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APEX CLASS ESTD 2018

IITJEE: Biot-Savart Magnetic Field Graphs, Solenoids & Cyclotron Motion

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Ashish Kumar Bahekar

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Apex class • IITJEE Magnetism & Lorentz Force Lab

Biot-Savart Magnetic Field Graphs, Solenoids & Cyclotron Motion

Master Ampere’s Law B(r) Graphs, Solenoid End Effects, and Charged Particle Trajectories in Magnetic Fields!

🧭 Select a Current-Carrying Conductor to Inspect Its Magnetic Field Profile:
Current I ↑ Perpendicular d θ₁ θ₂ ⊗ B (Into Page) Straight Wire Master Formulas B = (μ₀ I / 4πd) (sinθ₁ + sinθ₂) 1. Infinite Wire (θ₁ = θ₂ = 90°): B_inf = μ₀ I / (2π d) 2. Semi-Infinite Wire (θ₁ = 90°, θ₂ = 0°): B_semi = μ₀ I / (4π d) (Half of Infinite!) 🎯 JEE Polygon Shortcut: For a regular N-sided polygon of side ‘a’ carrying current I, check θ₁ = θ₂ = π/N at center! Angle φ (rad) B_arc = (μ₀ I / 4πR) × φ (Full Circle φ = 2π → B = μ₀I / 2R) B_axis(x) of Circular Coil B_max = μ₀NI / (2R) x = +R/2 x = -R/2 ⭕ Axial Field Formula: B(x) = μ₀NIR² / [2(R² + x²)³/²] | At x = R: B = B_center / (2√2)! R Uniform Current I (⊙ Out) B(r) r = R (Surface) Distance r → Inside: B ∝ r B_in = (μ₀Ir) / (2πR²) B_max = μ₀I / (2πR) Outside: B ∝ 1 / r 🪵 Hollow Pipe vs. Solid Wire Trap: Inside a HOLLOW pipe, B_in = 0; inside a SOLID wire, B_in rises linearly (B ∝ r)! LONG SOLENOID (n = N / L turns per meter, Current I) Center: B_in = μ₀ n I (100%) Left End: ½ μ₀nI Right End: ½ μ₀nI 🌀 Solenoid End Formula: Since B = ½ μ₀nI (cosθ₁ – cosθ₂), at the end of a long solenoid (θ₁=90°, θ₂=180°), B_end = ½ B_center!
Magnetic Constant (μ₀/4π)
10⁻⁷ T·m / A
μ₀ = 4π × 10⁻⁷ H/m
Peak Field Formula
B = μ₀I / (2πd) B_center = μ₀NI / (2R) B_surface = μ₀I / (2πR) B_center = μ₀ n I
Standard SI Expression
Special JEE Ratio
B_semi = ½ B_infinite B_center / B(R) = 2√2 B(R/2) = ½ B_surface B_end = ½ B_center
Instant MCQ Eliminator
Magnetic Dipole Moment
M = N · I · A
Gyromagnetic Ratio: M/L = q/(2m)
⚡ Lorentz Force F = qE + q(v × B). Select Particle Trajectory Scenario:
⊗⊗⊗⊗ ⊗⊗ ⊗⊗⊗⊗ +q v ⊥ B (Tangent) F_B = qvB ↑ 4 Equivalents of Cyclotron Radius (r) r = mv / (qB) = p / (qB) = √(2mK) / (qB) = √(2mV/q) / B • Time Period: T = 2πm / (qB) (Independent of v & r!) • Work Done by B Field: W_B = 0 J (Always!) • Proton : Deuteron : Alpha (Same K in same B): r_p : r_d : r_α = √1/1 : √2/1 : √4/2 = 1 : √2 : 1! 🏆 Most Repeated JEE Ratio: A Proton and an Alpha particle with the SAME Kinetic Energy have the EXACT SAME radius (1 : 1)! Magnetic Field B → ↔ Pitch p = (v cosθ) × T r = mv sinθ / (qB) Helical Motion Decomposition 1. Parallel Component (v_∥ = v cosθ): Zero Magnetic Force → Moves straight! 2. Perp Component (v_⊥ = v sinθ): Creates circle of radius r = mv sinθ/(qB) Pitch p = 2π m v cosθ / (qB) 🧬 Angle Summary: θ = 0° or 180° → Straight Line | θ = 90° → Circle | Any other angle θ → Helix! + + + Positive Plate (Electric Field E ↓, Magnetic B ⊗) – – – Negative Plate +q F_mag = qvB ↑ F_elec = qE ↓ Velocity Filter Condition (F_net = 0) q E = q v B ⇒ v = E / B! • Works for ALL charges (+q or -q) & masses! • If v > E/B: Magnetic force wins (bends ↑) • If v < E/B: Electric force wins (bends ↓) 🎯 J.J. Thomson e/m Experiment: Crossed E ⊥ B fields select particles with exact speed v = E / B! Case A: Parallel Currents (I₁ ↑↑ I₂) ATTRACT! I₁ ↑ ↑ I₂ Attract with F/L = μ₀I₁I₂ / (2πd) Case B: Anti-Parallel (I₁ ↑↓ I₂) REPEL! Repel with F/L = μ₀I₁I₂ / (2πd) ⚡ Definition of 1 Ampere: Two parallel wires 1 m apart carrying 1 A each experience exactly 2 × 10⁻⁷ N/m force!
Orbit Radius (r)
r = m v_⊥ / (q B)
Also r = √(2mK) / (qB)
Time Period (T) & Freq
T = 2π m / (q B)
Independent of Speed v!
Force on Current Wire
F = I (L_eff × B)
L_eff = Straight vector A to B!
Torque on Current Loop
τ = M × B = NIAB sinθ
θ is angle between Normal & B
🔥 2 High-Yield JEE Magnetism Shortcuts:
1. Arbitrary Curved Wire in Uniform B: The magnetic force on ANY curved wire carrying current I between points A and B is equal to the force on a straight wire connecting A to B: F = I (L_AB × B) (So for any closed loop in uniform B, F_net = 0!).
2. Revolving Charge (Gyromagnetic Ratio): For ANY body of charge q and mass m rotating with angular momentum L, the ratio of Magnetic Moment to Angular Momentum is constant: M / L = q / (2m)!

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