Magnetic Effects of Current – JEE Main Physics Formula Sheet & Class 12 Notes | Biot-Savart Law, Ampere's Law, Lorentz Force, Cyclotron, Galvanometer & All Formulas

Biot-Savart Law — Magnetic Field Due to Wire, Loop, Arc

Why B at Centre of Circular Loop μ₀I/2R and Long Wire μ₀I/2πr Are the Most-Tested Biot-Savart JEE Main Results

Biot-Savart Law (from Aakash PDF — Magnetic Effects of Current JEE Main):

The magnetic field dB produced by a small current element Idl at a point P at distance r, where θ is the angle between dl and the position vector:

dB = (μ₀/4π) × Idl sinθ / r²

Vector form: dB = (μ₀/4π) × (Idl × r̂)/r²

μ₀ = 4π×10⁻⁷ T·m/A (permeability of free space). Direction: right-hand screw rule (dB perpendicular to both dl and r̂).

B Due to Straight Wire (from Aakash PDF — Magnetic Effects of Current JEE Main):

For a finite straight wire carrying current I, at perpendicular distance d from the wire, subtending angles φ₁ and φ₂ from the two ends:

B = (μ₀I/4πd)(sinφ₁ + sinφ₂)

For an infinite straight wire (φ₁=φ₂=90°): sinφ₁+sinφ₂=2:

B = μ₀I/2πd (at perpendicular distance d)

For a semi-infinite wire (one end at the field point level, φ₁=90°, φ₂=0°): sinφ₁+sinφ₂=1:

B = μ₀I/4πd

B at Centre of Circular Loop (from Aakash PDF — Magnetic Effects of Current JEE Main):

A circular loop of radius R, carrying current I:

B_centre = μ₀I/2R (direction: right-hand rule — curl fingers in current direction, thumb points to B)

For a circular arc subtending angle θ (in radians) at the centre:

B_arc = μ₀Iθ/4πR (proportional to angle fraction; full circle θ=2π gives μ₀I/2R ✓)

B on Axis of Circular Loop (from Aakash PDF — Magnetic Effects of Current JEE Main):

At axial point distance x from centre of loop of radius R:

B_axis = μ₀IR²/[2(R²+x²)^(3/2)]

At x=0: B = μ₀I/2R (centre) ✓. At x>>R: B ≈ μ₀IR²/2x³ = μ₀(IA)/2πx³ = μ₀M/2πx³ (magnetic dipole field).

Maximum B at centre; B decreases as x increases.

Download the Free PDF for Magnetic Effects of Current for all Biot-Savart examples for JEE Main.

Ampere's Circuital Law — Solenoid, Toroid, and Long Conductor

Why Solenoid B=μ₀nI and Toroid B=μ₀NI/2πr Are the Most-Tested Ampere's Law JEE Main Results

Ampere's Circuital Law (from Aakash PDF — Magnetic Effects of Current JEE Main):

The line integral of magnetic field B around any closed path (Amperian loop) equals μ₀ times the net current enclosed:

∮B·dl = μ₀I_enc

Analogous to Gauss's law for electric field. Used when the current distribution has high symmetry.

Infinite Straight Wire (from Ampere's law) (from Aakash PDF — Magnetic Effects of Current JEE Main):

Circular Amperian loop of radius r coaxial with wire: B is constant on the loop and parallel to dl.

B × 2πr = μ₀I → B = μ₀I/2πr ✓ (same as Biot-Savart)

Inside a solid conductor of radius R carrying current I (uniform distribution, r

I_enc = I(r/R)² → B = μ₀Ir/2πR² (increases linearly with r inside)

At surface r=R: B = μ₀I/2πR. Outside (r>R): B = μ₀I/2πr.

Solenoid (from Aakash PDF — Magnetic Effects of Current JEE Main):

A solenoid is a long helix of wire (length L, total N turns, n=N/L turns per metre). Rectangular Amperian loop with one side inside, one outside.

B_inside = μ₀nI = μ₀NI/L (uniform, parallel to axis)

B_outside ≈ 0 (for ideal long solenoid). At the ends: B = μ₀nI/2 (half of central value).

Direction: right-hand rule — curl fingers in current direction, thumb → B direction inside.

Toroid (from Aakash PDF — Magnetic Effects of Current JEE Main):

A toroid is a solenoid bent into a doughnut shape (total N turns, mean radius r).

B_inside (circular Amperian loop of radius r) = μ₀NI/2πr = μ₀nI

where n = N/2πr (turns per unit length of the toroid).

B_outside toroid = 0 (Amperian loop outside encloses zero net current — equal upward and downward currents).

B_in hole of toroid = 0 (Amperian loop in the hole encloses zero current).

Download the Free PDF for Magnetic Effects of Current for all Ampere's law examples for JEE Main.

