Govt. Exams
Motional EMF ε = BLv = 2 × 0.5 × 10 = 10 V.
For undeflected motion, electric force equals magnetic force: qE = qvB, which simplifies to E = vB. This is the velocity selector condition.
m = IA = 2 × π × (0.05)² = 2 × π × 0.0025 ≈ 0.0157 A·m² ≈ 0.157 mA·m² or 0.157 × 10⁻³ A·m². Recalculating: m = 2 × 3.14 × 0.0025 ≈ 0.0157 A·m² or 1.57 × 10⁻² A·m².
B = μ₀I/(2πr). So I = 2πrB/μ₀ = 2π × 0.02 × 4 × 10⁻⁵/(4π × 10⁻⁷) = 2 A.
Torque τ = NIAB sin θ. When B is perpendicular to the plane of the loop, it is parallel to the normal of the loop area, so θ = 0° and τ = 0.
F/L = μ₀I₁I₂/(2πr) = (4π × 10⁻⁷ × 5 × 3)/(2π × 0.1) = 3 × 10⁻⁵ N/m. Same direction currents attract.
B = μ₀nI where n = N/L = 500/0.5 = 1000 turns/m. B = 4π × 10⁻⁷ × 1000 × 2 = 2.51 × 10⁻¹ T ≈ 0.251 T.
Self-inductance of solenoid is derived from L = NΦ/I where Φ = μ₀nIA. This gives L = μ₀N²A/L, proportional to N² and inversely proportional to length.
Torque τ = m × B has magnitude τ = mB sin(θ). Maximum occurs when sin(θ) = 1, i.e., θ = 90° (perpendicular orientation).
Period T = 2πm/eB is independent of velocity. This remarkable result means all particles with same m and q have same period regardless of speed in a given B field.
