NEET UG Comprehensive Theory
Magnetic Effects of Current & Magnetism
Biot-Savart • Ampere's Law • Lorentz Force • Galvanometer • Earth's Magnetism • Matter
1. Oersted's Discovery & Biot-Savart Law
Hans Christian Oersted (1820) established that a steady electric current produces a magnetic field in the surrounding space. Magnetic field lines around a long straight current form concentric circles whose direction is governed by the Right-Hand Thumb Rule.
Biot-Savart Law (Vector Formulation):
The magnetic field dB produced at a position vector r by an infinitesimally small current element I dl is:
• μ04π = 10-7 T·m/A.
• dB is perpendicular to both current element dl and radius vector r.
The magnetic field dB produced at a position vector r by an infinitesimally small current element I dl is:
dB = μ04π · I (dl × r̂)r2 = μ04π · I (dl × r)r3
Scalar Magnitude:
dB = μ04π · I dl sinθr2
• μ0 = 4π × 10-7 T·m/A (or N/A2, or H/m).• μ04π = 10-7 T·m/A.
• dB is perpendicular to both current element dl and radius vector r.
Figure 1: Biot-Savart Current Element, Straight Conductor Field, and Circular Loop Axis Field
High-Yield Applications for NEET
A. Finite Straight Wire Subtending Angles θ1, θ2:
B = μ0 I4π r (sinθ1 + sinθ2)
• Infinitely Long Wire: θ1 = θ2 = 90° &implies; B = μ0 I2π r.• Semi-Infinite Wire (At one end): θ1 = 90°, θ2 = 0° &implies; B = μ0 I4π r.
B. Circular Coil of N Turns at Center:
Bcenter = μ0 N I2 R
• Semicircular Arc: B = μ0 I4 R.• Circular Arc Subtending Angle φ (in radians): B = μ0 I4π R · φ.
C. On the Axis of a Circular Coil at Distance x:
Baxis = μ0 N I R22 (R2 + x2)3/2
• Ratio of axial to center field: BaxisBcenter = R3(R2 + x2)3/2.• When x = √3 R &implies; Baxis = 18 Bcenter (Classic NEET PYQ).
• Far away along axis (x ≫ R): B ≈ μ0 M2π x3 where M = N I A.
2. Ampere's Circuital Law, Solenoid & Toroid
Ampere's Circuital Law plays the same role in magnetostatics as Gauss's law does in electrostatics: it enables swift calculation of magnetic fields when high geometrical symmetry exists.
Statement of Ampere's Law:
The line integral of magnetic field B around any closed closed Amperian loop equals μ0 times the net algebraic current enclosed by that loop:
The line integral of magnetic field B around any closed closed Amperian loop equals μ0 times the net algebraic current enclosed by that loop:
∮ B · dl = μ0 Ienclosed
• Currents piercing the loop surface in the direction given by right-hand fingers curled along loop orientation are taken as positive; opposite currents are negative.
Figure 2: Amperian Loop, Field Inside Long Solenoid, and Toroidal Core Confinement
Solenoid & Toroid Summary
• Long Ideal Solenoid (Turns per unit length n = N / L):
- Inside Central Region: B = μ0 n I (Uniform & parallel to axis)
- Near the Ends of Solenoid: Bend = 12 μ0 n I
- Outside Solenoid: Boutside ≈ 0
• Toroid (Endless Solenoid bent into a circle of mean radius r):
- Inside Core (Mean circumference 2πr): B = μ0 N I2π r = μ0 n I
- In the open space interior (hollow region) and exterior: B = 0 because Ienc = 0.
- Inside Central Region: B = μ0 n I (Uniform & parallel to axis)
- Near the Ends of Solenoid: Bend = 12 μ0 n I
- Outside Solenoid: Boutside ≈ 0
• Toroid (Endless Solenoid bent into a circle of mean radius r):
- Inside Core (Mean circumference 2πr): B = μ0 N I2π r = μ0 n I
- In the open space interior (hollow region) and exterior: B = 0 because Ienc = 0.
3. Lorentz Force & Charged Particle Motion
When a charge q moves with velocity v in combined electric E and magnetic B fields, the total Lorentz force is: F = q (E + v × B).
Pure Magnetic Force: Fm = q (v × B)
• Magnitude: F = q v B sinθ (where θ is angle between v and B).
• Zero Work Done: Since Fm is strictly perpendicular to v at every instant:
• Magnitude: F = q v B sinθ (where θ is angle between v and B).
• Zero Work Done: Since Fm is strictly perpendicular to v at every instant:
Power P = F · v = 0 &implies; Work W = 0 &implies; ΔK = 0
A magnetic field changes ONLY the direction of motion of a charged particle, NEVER its speed or kinetic energy!Trajectory Regimes Depending on Angle θ
1. θ = 0° or 180° (Parallel/Antiparallel): F = 0 &implies; Particle travels in an undeviated straight line with constant velocity.
