What is Biot-Savart Law?

Biot-Savart Law gives the magnetic field produced by a small current element. The magnetic field dB at a point P due to current element Idl is: dB = (μ₀/4π) × (Idl × r̂)/r², where r is the distance from the element to point P. It's the magnetic equivalent of Coulomb's law.

What is Ampere's Circuital Law?

Ampere's Circuital Law states that the line integral of magnetic field around any closed loop equals μ₀ times the total current enclosed. Mathematically: ∮B⃗·dl⃗ = μ₀I_enclosed. It's used to find magnetic fields in symmetric current distributions like solenoids and toroids.

What is Lorentz Force?

Lorentz Force is the total electromagnetic force on a charged particle moving in both electric and magnetic fields. F = qE + q(v × B). The magnetic component F = qvBsinθ is perpendicular to both velocity and magnetic field, causing circular or helical motion.

How does a cyclotron work?

A cyclotron accelerates charged particles using alternating electric field between two D-shaped electrodes (dees) placed in a perpendicular magnetic field. The magnetic field keeps particles in circular paths while the electric field accelerates them. Time period T = 2πm/qB is independent of velocity, enabling resonance.

What is the force between two parallel current carrying conductors?

Two parallel conductors carrying currents I₁ and I₂ separated by distance d experience force F/L = μ₀I₁I₂/(2πd) per unit length. Parallel currents (same direction) attract, while antiparallel currents (opposite direction) repel. This defines the SI unit of current - Ampere.

What is the working principle of Moving Coil Galvanometer?

A Moving Coil Galvanometer works on the principle that a current-carrying coil in a magnetic field experiences torque τ = NIAB. This torque rotates the coil until balanced by restoring torque of spring. Deflection is proportional to current: θ = (NAB/k)I, where k is the spring constant.

Magnetic Effects of Current and Magnetism for JEE 2025-26 | Complete Notes with Formulas & Solved Examples

Magnetic Effects of Current and Magnetism JEE notes, Formulas, PYQs — Magnetic Effects of Current and Magnetism - JEE Notes, Formulas, PYQs

Magnetic Field - Fundamentals

A magnetic field is a region around a magnet or current-carrying conductor where magnetic effects can be experienced. Understanding magnetic fields is fundamental to electromagnetism and forms the basis for numerous JEE problems.

1.1 Sources of Magnetic Field

Natural Sources

Permanent magnets (lodestone)
Earth's magnetic field
Celestial bodies (Sun, planets)

Artificial Sources

Current-carrying conductors
Electromagnets
Moving charges

1.2 Magnetic Field (B)

Magnetic Field Definition

The magnetic field B (also called magnetic flux density or magnetic induction) is defined by the force it exerts on a moving charge.

\vec{F} = q(\vec{v} \times \vec{B})

SI Unit: Tesla (T) = Wb/m² = kg/(A·s²)

CGS Unit: Gauss (G), where 1 T = 10⁴ G

Dimension: [MT⁻²A⁻¹]

1.3 Properties of Magnetic Field Lines

💡 Key Properties

Closed curves: Magnetic field lines always form closed loops
Never intersect: Two field lines never cross each other
Direction: Outside magnet: N to S; Inside magnet: S to N
Tangent gives direction: Tangent at any point gives B direction
Density indicates strength: Closer lines = stronger field
No starting/ending point: Unlike electric field lines

⚠️ Important Distinction

Property	Electric Field Lines	Magnetic Field Lines
Nature	Open curves	Closed curves
Starting point	Positive charge	None (no monopole)
Ending point	Negative charge	None

1.4 Magnetic Flux

Magnetic Flux (Φ)

\Phi = \vec{B} \cdot \vec{A} = BA\cos\theta

Where θ is the angle between B and area vector (normal to surface)

SI Unit: Weber (Wb) = T·m² = V·s

Dimension: [ML²T⁻²A⁻¹]

Gauss's Law for Magnetism:

\oint \vec{B} \cdot d\vec{A} = 0

Net magnetic flux through any closed surface is zero (no magnetic monopoles exist)

Biot-Savart Law

Biot-Savart Law is the fundamental law for calculating magnetic fields due to current-carrying conductors. It's the magnetic equivalent of Coulomb's law in electrostatics and is essential for JEE Advanced.