Lorentz Force — Motion in Magnetic Field, Cyclotron, Velocity Selector

Why r=mv/qB, T=2πm/qB, and Cyclotron KE=q²B²R²/2m Are the Most-Tested Magnetic Force JEE Main Formulas

Lorentz Force (from Aakash PDF — Magnetic Effects of Current JEE Main):

Force on a charge q moving with velocity v in magnetic field B:

F = q(v×B); magnitude: F = qvBsinθ

where θ = angle between v and B. Direction: right-hand rule for cross product (or Fleming's left-hand rule for positive charge).

Key properties: (1) F ⊥ v always → work done by magnetic force = 0 → magnetic force cannot change the speed (KE) of charge, only its direction. (2) F ⊥ B always. (3) F = 0 when v ∥ B (θ=0°) or when v=0.

Circular Motion in Magnetic Field (from Aakash PDF — Magnetic Effects of Current JEE Main):

When v ⊥ B, the charge moves in a circle (magnetic force = centripetal force):

qvB = mv²/r → r = mv/(qB)

Also: r = p/qB (p=momentum); r = √(2mKE)/(qB)

Time period: T = 2πr/v = 2πm/(qB) (independent of speed v and radius r)

Cyclotron frequency: f = qB/(2πm); angular velocity: ω = qB/m

Helical Motion (from Aakash PDF — Magnetic Effects of Current JEE Main):

When v has a component parallel to B (v_∥) and perpendicular (v_⊥): the charge undergoes circular motion (due to v_⊥) combined with uniform motion (due to v_∥) → helical path.

Radius: r = mv_⊥/(qB). Time period: T = 2πm/(qB) [same as circular]. Pitch: p = v_∥ × T = 2πmv cosθ/(qB) where θ = angle between v and B.

Cyclotron (from Aakash PDF — Magnetic Effects of Current JEE Main):

A cyclotron accelerates positive ions using alternating electric field (frequency = cyclotron frequency f=qB/2πm) and magnetic field B for circular motion.

Resonance condition: applied frequency = cyclotron frequency: ω_applied = qB/m

At maximum radius R (dee radius): maximum velocity v_max = qBR/m

Maximum kinetic energy: KE_max = ½mv²_max = q²B²R²/(2m)

Cyclotron doesn't work for electrons (too light, relativistic effects become important at achievable energies).

Velocity Selector (from Aakash PDF — Magnetic Effects of Current JEE Main):

Electric and magnetic forces balance for one velocity: qE = qvB → v = E/B (selected velocity)

Only particles with this speed pass straight through; others are deflected. Download the Free PDF for Magnetic Effects of Current for all Lorentz force examples for JEE Main.

Force on Current-Carrying Conductor, Force Between Wires, Torque on Current Loop

Why Force Between Parallel Wires μ₀I₁I₂/2πd and Torque τ=NIBA Are Key Magnetic Effects JEE Main Results

Force on Current-Carrying Conductor (from Aakash PDF — Magnetic Effects of Current JEE Main):

A wire of length l carrying current I in magnetic field B at angle θ:

F = BIl sinθ = Il×B

Direction: Fleming's left-hand rule (FBI rule). F_max = BIl (when θ=90°); F=0 when θ=0° (wire parallel to B).

Force Between Two Parallel Current-Carrying Wires (from Aakash PDF — Magnetic Effects of Current JEE Main):

Two parallel wires separated by distance d, carrying currents I₁ and I₂:

F/l = μ₀I₁I₂/(2πd)

Parallel currents (same direction): attract each other. Anti-parallel currents: repel.

Definition of Ampere: 1 Ampere is the current that, when flowing through two infinitely long parallel wires 1 metre apart in vacuum, produces a force of 2×10⁻⁷ N/m on each wire.

Torque on Current Loop in Magnetic Field (from Aakash PDF — Magnetic Effects of Current JEE Main):

A rectangular loop of N turns, area A, carrying current I, placed in uniform magnetic field B at angle α between the plane of the loop and B (equivalently, angle (90°–α) between B and normal to the loop):

τ = NIBA sinα

Or equivalently: τ = MB sinα = M×B

where M = NIA is the magnetic moment of the loop.

If α is the angle between the magnetic moment M and B: τ = MB sinα. τ = 0 when M ∥ B (stable equilibrium) or M anti∥ B (unstable). τ = MB (maximum) when M ⊥ B (α=90°).

Magnetic Moment (from Aakash PDF — Magnetic Effects of Current JEE Main):

M = NIA (N=turns, I=current, A=area). Unit: A·m².

Direction: normal to the plane of the loop (right-hand rule).

Potential energy: U = –MB cosα = –M·B

Stable equilibrium: α=0°, U = –MB (minimum). Unstable: α=180°, U = +MB.

Download the Free PDF for Magnetic Effects of Current for all torque and force examples for JEE Main.