2. θ = 90° (Perpendicular): F = qvB provides required centripetal force m v2r &implies; Circular Orbit.
• Orbital Radius:
3. 0° < θ < 90° (Oblique Angle): Velocity splits into:
• v⊥ = v sinθ produces circular rotation with radius r = m v sinθq B.
• v∥ = v cosθ pulls particle along field &implies; Helical Trajectory.
• Pitch of Helix: Linear distance traveled during one time period:
2. θ = 90° (Perpendicular): F = qvB provides required centripetal force m v2r &implies; Circular Orbit.
• Orbital Radius:
r = m vq B = pq B = √2 m Kq B = √2 m q Vq B
• Time Period & Cyclotron Frequency:
T = 2π mq B, f = q B2π m, ω = q Bm
Golden Rule: T and f are completely independent of speed v and orbit radius r!3. 0° < θ < 90° (Oblique Angle): Velocity splits into:
• v⊥ = v sinθ produces circular rotation with radius r = m v sinθq B.
• v∥ = v cosθ pulls particle along field &implies; Helical Trajectory.
• Pitch of Helix: Linear distance traveled during one time period:
p = v∥ · T = 2π m v cosθq B
Figure 3: Circular Orbit (v ⊥ B), Helical Trajectory (Pitch & Radius), and Cyclotron Dees Resonance
Cyclotron Core NEET Notes:
• Principle: Electric field between Dees accelerates ions; magnetic field bends them in semicircular arcs of increasing radius.
• Resonance Condition: Frequency of RF oscillator must match cyclotron frequency: fa = q B2π m.
• Maximum Kinetic Energy (At Dee exit radius R):
• Principle: Electric field between Dees accelerates ions; magnetic field bends them in semicircular arcs of increasing radius.
• Resonance Condition: Frequency of RF oscillator must match cyclotron frequency: fa = q B2π m.
• Maximum Kinetic Energy (At Dee exit radius R):
Kmax = q2 B2 R22 m
• Crucial Limitations: Cannot accelerate neutral particles (neutrons, q=0) or electrons (due to small mass, velocity quickly approaches speed of light c, increasing relativistic mass m and throwing them out of resonance).4. Force on Conductors & Moving Coil Galvanometer
Force on a Current-Carrying Conductor:
Force Between Two Parallel Current-Carrying Conductors:
• Current in OPPOSITE direction → REPEL each other.
• Definition of 1 Ampere: Current which, flowing in two infinitely long parallel wires separated by 1 meter in vacuum, produces an attractive/repulsive force of 2 × 10-7 N/m.
F = I (L × B) &implies; F = I L B sinθ
Direction given by Fleming's Left-Hand Rule (Forefinger = B, Middle finger = I, Thumb = Force F).Force Between Two Parallel Current-Carrying Conductors:
FL = μ0 I1 I22π d
• Current in SAME direction → ATTRACT each other.• Current in OPPOSITE direction → REPEL each other.
• Definition of 1 Ampere: Current which, flowing in two infinitely long parallel wires separated by 1 meter in vacuum, produces an attractive/repulsive force of 2 × 10-7 N/m.
Torque on Coil & Moving Coil Galvanometer
• Torque on Current Loop: τ = N I A B sinθ = M × B, where magnetic dipole moment M = N I A.
• Radial Field Construction: Concave magnetic poles and a soft iron core make field lines radial, ensuring plane of the coil is always parallel to B (θ = 90° &implies; sinθ = 1).
• Equilibrium Condition: Deflecting torque = Restoring torque &implies; N I A B = k θ &implies;
• Voltage Sensitivity: Vs = θV = θI R = N A Bk R.
NEET Trap: Increasing N doubles Is, but wire length doubles so resistance R also doubles, leaving Vs unchanged!
• Radial Field Construction: Concave magnetic poles and a soft iron core make field lines radial, ensuring plane of the coil is always parallel to B (θ = 90° &implies; sinθ = 1).
• Equilibrium Condition: Deflecting torque = Restoring torque &implies; N I A B = k θ &implies;
θ = N A Bk · I
• Current Sensitivity: Is = θI = N A Bk.• Voltage Sensitivity: Vs = θV = θI R = N A Bk R.
NEET Trap: Increasing N doubles Is, but wire length doubles so resistance R also doubles, leaving Vs unchanged!
Figure 4: Parallel Currents Force, Galvanometer Radial Field, Earth's Magnetic Elements, and B-H Hysteresis Loop
Conversion to Ammeter (Measures Current in Series):
Connect very small resistance (Shunt S) in PARALLEL with Galvanometer of resistance G:
S = Ig GI - Ig
Equivalent Ammeter resistance: RA = G SG + S ≈ S (Extremely low, ideal ammeter = 0).