2.1 Statement of Biot-Savart Law

Biot-Savart Law

"The magnetic field dB at a point P due to a small current element Idl is directly proportional to the current I, the length element dl, sine of the angle between dl and r, and inversely proportional to the square of distance r."

d\vec{B} = \frac{\mu_0}{4\pi} \frac{I(d\vec{l} \times \hat{r})}{r^2}

OR in magnitude form:

dB = \frac{\mu_0}{4\pi} \frac{I \cdot dl \cdot \sin\theta}{r^2}

Where:

• μ₀ = Permeability of free space = 4π × 10⁻⁷ T·m/A

• I = Current through conductor

• dl = Length of current element

• r = Distance from element to point P

• θ = Angle between dl and r

💡 Direction of Magnetic Field

The direction of dB is given by the Right Hand Rule:

Point fingers along current direction (dl)
Curl fingers toward r (position vector)
Thumb points in direction of dB

Remember: dB is perpendicular to both dl and r (comes out of the plane containing them)

2.2 Magnetic Field Due to Straight Wire

Finite Straight Wire

B = \frac{\mu_0 I}{4\pi d}(\sin\phi_1 + \sin\phi_2)

Where d is perpendicular distance, φ₁ and φ₂ are angles at point P

Infinite Straight Wire:

(φ₁ = φ₂ = 90°)

B = \frac{\mu_0 I}{2\pi d}

Semi-infinite Wire:

(φ₁ = 90°, φ₂ = 0°)

B = \frac{\mu_0 I}{4\pi d}

2.3 Magnetic Field Due to Circular Loop

Circular Current Loop

At Center (x = 0):

B_{\text{center}} = \frac{\mu_0 I}{2R}

At Axial Point:

B_{\text{axis}} = \frac{\mu_0 IR^2}{2(R^2 + x^2)^{3/2}}

For N turns (coil):

B_{\text{center}} = \frac{\mu_0 NI}{2R}

Far from loop (x >> R):

B \approx \frac{\mu_0 IR^2}{2x^3} = \frac{\mu_0}{4\pi} \cdot \frac{2M}{x^3}

where M = IA = Magnetic dipole moment

2.4 Magnetic Field Due to Arc

Arc of Circle at Center

B = \frac{\mu_0 I \theta}{4\pi R}

Where θ is the angle subtended in radians

Full Circle (θ = 2π):

B = \frac{\mu_0 I}{2R}

Semicircle (θ = π):

B = \frac{\mu_0 I}{4R}

Quarter circle (θ = π/2):

B = \frac{\mu_0 I}{8R}

📝 Solved Example 1

Question: A circular coil of 100 turns and radius 10 cm carries a current of 1 A. Calculate the magnetic field at (a) the center (b) a point on axis 10 cm from center.

Solution:

Given: N = 100, R = 0.1 m, I = 1 A, x = 0.1 m

(a) At center:

B_{\text{center}} = \frac{\mu_0 NI}{2R} = \frac{4\pi \times 10^{-7} \times 100 \times 1}{2 \times 0.1}

B_{\text{center}} = \frac{4\pi \times 10^{-5}}{0.2} = 2\pi \times 10^{-4}

B_{\text{center}} = 6.28 \times 10^{-4} \text{ T} = 0.628 \text{ mT}

(b) At axial point (x = R):

B_{\text{axis}} = \frac{\mu_0 NIR^2}{2(R^2 + x^2)^{3/2}}

Since x = R:

B_{\text{axis}} = \frac{\mu_0 NIR^2}{2(2R^2)^{3/2}} = \frac{\mu_0 NI}{2R \cdot 2\sqrt{2}}

B_{\text{axis}} = \frac{B_{\text{center}}}{2\sqrt{2}} = \frac{6.28 \times 10^{-4}}{2.83}

B_{\text{axis}} = 2.22 \times 10^{-4} \text{ T} = 0.222 \text{ mT}

Ampere's Circuital Law

Ampere's Circuital Law provides an elegant method to calculate magnetic fields when there is high symmetry. It's the magnetic equivalent of Gauss's law in electrostatics.

3.1 Statement of Ampere's Law

Ampere's Circuital Law

"The line integral of magnetic field B around any closed loop (Amperian loop) is equal to μ₀ times the total current enclosed by that loop."