Moving Coil Galvanometer, Ammeter, Voltmeter Conversions

Why Galvanometer θ=NABI/k and Ammeter Shunt S=IgRg/(I–Ig) Are Direct Magnetic Effects JEE Main Questions

Moving Coil Galvanometer (from Aakash PDF — Magnetic Effects of Current JEE Main):

A galvanometer consists of a coil of N turns, area A, suspended in a radial magnetic field B (using curved pole pieces so that B is always parallel to the plane of the coil → always maximum torque). The coil is suspended by a spring with torsional constant k.

Torque by magnetic force: τ_mag = NABI (since sinα=1 for radial field always)

Restoring torque by spring: τ_spring = kθ

Equilibrium: NABI = kθ → θ = NABI/k

Deflection θ ∝ I (linear scale — advantage of radial field).

Current sensitivity = θ/I = NAB/k (deflection per unit current)

Voltage sensitivity = θ/V = NAB/(kR_g) where R_g = coil resistance

Full-scale deflection current I_g = kθ_max/(NAB).

Conversion to Ammeter (from Aakash PDF — Magnetic Effects of Current JEE Main):

Connect a low resistance shunt S in PARALLEL with galvanometer to bypass excess current.

At full scale: I_g flows through galvanometer, (I–I_g) through shunt. Same voltage across both:

I_g R_g = (I–I_g)S → S = I_g R_g/(I–I_g)

Effective resistance of ammeter: R_ammeter = SR_g/(S+R_g) ≈ S (very small, since shunt is low)

Ideal ammeter: zero resistance (doesn't disturb circuit).

Conversion to Voltmeter (from Aakash PDF — Magnetic Effects of Current JEE Main):

Connect a high resistance R_s in SERIES with galvanometer.

Full scale voltage V must produce full-scale current I_g through (R_g+R_s):

V = I_g(R_g+R_s) → R_s = V/I_g – R_g = (V–I_gR_g)/I_g

Effective resistance of voltmeter: R_voltmeter = R_g + R_s = V/I_g (very large)

Ideal voltmeter: infinite resistance (draws no current). Download the Free PDF for Magnetic Effects of Current for all galvanometer examples for JEE Main.

Magnetism — Bar Magnet, Earth's Field, Diamagnetic, Paramagnetic, Ferromagnetic

Why Bar Magnet Field Formulas and Dia/Para/Ferromagnetic Classification Are Key Magnetism JEE Main Topics

Bar Magnet as Magnetic Dipole (from Aakash PDF — Magnetic Effects of Current JEE Main):

A bar magnet of magnetic moment M = m×2l (m=pole strength, 2l=distance between poles).

Field on axial line (along the axis, at distance r from centre, r>>l):

B_axial = (μ₀/4π) × 2M/r³ (direction: from S to N, i.e., along M)

Field on equatorial line (perpendicular bisector, at distance r, r>>l):

B_equatorial = (μ₀/4π) × M/r³ (direction: antiparallel to M)

B_axial = 2 × B_equatorial at same distance r (analogy with electric dipole).

Earth's Magnetism (from Aakash PDF — Magnetic Effects of Current JEE Main):

Earth acts like a huge bar magnet. The total field B at any location has components:

H = horizontal component (pointing toward magnetic north)

V = vertical component (pointing downward in northern hemisphere)

Total field: B = √(H²+V²)

Angle of dip (δ): tanδ = V/H

Angle of declination: angle between geographic north and magnetic north at a location.

At magnetic equator: δ=0°, V=0, B=H. At magnetic poles: δ=90°, H=0, B=V.

Tangent Law (from Aakash PDF — Magnetic Effects of Current JEE Main):

When a bar magnet is placed perpendicular to horizontal component H in a deflection magnetometer, the deflection θ satisfies:

B = H tanθ

Classification of Magnetic Materials (from Aakash PDF — Magnetic Effects of Current JEE Main):

Diamagnetic: magnetic susceptibility χ < 0 (slightly negative). Weakly repelled by magnets. Examples: Cu, Au, Bi, water. No permanent dipole moment; B inside slightly less than B₀.

Paramagnetic: χ > 0 (small positive). Weakly attracted by magnets. Examples: Al, Mn, O₂, Na. Has permanent dipole moments but randomly oriented; Curie law: χ = C/T.

Ferromagnetic: χ >> 1 (very large positive). Strongly attracted. Examples: Fe, Co, Ni. Permanent domains; very strong magnetisation. Curie-Weiss law: χ = C/(T–T_C) where T_C = Curie temperature.

Above Curie temperature: ferromagnetic → paramagnetic.

Hysteresis: B-H curve for ferromagnets. Retentivity = B when H=0 after magnetisation. Coercivity = H required to demagnetise. Area of loop = energy dissipated per cycle. Download the Free PDF for Magnetic Effects of Current for all magnetism examples for JEE Main.