Conversion to Voltmeter (Measures Potential Difference in Parallel):
Connect very high resistance R in SERIES with Galvanometer:
R = VIg - G
Equivalent Voltmeter resistance: RV = G + R (Extremely high, ideal voltmeter = ∞).5. Magnetism & Matter
Earth's Magnetic Elements:
Earth acts as a giant magnetic dipole tilted at ≈ 11.3° to its geographic rotation axis.
1. Magnetic Declination (α): Angle between Geographic Meridian and Magnetic Meridian.
2. Angle of Dip or Inclination (δ): Angle made by total magnetic field B with the horizontal.
• Horizontal Component: BH = B cosδ
• Vertical Component: BV = B sinδ
• Relation: BVBH = tanδ, B = √BH2 + BV2.
• At Magnetic Poles: δ = 90° &implies; BH = 0, B = BV.
• At Magnetic Equator: δ = 0° &implies; BV = 0, B = BH.
• Apparent Dip in two mutually perpendicular planes:
Earth acts as a giant magnetic dipole tilted at ≈ 11.3° to its geographic rotation axis.
1. Magnetic Declination (α): Angle between Geographic Meridian and Magnetic Meridian.
2. Angle of Dip or Inclination (δ): Angle made by total magnetic field B with the horizontal.
• Horizontal Component: BH = B cosδ
• Vertical Component: BV = B sinδ
• Relation: BVBH = tanδ, B = √BH2 + BV2.
• At Magnetic Poles: δ = 90° &implies; BH = 0, B = BV.
• At Magnetic Equator: δ = 0° &implies; BV = 0, B = BH.
• Apparent Dip in two mutually perpendicular planes:
cot2δ = cot2δ1 + cot2δ2
Comparison of Magnetic Materials
• Diamagnetic: Repelled feebly by magnets (moves stronger to weaker field). No unpaired electrons. Susceptibility χ is small and negative (-1 ≤ χ < 0), independent of temperature. μr < 1. Superconductors are perfect diamagnets (χ = -1, Meissner effect). Examples: Bi, Cu, H2O, Pb, N2.
• Paramagnetic: Attracted feebly (moves weaker to stronger field). Atoms possess permanent dipole moments. Susceptibility χ is small and positive, inversely proportional to absolute temperature (Curie's Law: χ = C / T). μr > 1. Examples: Al, Na, O2, Pt, CuCl2.
• Ferromagnetic: Strongly attracted by magnets. Has domain structure. Susceptibility χ is very large and positive (χ ∼ 103 - 105), μr ≫ 1. Above Curie Temperature TC, domains break down and it transforms into a paramagnetic material (Curie-Weiss Law: χ = CT - TC). Examples: Fe, Co, Ni, Alnico, Gadolinium.
• Paramagnetic: Attracted feebly (moves weaker to stronger field). Atoms possess permanent dipole moments. Susceptibility χ is small and positive, inversely proportional to absolute temperature (Curie's Law: χ = C / T). μr > 1. Examples: Al, Na, O2, Pt, CuCl2.
• Ferromagnetic: Strongly attracted by magnets. Has domain structure. Susceptibility χ is very large and positive (χ ∼ 103 - 105), μr ≫ 1. Above Curie Temperature TC, domains break down and it transforms into a paramagnetic material (Curie-Weiss Law: χ = CT - TC). Examples: Fe, Co, Ni, Alnico, Gadolinium.
Hysteresis & Material Selection:
• Retentivity: Value of magnetic field B remaining when magnetizing field H is reduced to zero.
• Coercivity: Reverse magnetizing field -H required to wipe out residual magnetism.
• Area of Hysteresis Loop: Represents energy lost as heat per unit volume per magnetization cycle.
• Electromagnets & Transformer Cores: Require Soft Iron — High permeability μr, low retentivity, low coercivity, and very narrow loop (minimum heat loss).
• Permanent Magnets: Require Steel / Alnico — High retentivity, high coercivity, and broad hysteresis loop (resists demagnetization from stray fields).
• Retentivity: Value of magnetic field B remaining when magnetizing field H is reduced to zero.
• Coercivity: Reverse magnetizing field -H required to wipe out residual magnetism.
• Area of Hysteresis Loop: Represents energy lost as heat per unit volume per magnetization cycle.
• Electromagnets & Transformer Cores: Require Soft Iron — High permeability μr, low retentivity, low coercivity, and very narrow loop (minimum heat loss).
• Permanent Magnets: Require Steel / Alnico — High retentivity, high coercivity, and broad hysteresis loop (resists demagnetization from stray fields).