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}

Physical Basis:

Based on the fact that magnetic field lines form closed loops around current-carrying conductors.

💡 When to Use Ampere's Law

Ampere's law is most useful when:

Symmetry exists such that B is constant along the Amperian loop
B is parallel or perpendicular to dl at all points
Common applications: Infinite wire, solenoid, toroid

3.2 Applications of Ampere's Law

Configuration	Magnetic Field	Where
Infinite Straight Wire	B = μ₀I/(2πr)	At distance r from wire
Solenoid (inside)	B = μ₀nI	n = N/L = turns per unit length
Solenoid (outside)	B ≈ 0	For ideal solenoid
Toroid (inside)	B = μ₀NI/(2πr)	r = radius of Amperian loop
Toroid (outside & center)	B = 0	No enclosed current
Thick Cylindrical Wire (r < R)	B = μ₀Ir/(2πR²)	Inside the wire
Thick Cylindrical Wire (r > R)	B = μ₀I/(2πr)	Outside the wire

3.3 Solenoid

Magnetic Field in Solenoid

A solenoid is a long coil of wire with many turns wound closely together.

Inside (ideal solenoid):

B = \mu_0 nI = \frac{\mu_0 NI}{L}

Uniform field parallel to axis

At End (finite solenoid):

B_{\text{end}} = \frac{\mu_0 nI}{2} = \frac{B_{\text{center}}}{2}

Half of center value

Properties of Solenoid:

Field inside is nearly uniform
Field outside is nearly zero (for ideal solenoid)
Acts like a bar magnet
End from which field lines emerge is North pole

3.4 Toroid

Magnetic Field in Toroid

A toroid is a solenoid bent into a circular ring shape.

Inside the toroid (within windings):

B = \frac{\mu_0 NI}{2\pi r}

where r is distance from center of toroid

Outside toroid:

\[B = 0\]

In central cavity:

\[B = 0\]

For thin toroid (R >> a):

B = \mu_0 nI \text{ where } n = \frac{N}{2\pi R}

📝 Solved Example 2

Question: A solenoid 50 cm long has 500 turns and carries a current of 2 A. Calculate: (a) Magnetic field at the center (b) Magnetic field at one end.

Solution:

Given: L = 0.5 m, N = 500, I = 2 A

n = N/L = 500/0.5 = 1000 turns/m

(a) At center:

B_{\text{center}} = \mu_0 nI = 4\pi \times 10^{-7} \times 1000 \times 2

B_{\text{center}} = 8\pi \times 10^{-4} = 2.51 \times 10^{-3} \text{ T} = 2.51 \text{ mT}

(b) At one end:

B_{\text{end}} = \frac{B_{\text{center}}}{2} = \frac{2.51}{2}

B_{\text{end}} = 1.26 \text{ mT}

Lorentz Force

The Lorentz Force is the total electromagnetic force experienced by a charged particle moving in electric and magnetic fields. This concept is fundamental for understanding particle motion and forms the basis of many JEE questions.

4.1 Lorentz Force Equation

Complete Lorentz Force

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

Where:

qE = Electric force (along E)
q(v × B) = Magnetic force (perpendicular to both v and B)

4.2 Magnetic Force on Moving Charge

Magnetic Force

\vec{F} = q(\vec{v} \times \vec{B})

In magnitude:

F = qvB\sin\theta

Where θ is angle between v and B

Special Cases:

θ = 0° or 180°: F = 0 (v parallel or antiparallel to B)
θ = 90°: F = qvB (maximum force)

⚠️ Key Properties of Magnetic Force

Always perpendicular to both v and B
Does no work on the charged particle (W = 0)
Cannot change speed (only direction changes)
Kinetic energy remains constant
Zero force on stationary charge (v = 0 → F = 0)

4.3 Direction of Magnetic Force

💡 Right Hand Rules

For Positive Charge:

Point fingers along v, curl toward B, thumb gives F direction

For Negative Charge:

Use left hand OR right hand rule gives opposite direction

Motion of Charged Particle in Magnetic Field

The motion of charged particles in magnetic fields leads to circular or helical paths, depending on the angle of entry. This concept is crucial for understanding devices like cyclotrons and mass spectrometers.

