10–3 Weber. It is placed in a magnetic field of induction 4π × 10–3 Tesla in a direction making an angle of 30o with the field direction. The value of torque acting on the magnet will be-
(A) 2π × 10–7 N-m
(B) 2π × 10–5 N-m
(C) 0.5 × 102 N-m
(D)0.5N-m.
Sol. The length of the bar magnet is 10cm.
Pole strength is given to be 10-3 Weber. The torque is given by
t = MB sin θ = m / B sin θ
= 10–3 × 0.1 4π × 10–3 × 0.5 = 2π × 10–7 N – m
Question. The ratio of total intensities of magnetic field at the equator and
the poles will be-(A) 1 : 1
(B) 1 : 2
(C) 2 : 1
(D) 1 : 4
Sol. Since we know that the total intensity of magnetic field remains
constant, hence, Be /Bp = 1 : 1.
Q. Two magnets A and B are equal in length, breadth and mass, but their magnetic moments are different. If the time period of B in a vibration magnetometer is twice that of A, then the ratio of magnetic moments will be-
(A) 1/2
(B) 2
(C) 4
(D) 12
Sol. The question is demanding the relation between the time periods and magnetic moments. So we use the given formula
T = 2π √I / MB M μ 1/T2
This gives (C) as the correct option.
Question. The period of oscillation of a freely suspended bar magnet is 4 second. If it is cut into two equal parts lengthwise then the time period of each part will be-
(A) 4 sec.
(B) 2 sec.
(C) 0.5 sec.
(D) 0.25 sec.
Sol. The period of oscillation is 4 sec.
The time period is calculated with the help of the formula
T = 2π √I / MB = 2π√ml2/ 12 × m/b= 4 sec.
Question. A current of 2 ampere is flowing in a coil of radius 50 cm and number of turns 20. The magnetic moment of the coil will be-
(A) 3.14 amp-m2
(B) 31.4 amp-m2
(C) 314 amp-m2
(D) 0.314amp-m2
Sol. M = μ R2Ni = 3.14 × 0.25 × 20 × 2 = 3.14 amp / m2.
Question. The magnetic induction inside a solenoid is 6.5×10–4T. When it is filled with iron medium then the induction becomes 1.4T. The relative permeability of iron will be-
(A) 1578
(B) 2355
(C) 1836
(D) 2154
Sol. The given induction inside the solenoid is 6.5×10–4T.
It changes to 1.4T when filled with iron medium. Hence, the relative
permeability is given by the relation
μ1 = B/B0 = 1.4 / 6.5 × 10–4
= 2154.
JEE Main Magnetism FAQs
Question: How much weightage is carried in JEE Main by the topic of Magnetism?
Answer: The topic of Magnetism carries 11%-12% of the weightage in JEE Main. The question paper has around 8-10 marks on this topic.
Question: What is the marking scheme for the questions on the topic of Magnetism in JEE Main?
Answer: The marking scheme is similar to the other topics of JEE Main Physics Syllabus, Candidates will be awarded 4 marks for every correct answer and 1 mark will be deducted for every incorrect answer. Check JEE Main Exam Pattern
Question: What are the topics covered in JEE Main Magnetism Study Notes?
Answer: JEE Main Study Notes for Magnetism cover some important topics such as Motion in a magnetic field, Torque on a current loop, Magnetic dipole, magnetic field due to the current element - Biot Savart Law, The Solenid & Toroid, and The moving Coil Galvanometer.
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