NEET UG High-Speed Revision
Master Formula Matrix & 20 Golden Facts
Magnetic Effects of Current • Moving Charges • Magnetism and Matter
âš¡ Master Formula Matrix
1. Biot-Savart Law (Current Element)
dB = μ04π · I (dl × r)r3,
dB = μ04π · I dl sinθr2
2. Straight Wire of Infinite Length
B = μ0 I2π r (Semi-infinite at end: B = μ0 I4π r)
3. Circular Loop of Radius R (Center & Axis)
Bcenter = μ0 N I2 R,
Baxis = μ0 N I R22 (R2 + x2)3/2
4. Ampere's Law, Solenoid & Toroid
∮ B · dl = μ0 Ienc,
Bsolenoid = μ0 n I,
Bend = 12 μ0 n I,
Btoroid = μ0 N I2π r
5. Magnetic Lorentz Force (Moving Charge)
F = q (v × B) &implies; F = q v B sinθ, Work W = 0, ΔK = 0
6. Circular Orbit & Cyclotron Motion (θ = 90°)
r = m vq B = √2 m Kq B = √2 m q Vq B,
T = 2π mq B,
f = q B2π m
7. Helical Motion & Pitch (0 < θ < 90°)
r = m v sinθq B,
Pitch p = v∥ · T = 2π m v cosθq B
8. Force on Wire & Force Between Parallel Wires
F = I L B sinθ,
FL = μ0 I1 I22π d (Same → Attract, Opposite → Repel)
9. Magnetic Dipole Moment & Torque on Coil
M = N I A,
τ = M × B = M B sinθ,
U = -M · B = -M B cosθ,
W = M B (cosθ1 - cosθ2)
10. Moving Coil Galvanometer & Sensitivities
θ = N A Bk · I,
Is = θI = N A Bk,
Vs = θV = N A Bk R
11. Ammeter Shunt & Voltmeter Multiplier
Shunt S = Ig GI - Ig (in parallel),
Series R = VIg - G (in series)
12. Earth's Magnetic Field Elements
BH = B cosδ, BV = B sinδ,
tanδ = BVBH,
cot2δ = cot2δ1 + cot2δ2
13. Bohr Magneton & Gyromagnetic Ratio
μB = e h4π me = 9.27 × 10-24 A·m2,
ML = e2 me = 8.8 × 1010 C/kg
14. Magnetic Susceptibility & Curie's Law
B = μ0 (H + I),
χ = IH,
μr = 1 + χ,
χ = CT (Paramagnetic),
χ = CT - TC (Ferromagnetic)
🎯 20 Golden NEET Must-Know Facts & Traps
1. Zero Work by Magnetic Force: Fm is always perpendicular to instantaneous velocity v. Hence, W = ∫ F · ds = 0. Speed and Kinetic Energy NEVER change in a pure magnetic field; only the trajectory curves.
2. Velocity Selector Condition: When electric and magnetic fields are mutually perpendicular (crossed fields, E ⊥ B), particles with speed v = EB pass straight through completely undeflected regardless of their mass or charge!
3. Radius Scaling Shortcut: When charged particles are accelerated through the same potential difference V: r ∝ √mq. Thus for Proton (1, 1), Deuteron (2, 1), and α-particle (4, 2): rp : rd : rα = 1 : √2 : 1!
4. Cyclotron Isochronism: Time period T = 2π mq B and frequency f do NOT depend on speed v or orbit radius r. As the particle speeds up, it describes a proportionally larger circle in the exact same time!
5. Why Cyclotron Fails for Electrons & Neutrons: Neutrons have zero charge (q=0), so they experience no electric acceleration or magnetic bending. Electrons have tiny mass, so high speed makes relativistic mass increase (m = m0√1 - v2/c2), ruining resonance!
6. Parallel Currents Opposite to Electrostatics: In electrostatics, like charges repel and unlike attract. But in magnetostatics, parallel currents in the same direction ATTRACT and opposite currents REPEL!
7. Solenoid Ends vs Center: The magnetic field at the extreme ends of an open semi-infinite solenoid is exactly half that of the interior center: Bend = 12 μ0 n I.
8. Toroid Zero-Field Regions: Magnetic field is strictly confined within the core of a toroid. In the inner hollow region (center hole) and anywhere outside the outer casing, B = 0 because the enclosed current is zero!
9. Radial Field Magic: In a Moving Coil Galvanometer, concave pole pieces + soft iron cylinder ensure the magnetic field is always radial. Therefore, the angle between the coil's normal and B is always 90°, producing maximum linear torque (τ = NIAB) and a strictly uniform linear scale!
10. Sensitivity Fallacy Trap: Increasing coil turns N increases Current Sensitivity Is = N A Bk, but it does NOT automatically increase Voltage Sensitivity Vs = N A Bk R because adding turns increases coil wire length and resistance R in the exact same proportion!