5.1 Circular Motion (v ⊥ B)

When Velocity is Perpendicular to Magnetic Field

The charged particle moves in a circular path.

Radius of circular path:

r = \frac{mv}{qB}

Time period:

T = \frac{2\pi m}{qB}

Frequency:

f = \frac{qB}{2\pi m}

Angular frequency:

\omega = \frac{qB}{m}

Key Insight:

Time period T and frequency f are independent of velocity and radius!

This property is used in cyclotrons.

5.2 Helical Motion (v at angle θ to B)

When Velocity is at Angle θ to B

The particle moves in a helical (spiral) path.

Velocity components:

v_∥ = v cos θ (along B) - causes linear motion
v_⊥ = v sin θ (perpendicular to B) - causes circular motion

Radius of helix:

r = \frac{mv\sin\theta}{qB}

Pitch of helix:

p = v_\parallel \times T = \frac{2\pi mv\cos\theta}{qB}

5.3 Cyclotron

Cyclotron - Particle Accelerator

A cyclotron accelerates charged particles using alternating electric field and magnetic field.

Working Principle:

Magnetic field keeps particles in circular paths (inside dees)
Electric field accelerates particles (in the gap between dees)
Time period is constant → Resonance condition

Cyclotron frequency:

f_c = \frac{qB}{2\pi m}

Maximum KE:

KE_{max} = \frac{q^2B^2r_{max}^2}{2m}

Maximum velocity:

v_{max} = \frac{qBr_{max}}{m}

⚠️ Limitations of Cyclotron

Cannot accelerate neutrons (uncharged)
Cannot accelerate electrons efficiently (too light, become relativistic quickly)
Relativistic effects: At high speeds, mass increases, breaking resonance
Solution: Synchrotron (varies B and/or f)

5.4 Velocity Selector

Velocity Selector (Wien Filter)

Uses crossed electric and magnetic fields to select particles of specific velocity.

Condition for undeflected path:

Electric force = Magnetic force

\[qE = qvB\]

Selected velocity:

v = \frac{E}{B}

Independent of charge and mass!

📝 Solved Example 3

Question: A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C) enters a magnetic field of 0.1 T perpendicular to the field with velocity 10⁶ m/s. Find: (a) Radius of path (b) Time period (c) Frequency.

Solution:

Given: m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C, B = 0.1 T, v = 10⁶ m/s

(a) Radius:

r = \frac{mv}{qB} = \frac{1.67 \times 10^{-27} \times 10^6}{1.6 \times 10^{-19} \times 0.1}

r = \frac{1.67 \times 10^{-21}}{1.6 \times 10^{-20}} = 0.104 \text{ m} = 10.4 \text{ cm}

(b) Time period:

T = \frac{2\pi m}{qB} = \frac{2\pi \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 0.1}

T = 6.56 \times 10^{-7} \text{ s} = 0.656 \text{ μs}

(c) Frequency:

f = \frac{1}{T} = \frac{1}{6.56 \times 10^{-7}}

f = 1.52 \times 10^{6} \text{ Hz} = 1.52 \text{ MHz}

Force on Current Carrying Conductor

When a current-carrying conductor is placed in a magnetic field, it experiences a force. This principle is the basis of electric motors and many electromagnetic devices.

6.1 Force on Straight Conductor

Force Formula

\vec{F} = I(\vec{L} \times \vec{B})

In magnitude:

F = BIL\sin\theta

Where:

• B = Magnetic field (T)

• I = Current (A)

• L = Length of conductor in field (m)

• θ = Angle between L and B

Maximum force:

When θ = 90° (conductor perpendicular to field)

F_{max} = BIL

6.2 Force Between Two Parallel Conductors

Force Between Parallel Wires

Two parallel conductors carrying currents exert forces on each other.

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

Force per unit length between two infinite parallel wires

Same Direction Currents:

ATTRACT

Opposite Direction Currents:

REPEL

Definition of Ampere:

One ampere is that constant current which, when flowing through two parallel infinite conductors placed 1 meter apart in vacuum, produces a force of 2 × 10⁻⁷ N per meter length on each conductor.

💡 Memory Trick

"Like currents attract, unlike currents repel"

This is opposite to charges! (Like charges repel, unlike charges attract)

Torque on Current Loop & Moving Coil Galvanometer

A current-carrying loop in a magnetic field experiences torque, which is the working principle of electric motors and galvanometers. This is a high-weightage topic in JEE.

