Magnetism and Magnetic Effect of Electric Current

Helical path with radius = mv sin30°/(eB) ≈ 2.25×10⁻⁴ m

Step 1: When a charged particle moves at an angle θ ≠ 0° and θ ≠ 90° to a magnetic field, it follows a helical path. Step 2: The velocity component perpendicular to B is v⊥ = v sin30° = 4×10⁷ × 0.5 = 2×10⁷ m/s. This causes circular motion. Step 3: The radius of the helix: R = mv⊥/(eB) = (9×10⁻³¹ × 2×10⁷)/(1.6×10⁻¹⁹ × 0.5) = 1.8×10⁻²³ / 8×10⁻²⁰ = 2.25×10⁻⁴ m. Step 4: The velocity component parallel to B (v∥ = v cos30°) causes translation along B, producing the helical path. Step 5: Option B is wrong — circular path only occurs for θ = 90°. Option C is wrong — a particle never follows a straight

B = μ₀(n₁I₁ – n₂I₂) = μ₀ × 1200 × (5 – 3) = 2400μ₀ ≈ 3.02×10⁻³ T

Step 1: Number of turns per unit length for each solenoid: n = (3 layers × 400 turns) / 1 m = 1200 turns/m. Step 2: Field due to outer solenoid: B₁ = μ₀nI₁ = 4π×10⁻⁷ × 1200 × 5 = 7.54×10⁻³ T (along +x). Step 3: Field due to inner solenoid (opposite current): B₂ = μ₀nI₂ = 4π×10⁻⁷ × 1200 × 3 = 4.52×10⁻³ T (along –x). Step 4: Net field = B₁ – B₂ = μ₀ × 1200 × (5 – 3) = μ₀ × 1200 × 2 = 2400μ₀ = 4π×10⁻⁷ × 2400 ≈ 3.02×10⁻³ T. Step 5: Option B incorrectly adds the currents. Option C would be true only if I₁ = I₂. Option D ignores the inner solenoid's contribution.

0.6 × 10⁻⁴ T

Step 1: The relationship between BH (horizontal component) and total field B is: BH = B cos δ. Step 2: Therefore: B = BH / cos δ = (0.3 × 10⁻⁴) / cos 60°. Step 3: cos 60° = 0.5, so B = 0.3×10⁻⁴ / 0.5 = 0.6×10⁻⁴ T. Step 4: Vertical component would be BV = B sin δ = 0.6×10⁻⁴ × sin 60° = 0.6×10⁻⁴ × (√3/2) = 0.3√3×10⁻⁴ T — this is Option B but it represents BV, not B. Step 5: Option C would apply if δ = 45°. Option D is BH × cos60° which is wrong. Option A is correct.

0.4157 N·m

Step 1: Area of loop: A = 4×10⁻² × 6×10⁻² = 24×10⁻⁴ = 2.4×10⁻³ m². Step 2: Magnetic moment: M = NIA = 50 × 5 × 2.4×10⁻³ = 0.6 A·m². Step 3: Torque formula: τ = MB sinθ, where θ is the angle between M (normal to loop) and B. Here θ = 60°. Step 4: τ = 0.6 × 0.8 × sin60° = 0.48 × (√3/2) = 0.48 × 0.866 = 0.4157 N·m. Step 5: A common mistake is using cos60° instead of sin60°. Note that θ is the angle between the normal to the coil and B, so we use sinθ. If θ were between the plane of coil and B, we would use cosθ. Here the angle is with the normal, so sin60° is correct.