11. Ideal Meters: An Ideal Ammeter has zero resistance (connected in series so it causes zero voltage drop). An Ideal Voltmeter has infinite resistance (connected in parallel so it draws zero current).
12. Dip Angle at Poles and Equator: At Earth's Magnetic Poles, the dip angle δ = 90° &implies; BH = 0 (horizontal compass needle spins uselessly). At Earth's Magnetic Equator, δ = 0° &implies; BV = 0 and B = BH.
13. Apparent Dip Formula: If dip angles measured in two mutually perpendicular vertical planes are δ1 and δ2, the true dip δ is given by: cot2δ = cot2δ1 + cot2δ2.
14. Gauss's Law for Magnetism: ∮ B · dS = 0. This proves that isolated magnetic monopoles do NOT exist! Magnetic field lines are always continuous closed loops without start or end.
15. Superconductors & Meissner Effect: Superconductors exhibit perfect diamagnetism with magnetic susceptibility χ = -1 and relative permeability μr = 0. They expel all magnetic field lines from their interior.
16. Temperature Dependence: Diamagnetic susceptibility is independent of temperature. Paramagnetic susceptibility obeys Curie's law (χ ∝ 1/T). Ferromagnetic materials transform into paramagnetics above the Curie temperature TC.
17. Soft Iron vs Steel for Electromagnets: Soft iron is selected for electromagnets and transformer cores because it has high permeability, low retentivity, low coercivity, and a narrow hysteresis loop (minimal energy loss per cycle).
18. Steel / Alnico for Permanent Magnets: Permanent magnets require materials that do not easily demagnetize from stray fields. Hence they must have high retentivity and high coercivity (large area under loop).
19. Neutral Points in Earth's Field: When a bar magnet has its North pole pointing North, neutral points (where Bmagnet = BH) lie on the broadside-on (equatorial) line. When its North pole points South, neutral points lie on the end-on (axial) line!
20. Torque & Potential Energy of Dipole: Stable equilibrium occurs at θ = 0° (τ = 0, U = -MB is minimum). Unstable equilibrium occurs at θ = 180° (τ = 0, U = +MB is maximum). Work required to rotate from stable to unstable is ΔU = 2MB!
Step-by-Step Pedagogical Solutions
Solved Numericals: Magnetic Effects
8 High-Yield NEET Problems Solved with Strict 3-Box Pedagogy
Problem 1 • Circular Loop Field
NEET 2022 PYQ
A circular coil of radius R carries a steady current I. At what distance x along the axis from the center of the coil does the magnetic field become 18 of its value at the center?
Given & Core Concept:
• Bcenter = μ0 I2 R
• Baxis = μ0 I R22 (R2 + x2)3/2
• Condition given: Baxis = 18 Bcenter
• Bcenter = μ0 I2 R
• Baxis = μ0 I R22 (R2 + x2)3/2
• Condition given: Baxis = 18 Bcenter
Step-by-Step Mathematical Calculation:
Equate the two formulas:
R2 + x2 = 4 R2 &implies; x2 = 3 R2 &implies; x = √3 R.
Equate the two formulas:
μ0 I R22 (R2 + x2)3/2 = 18 · μ0 I2 R
Cancel common factor μ0 I2 on both sides:
R2(R2 + x2)3/2 = 18 R &implies; R3(R2 + x2)3/2 = 18
Take the reciprocal and take the cube root on both sides:
(R2 + x2)1/2R = (8)1/3 = 2
Square both sides:R2 + x2 = 4 R2 &implies; x2 = 3 R2 &implies; x = √3 R.
NEET Speed Formula:
Whenever Baxis = 1n Bcenter, the distance is:
Whenever Baxis = 1n Bcenter, the distance is:
x = R · √n2/3 - 1
Here n = 8 &implies; n2/3 = (81/3)2 = 4 &implies; x = R √4 - 1 = √3 R in 5 seconds flat!
Problem 2 • Composite Current Wire
NEET 2021 PYQ
A long straight wire carrying a current of I = 10 A is bent into a hairpin shape: a semi-infinite straight section connected to a semicircular arc of radius R = 5 cm, followed by another semi-infinite straight section. Find the magnitude of magnetic field at the center of curvature O.
Given & Core Concept:
• Current I = 10 A, Radius R = 5 cm = 0.05 m.
• Straight segment 1: Line of current passes through O &implies; B1 = 0.
• Semicircular arc: Barc = μ0 I4 R (Directed into the page ⊗).
• Straight segment 2: Semi-infinite wire at perpendicular distance R &implies; B2 = μ0 I4π R (Directed into the page ⊗).
• Current I = 10 A, Radius R = 5 cm = 0.05 m.