7.1 Magnetic Dipole Moment

Magnetic Dipole Moment (M)

\vec{M} = NI\vec{A}

Where:

• N = Number of turns

• I = Current

• A = Area of loop

SI Unit: A·m² or J/T

Direction: Perpendicular to plane of loop (by right-hand rule)

7.2 Torque on Current Loop

Torque Formula

\vec{\tau} = \vec{M} \times \vec{B} = NI\vec{A} \times \vec{B}

In magnitude:

\tau = NIAB\sin\theta

Where θ is angle between area vector (M) and B

Maximum torque:

When θ = 90° (plane of loop parallel to B)

\tau_{max} = NIAB

Zero torque:

When θ = 0° or 180° (plane perpendicular to B)

\tau = 0

7.3 Moving Coil Galvanometer

Working Principle

A moving coil galvanometer measures small currents using the torque on a current loop.

At Equilibrium:

Magnetic torque = Restoring torque of spring

NIAB = k\theta

Deflection:

\theta = \frac{NAB}{k}I = GI

where G = NAB/k is galvanometer constant

Current Sensitivity:

S_I = \frac{\theta}{I} = \frac{NAB}{k}

Voltage Sensitivity:

S_V = \frac{\theta}{V} = \frac{NAB}{kR}

7.4 Conversion of Galvanometer

To Ammeter

Connect low resistance (shunt S) in parallel

S = \frac{I_g G}{I - I_g} = \frac{G}{n-1}

where n = I/I_g (multiplication factor)

Range increases, resistance decreases

To Voltmeter

Connect high resistance R in series

R = \frac{V}{I_g} - G = G(n-1)

where n = V/(I_gG) (multiplication factor)

Range increases, resistance increases

⚠️ Important Points for JEE

Ideal Ammeter: Zero resistance (shunt S → 0)
Ideal Voltmeter: Infinite resistance (R → ∞)
Radial magnetic field: Used for uniform torque at all angles
Increasing sensitivity: Increase N, A, B or decrease k

Bar Magnets & Magnetic Dipoles

A bar magnet behaves like a magnetic dipole, similar to how a current loop creates a magnetic dipole moment. Understanding the analogy between electric and magnetic dipoles is important for JEE.

8.1 Magnetic Field Due to Bar Magnet

Magnetic Field at Various Points

On Axial Line:

B_{\text{axial}} = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3}

(for r >> l)

On Equatorial Line:

B_{\text{eq}} = \frac{\mu_0}{4\pi} \cdot \frac{M}{r^3}

(for r >> l)

Relationship:

B_{\text{axial}} = 2B_{\text{eq}}

Axial field is twice the equatorial field at same distance

At General Point (r, θ):

B = \frac{\mu_0 M}{4\pi r^3}\sqrt{1 + 3\cos^2\theta}

8.2 Bar Magnet in Uniform Magnetic Field

Torque and Energy

Torque:

\tau = MB\sin\theta

Potential Energy:

U = -MB\cos\theta = -\vec{M} \cdot \vec{B}

Work done in rotating:

W = MB(\cos\theta_1 - \cos\theta_2)

8.3 Analogy: Electric vs Magnetic Dipole

Property	Electric Dipole	Magnetic Dipole
Dipole Moment	p = qd	M = NIA or M = ml
Axial Field	E = 2kp/r³	B = (μ₀/4π)(2M/r³)
Equatorial Field	E = kp/r³	B = (μ₀/4π)(M/r³)
Torque	τ = pE sinθ	τ = MB sinθ
Potential Energy	U = -p·E	U = -M·B

Earth's Magnetism

Earth behaves like a giant bar magnet with its magnetic south pole near the geographic north pole. Understanding Earth's magnetism and related terms is important for JEE.

9.1 Elements of Earth's Magnetism

1. Magnetic Declination (θ)

Angle between geographic meridian and magnetic meridian at a place.

It varies from place to place

2. Angle of Dip (δ or I)

Angle between Earth's magnetic field and horizontal at a place.

At equator: δ = 0°
At poles: δ = 90°

3. Horizontal Component (B_H)

Component of Earth's field in horizontal direction.

B_H = B\cos\delta

Relations between Components

Horizontal component:

B_H = B\cos\delta

Vertical component:

B_V = B\sin\delta

Total field:

B = \sqrt{B_H^2 + B_V^2}

Angle of dip:

\tan\delta = \frac{B_V}{B_H}

💡 Key Facts about Earth's Magnetism

Geographic North: Near Earth's magnetic South pole
Earth's field: Approximately 25-65 μT (varies with location)
Magnetic equator: Line where dip = 0°
Isogonic lines: Lines of equal declination
Isoclinic lines: Lines of equal dip
Neutral points: Where B_H = 0 (field of magnet cancels B_H)

Magnetic Properties of Materials

Materials respond differently to external magnetic fields. Understanding diamagnetic, paramagnetic, and ferromagnetic materials is crucial for JEE, especially the concept of magnetic susceptibility and permeability.

10.1 Important Magnetic Terms

Definitions

Magnetization (I or M):

I = \frac{\text{Magnetic moment}}{\text{Volume}} = \frac{M}{V}

Unit: A/m

Magnetic Intensity (H):

H = \frac{B}{\mu_0} - I

Unit: A/m

Magnetic Susceptibility (χ):

\chi = \frac{I}{H}

Dimensionless

Relative Permeability (μ_r):

\mu_r = 1 + \chi

Dimensionless

Relation between B, H, and I:

B = \mu_0(H + I) = \mu_0 H(1 + \chi) = \mu_0 \mu_r H = \mu H

10.2 Classification of Magnetic Materials

Property	Diamagnetic	Paramagnetic	Ferromagnetic
Susceptibility (χ)	Negative, small (-10⁻⁵ to -10⁻⁹)	Positive, small (10⁻⁵ to 10⁻³)	Positive, large (10² to 10⁵)
Relative Permeability (μ_r)	Slightly < 1	Slightly > 1	>> 1
Behavior in field	Weakly repelled (moves to weaker field)	Weakly attracted (moves to stronger field)	Strongly attracted
Temperature dependence	Independent	χ ∝ 1/T (Curie's law)	Above Curie point → paramagnetic
Examples	Bi, Cu, Ag, Au, H₂O, N₂	Al, Na, Ca, O₂, Pt	Fe, Co, Ni, Gd
Atomic property	All electrons paired	Unpaired electrons present	Domains present

10.3 Hysteresis

Hysteresis Loop

When a ferromagnetic material is magnetized and demagnetized cyclically, the B-H curve forms a loop called hysteresis loop.

Retentivity (B_r):

Value of B when H = 0 (magnetic memory)

Coercivity (H_c):

Value of H needed to reduce B to zero

Hysteresis Loss:

Energy lost per cycle = Area of loop

Applications:

Permanent magnets: High retentivity, high coercivity (steel, alnico)
Electromagnet cores: Low retentivity, low coercivity (soft iron)
Transformer cores: Low hysteresis loss (soft iron)

⚠️ Curie's Law & Temperature

Paramagnetic materials:

\chi = \frac{C}{T}

C = Curie constant

Ferromagnetic materials:

\chi = \frac{C}{T - T_c}

T_c = Curie temperature (above which → paramagnetic)

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

✓ Biot-Savart Law Applications: 30%
✓ Force on Moving Charge/Conductor: 25%
✓ Cyclotron & Motion in B: 20%
✓ Torque on Current Loop: 15%
✓ Magnetism & Materials: 10%

JEE Advanced (Last 5 Years)

✓ Complex B-field Problems: 35%
✓ Motion of Charged Particles: 25%
✓ Ampere's Law Applications: 20%
✓ Force Between Conductors: 12%
✓ Earth's Magnetism: 8%

Weightage Analysis

JEE Main: 16-20 marks (4-5 questions)

JEE Advanced: 18-24 marks (5-6 questions)

Difficulty Level: Medium to Hard

Time Required: 6-8 hours practice

Year-wise Question Distribution

Year	JEE Main	JEE Advanced	Key Topics
2024	5 Questions (20 marks)	6 Questions (22 marks)	Cyclotron, Biot-Savart, Torque
2023	4 Questions (16 marks)	5 Questions (18 marks)	Helical motion, Solenoid, Galvanometer
2022	5 Questions (20 marks)	6 Questions (24 marks)	Force on conductor, Ampere's law
2021	4 Questions (16 marks)	5 Questions (20 marks)	Circular motion in B, Bar magnet