• Straight segment 1: Line of current passes through O &implies; B1 = 0.
• Semicircular arc: Barc = μ0 I4 R (Directed into the page ⊗).
• Straight segment 2: Semi-infinite wire at perpendicular distance R &implies; B2 = μ0 I4π R (Directed into the page ⊗).
Step-by-Step Mathematical Calculation:
Both non-zero fields point into the page (⊗), so they add directly:
Both non-zero fields point into the page (⊗), so they add directly:
Bnet = Barc + B2 = μ0 I4 R + μ0 I4π R = μ0 I4π R (π + 1)
Substitute μ0 / 4π = 10-7 T·m/A, I = 10 A, R = 0.05 m:
Bnet = 10-7 × 100.05 × (3.14 + 1) = 10-60.05 × 4.14 = 20 × 10-6 × 4.14
Bnet = 8.28 × 10-5 T = 82.8 μT (into the page).
NEET High-Speed Rule:
Whenever any straight section of wire has its line of action passing through the point of observation, its magnetic field contribution is strictly ZERO because the angle between dl and r is 0° or 180° (sinθ = 0)! Ignore it instantly.
Whenever any straight section of wire has its line of action passing through the point of observation, its magnetic field contribution is strictly ZERO because the angle between dl and r is 0° or 180° (sinθ = 0)! Ignore it instantly.
Problem 3 • Accelerated Particle Trajectory
NEET 2020 PYQ
A proton, a deuteron, and an alpha particle are all accelerated from rest through the same potential difference V. They subsequently enter a uniform magnetic field B perpendicular to their velocities. Find the ratio of their orbital radii rp : rd : rα.
Given & Core Concept:
• Proton: mass m, charge e.
• Deuteron: mass 2m, charge e.
• Alpha particle (α): mass 4m, charge 2e.
• Kinetic energy gained: K = q V.
• Circular radius: r = √2 m Kq B = 1B √2 V mq &implies; r ∝ √mq
• Proton: mass m, charge e.
• Deuteron: mass 2m, charge e.
• Alpha particle (α): mass 4m, charge 2e.
• Kinetic energy gained: K = q V.
• Circular radius: r = √2 m Kq B = 1B √2 V mq &implies; r ∝ √mq
Step-by-Step Mathematical Calculation:
Apply the proportionality r ∝ √m√q:
1. For Proton:
Ratio = rp : rd : rα = 1 : √2 : √2.
Apply the proportionality r ∝ √m√q:
1. For Proton:
rp ∝ √m√e ∝ 1
2. For Deuteron:
rd ∝ √2m√e ∝ √2
3. For Alpha Particle:
rα ∝ √4m√2e = 2√2 = √2
Ratio = rp : rd : rα = 1 : √2 : √2.
NEET Pro Trap Alert:
If particles have the SAME MOMENTUM: r ∝ 1q &implies; rp : rd : rα = 2 : 2 : 1.
If particles have the SAME KINETIC ENERGY: r ∝ √mq &implies; 1 : √2 : 1.
If accelerated through the SAME POTENTIAL: r ∝ √mq &implies; 1 : √2 : √2. Read question carefully!
If particles have the SAME MOMENTUM: r ∝ 1q &implies; rp : rd : rα = 2 : 2 : 1.
If particles have the SAME KINETIC ENERGY: r ∝ √mq &implies; 1 : √2 : 1.
If accelerated through the SAME POTENTIAL: r ∝ √mq &implies; 1 : √2 : √2. Read question carefully!
Problem 4 • Helical Path of Electron
NEET 2019 PYQ
An electron with mass m = 9.1 × 10-31 kg and charge e = 1.6 × 10-19 C enters a uniform magnetic field B = 0.2 T with speed v = 4 × 106 m/s at an angle θ = 30° with the field lines. Calculate (a) the radius of the helix and (b) the pitch of the helical path.
Given & Core Concept:
• v⊥ = v sin 30° = 4 × 106 × 0.5 = 2 × 106 m/s.
• v∥ = v cos 30° = 4 × 106 × √32 = 2√3 × 106 ≈ 3.46 × 106 m/s.
• Radius r = m v⊥q B, Pitch p = v∥ · T = 2π m v∥q B.
• v⊥ = v sin 30° = 4 × 106 × 0.5 = 2 × 106 m/s.
• v∥ = v cos 30° = 4 × 106 × √32 = 2√3 × 106 ≈ 3.46 × 106 m/s.
• Radius r = m v⊥q B, Pitch p = v∥ · T = 2π m v∥q B.
Step-by-Step Mathematical Calculation:
(a) Radius of the Helix:
(b) Pitch of the Helix:
First, time period:
(a) Radius of the Helix:
r = (9.1 × 10-31) × (2 × 106)(1.6 × 10-19) × 0.2 = 18.2 × 10-250.32 × 10-19 = 5.69 × 10-5 m = 56.9 μm.