Download Complete PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

Find the magnetic field at the center of a circular loop of radius 5 cm carrying current 2 A.
A wire carrying 10 A current is placed perpendicular to a magnetic field of 0.5 T. Find force per unit length.
An electron moves with velocity 10⁶ m/s perpendicular to B = 0.1 T. Find the radius of circular path.
Calculate the magnetic field at a distance of 10 cm from an infinite straight wire carrying 5 A current.
A solenoid has 500 turns and length 25 cm. Find magnetic field inside when current is 4 A.
Two parallel wires 10 cm apart carry currents 5 A and 10 A in same direction. Find force per meter.
A rectangular coil of 100 turns, area 20 cm² carries 0.5 A in field 0.2 T. Find maximum torque.
Find the cyclotron frequency for a proton in magnetic field of 1 T.

Level 2: Intermediate (JEE Main/Advanced)

A charged particle enters a uniform magnetic field at 30° to the field direction. Describe its motion and find pitch if v = 10⁶ m/s, B = 0.5 T, q/m = 10⁸ C/kg.
Find the magnetic field at a point on the axis of a circular coil of radius R at distance x from center. Show that at x >> R, B ∝ 1/x³.
A galvanometer of resistance 50Ω shows full-scale deflection for 1 mA. Convert it to (a) ammeter of range 0-5 A (b) voltmeter of range 0-50 V.
Using Ampere's law, derive expression for magnetic field inside and outside a thick cylindrical wire of radius R carrying uniform current I.
A semicircular wire of radius 10 cm carries current 5 A. Find magnetic field at the center due to (a) semicircle (b) straight portion.
An electron enters a region with E = 10⁴ V/m and B = 10⁻² T (perpendicular). Find the velocity for which it passes undeflected.
A toroid has 1000 turns, mean radius 20 cm, and carries 2 A. Find B at a point (a) inside windings (b) in central cavity (c) outside.
The horizontal component of Earth's field is 30 μT. A bar magnet of moment 0.3 Am² is placed in magnetic meridian. Find null points.

Level 3: Advanced (JEE Advanced/Olympiad)

A charged particle of mass m and charge q starts from rest in combined electric (E) and magnetic (B) fields perpendicular to each other. Derive the equation of its trajectory.
Find the magnetic field at the center of a square loop of side 'a' carrying current I. Express in terms of μ₀, I, and a.
Two concentric circular loops of radii R and 2R lie in the same plane and carry currents I and 2I in opposite directions. Find net magnetic field at common center.
A wire is bent into a regular hexagon of side 'a'. Find magnetic field at center if it carries current I.
Derive the time period of oscillation of a bar magnet suspended in Earth's magnetic field. What happens if the magnet is heated above Curie temperature?
In a cyclotron, protons are accelerated using 50 kV potential difference and 1.5 T magnetic field. Find (a) frequency (b) radius after 50 revolutions (c) kinetic energy gained.
A charged particle moves in helical path in a non-uniform magnetic field (B increasing along axis). Explain the magnetic mirror effect.
A current loop of magnetic moment M is placed in a non-uniform magnetic field. Derive expression for force on the loop.

Numerical Answer Type (NAT) Practice

A proton (m = 1.67 × 10⁻²⁷ kg) moves in a circle of radius 20 cm in B = 0.5 T. Find its kinetic energy in MeV. [Answer: ___]
A wire carrying 8 A is bent into a circle of radius 4 cm. Find B at center in mT. [Answer: ___]
At what angle should a charge enter a magnetic field to have pitch equal to circumference of helical path? [Answer: ___ degrees]
A galvanometer (G = 100Ω, Ig = 2mA) is converted to ammeter of range 2 A. Find shunt resistance in Ω. [Answer: ___]
If χ = 0.0015 for a paramagnetic material, find μr. [Answer: ___]

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📋 Quick Formula Sheet - Magnetic Effects

Biot-Savart Law

dB = (μ₀/4π)(Idl×r̂)/r²

Straight wire: B = (μ₀I/4πd)(sinφ₁+sinφ₂)

Infinite wire: B = μ₀I/2πd

Circular loop center: B = μ₀I/2R

Arc at center: B = μ₀Iθ/4πR

Ampere's Law

∮B⃗·dl⃗ = μ₀I_enclosed

Solenoid: B = μ₀nI

Toroid: B = μ₀NI/2πr

Cylinder (r

Cylinder (r>R): B = μ₀I/2πr

Lorentz Force

F⃗ = q(E⃗ + v⃗×B⃗)

Magnetic: F = qvBsinθ

On conductor: F = BILsinθ

Parallel wires: F/L = μ₀I₁I₂/2πd

Motion in B Field

Radius: r = mv/qB

Time period: T = 2πm/qB

Frequency: f = qB/2πm

Pitch: p = 2πmvcosθ/qB

Velocity selector: v = E/B

Torque & Galvanometer

Magnetic moment: M = NIA

Torque: τ = MBsinθ = NIABsinθ

Energy: U = -M⃗·B⃗

Shunt (ammeter): S = IgG/(I-Ig)

Resistance (voltmeter): R = V/Ig - G

Magnetism

Axial: B = (μ₀/4π)(2M/r³)

Equatorial: B = (μ₀/4π)(M/r³)

BH = Bcosδ, BV = Bsinδ

tanδ = BV/BH

χ = I/H, μr = 1 + χ

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📑 Quick Navigation - Magnetic Effects of Current and Magnetism

Magnetic Effects of Current & Magnetism | JEE 2025-26

Magnetic Field - Fundamentals

1.1 Sources of Magnetic Field

Natural Sources

Artificial Sources

1.2 Magnetic Field (B)

Magnetic Field Definition

1.3 Properties of Magnetic Field Lines

💡 Key Properties

⚠️ Important Distinction

1.4 Magnetic Flux

Magnetic Flux (Φ)

Biot-Savart Law

2.1 Statement of Biot-Savart Law

Biot-Savart Law

💡 Direction of Magnetic Field

2.2 Magnetic Field Due to Straight Wire

Finite Straight Wire

2.3 Magnetic Field Due to Circular Loop

Circular Current Loop

2.4 Magnetic Field Due to Arc

Arc of Circle at Center

📝 Solved Example 1

Ampere's Circuital Law

3.1 Statement of Ampere's Law

Ampere's Circuital Law

💡 When to Use Ampere's Law

3.2 Applications of Ampere's Law

3.3 Solenoid

Magnetic Field in Solenoid

3.4 Toroid

Magnetic Field in Toroid

📝 Solved Example 2

Lorentz Force

4.1 Lorentz Force Equation

Complete Lorentz Force

4.2 Magnetic Force on Moving Charge

Magnetic Force

⚠️ Key Properties of Magnetic Force

4.3 Direction of Magnetic Force

💡 Right Hand Rules

Motion of Charged Particle in Magnetic Field

5.1 Circular Motion (v ⊥ B)

When Velocity is Perpendicular to Magnetic Field

5.2 Helical Motion (v at angle θ to B)

When Velocity is at Angle θ to B

5.3 Cyclotron

Cyclotron - Particle Accelerator

⚠️ Limitations of Cyclotron

5.4 Velocity Selector

Velocity Selector (Wien Filter)

📝 Solved Example 3

Force on Current Carrying Conductor

6.1 Force on Straight Conductor

Force Formula

6.2 Force Between Two Parallel Conductors

Force Between Parallel Wires

💡 Memory Trick

Torque on Current Loop & Moving Coil Galvanometer

7.1 Magnetic Dipole Moment

Magnetic Dipole Moment (M)

7.2 Torque on Current Loop

Torque Formula

7.3 Moving Coil Galvanometer

Working Principle

7.4 Conversion of Galvanometer

To Ammeter

To Voltmeter

⚠️ Important Points for JEE

Bar Magnets & Magnetic Dipoles

8.1 Magnetic Field Due to Bar Magnet

Magnetic Field at Various Points

8.2 Bar Magnet in Uniform Magnetic Field

Torque and Energy

8.3 Analogy: Electric vs Magnetic Dipole

Earth's Magnetism

9.1 Elements of Earth's Magnetism

1. Magnetic Declination (θ)

2. Angle of Dip (δ or I)

3. Horizontal Component (B_H)