(b) Pitch of the Helix:
First, time period:
T = 2 × 3.1416 × 9.1 × 10-311.6 × 10-19 × 0.2 = 57.18 × 10-310.32 × 10-19 ≈ 1.787 × 10-10 s.
Then, pitch:
p = v∥ · T = (3.464 × 106) × (1.787 × 10-10) = 6.19 × 10-4 m = 0.62 mm.
NEET Quick Check:
Notice that PitchRadius = 2π v∥v⊥ = 2π cotθ! For θ = 30°, pr = 2π √3 ≈ 10.88. Checking: 619 μm / 56.9 μm = 10.88. Instant verification!
Notice that PitchRadius = 2π v∥v⊥ = 2π cotθ! For θ = 30°, pr = 2π √3 ≈ 10.88. Checking: 619 μm / 56.9 μm = 10.88. Instant verification!
Problem 5 • Three Parallel Conductors
NEET 2018 PYQ
Three long parallel straight wires A, B, and C carry currents of 10 A, 20 A, and 30 A respectively in the same upward direction. Wire B is equidistant from A and C, with separation d = 10 cm. Determine the net magnetic force per unit length on wire B.
Given & Core Concept:
• IA = 10 A, IB = 20 A, IC = 30 A.
• Distance between A and B = d = 0.1 m; between B and C = d = 0.1 m.
• Since all currents flow in the same direction, both A and C exert attractive forces on B in opposite directions.
• IA = 10 A, IB = 20 A, IC = 30 A.
• Distance between A and B = d = 0.1 m; between B and C = d = 0.1 m.
• Since all currents flow in the same direction, both A and C exert attractive forces on B in opposite directions.
Step-by-Step Mathematical Calculation:
1. Force on wire B due to wire A (towards left, towards A):
2. Force on wire B due to wire C (towards right, towards C):
3. Net force per unit length on wire B:
1. Force on wire B due to wire A (towards left, towards A):
FBAL = μ0 IA IB2π d = (2 × 10-7) × 10 × 200.1 = 4 × 10-4 N/m (Left).
2. Force on wire B due to wire C (towards right, towards C):
FBCL = μ0 IC IB2π d = (2 × 10-7) × 30 × 200.1 = 12 × 10-4 N/m (Right).
3. Net force per unit length on wire B:
FnetL = 12 × 10-4 - 4 × 10-4 = 8 × 10-4 N/m (Towards Wire C / Right).
NEET Pro 1-Line Shortcut:
Since distances are equal, directly factor out common terms:
FnetL = μ0 IB2π d (IC - IA) = 2 × 10-7 × 200.1 × (30 - 10) = 4 × 10-5 × 20 = 8 × 10-4 N/m!
Since distances are equal, directly factor out common terms:
FnetL = μ0 IB2π d (IC - IA) = 2 × 10-7 × 200.1 × (30 - 10) = 4 × 10-5 × 20 = 8 × 10-4 N/m!
Problem 6 • Ammeter & Voltmeter Design
NEET 2022 PYQ
A galvanometer has an internal resistance G = 25 Ω and gives a full-scale deflection for a current of Ig = 4 mA. Calculate: (a) the shunt resistance required to convert it into an ammeter of range 0 to 5 A, and (b) the series resistance required to convert it into a voltmeter of range 0 to 10 V.
Given & Core Concept:
• Galvanometer resistance G = 25 Ω.
• Full-scale deflection current Ig = 4 mA = 0.004 A.
• Desired ammeter range I = 5 A; desired voltmeter range V = 10 V.
• Ammeter: Shunt in parallel, S = Ig GI - Ig.
• Voltmeter: Multiplier in series, R = VIg - G.
• Galvanometer resistance G = 25 Ω.
• Full-scale deflection current Ig = 4 mA = 0.004 A.
• Desired ammeter range I = 5 A; desired voltmeter range V = 10 V.
• Ammeter: Shunt in parallel, S = Ig GI - Ig.
• Voltmeter: Multiplier in series, R = VIg - G.
Step-by-Step Mathematical Calculation:
(a) Shunt for Ammeter (0 - 5 A):
(b) Series Resistance for Voltmeter (0 - 10 V):
(a) Shunt for Ammeter (0 - 5 A):
S = 0.004 × 255 - 0.004 = 0.14.996 ≈ 0.0200 Ω = 0.02 Ω in parallel.
(b) Series Resistance for Voltmeter (0 - 10 V):
R = 100.004 - 25 = 100004 - 25 = 2500 - 25 = 2475 Ω in series.
NEET Range Multiplier Shortcut:
For Ammeter: n = IIg = 50.004 = 1250 &implies; S = Gn - 1 ≈ 251250 = 0.02 Ω.
For Voltmeter: n = VIg G = 100.1 = 100 &implies; R = (n - 1) G = 99 × 25 = 2475 Ω.
For Ammeter: n = IIg = 50.004 = 1250 &implies; S = Gn - 1 ≈ 251250 = 0.02 Ω.
For Voltmeter: n = VIg G = 100.1 = 100 &implies; R = (n - 1) G = 99 × 25 = 2475 Ω.
Problem 7 • Torque & Dipole Work
NEET 2023 PYQ
A circular coil of 100 turns and radius 10 cm carries a current of 0.5 A. It is suspended in a uniform magnetic field B = 0.4 T. Find: (a) the magnetic dipole moment of the coil, (b) the torque when the plane of the coil makes an angle of 30° with the field, and (c) the work required to rotate the coil from stable equilibrium to unstable equilibrium.
Given & Core Concept:
• N = 100, I = 0.5 A, R = 0.1 m &implies; Area A = π R2 = π (0.1)2 = 0.01π m2.
• B = 0.4 T.
• Magnetic moment: M = N I A.
• Angle of dipole normal with field: θ = 90° - 30° = 60°.
• Work done: W = M B (cos 0° - cos 180°) = 2 M B.
• N = 100, I = 0.5 A, R = 0.1 m &implies; Area A = π R2 = π (0.1)2 = 0.01π m2.
• B = 0.4 T.
• Magnetic moment: M = N I A.
• Angle of dipole normal with field: θ = 90° - 30° = 60°.
• Work done: W = M B (cos 0° - cos 180°) = 2 M B.
Step-by-Step Mathematical Calculation:
(a) Magnetic Moment:
(b) Torque on Coil:
(c) Work Done (Stable to Unstable Equilibrium):
(a) Magnetic Moment:
M = 100 × 0.5 × (0.01 × 3.1416) = 50 × 0.031416 = 1.57 A·m2 (or J/T).
(b) Torque on Coil:
τ = M B sin 60° = 1.57 × 0.4 × √32 = 0.628 × 0.866 = 0.544 N·m.
(c) Work Done (Stable to Unstable Equilibrium):
W = M B [cos 0° - cos 180°] = M B [1 - (-1)] = 2 M B
W = 2 × 1.57 × 0.4 = 1.256 J.
W = 2 × 1.57 × 0.4 = 1.256 J.
NEET Angle Trap Alert:
Watch the phrasing! "Plane of coil makes 30° with B" means the dipole vector (normal to plane) makes θ = 90° - 30° = 60° with B! If you plug in 30° directly, you will pick the distractor option and lose 5 marks!
Watch the phrasing! "Plane of coil makes 30° with B" means the dipole vector (normal to plane) makes θ = 90° - 30° = 60° with B! If you plug in 30° directly, you will pick the distractor option and lose 5 marks!
Problem 8 • Earth's Magnetic Field & Apparent Dip
NEET 2017 PYQ
At a certain place, the horizontal component of Earth's magnetic field is BH = 0.35 G and the true angle of dip is δ = 45°. (a) Find the total magnetic field B of the Earth. (b) If the dip circle is set in a vertical plane making an angle of 60° with the magnetic meridian, find the apparent dip angle δ'.
Given & Core Concept:
• BH = 0.35 G, δ = 45° &implies; BV = BH tan 45° = 0.35 G.
• Total field: B = BHcosδ.
• Apparent Dip in plane rotated by angle α = 60° from meridian:
• BH = 0.35 G, δ = 45° &implies; BV = BH tan 45° = 0.35 G.
• Total field: B = BHcosδ.
• Apparent Dip in plane rotated by angle α = 60° from meridian:
tanδ' = tanδcosα
Step-by-Step Mathematical Calculation:
(a) Total Magnetic Field:
(b) Apparent Dip Angle δ':
(a) Total Magnetic Field:
B = BHcos 45° = 0.351/√2 = 0.35 × 1.414 = 0.495 G = 4.95 × 10-5 T.
(b) Apparent Dip Angle δ':
tanδ' = tan 45°cos 60° = 10.5 = 2
δ' = tan-1(2) ≈ 63.4°.
δ' = tan-1(2) ≈ 63.4°.
Golden NEET Property:
Because cosα ≤ 1 for all real angles, tanδ' ≥ tanδ! Thus, apparent dip is ALWAYS greater than or equal to true dip (δ' ≥ δ). If any exam option gives an apparent dip smaller than the true dip, eliminate it instantly!
Because cosα ≤ 1 for all real angles, tanδ' ≥ tanδ! Thus, apparent dip is ALWAYS greater than or equal to true dip (δ' ≥ δ). If any exam option gives an apparent dip smaller than the true dip, eliminate it instantly!
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