University:
Santa Fe College
Course:
ETS 2527C | Electromechanical Components and Mechanisms
Academic year:

2024
Views:

254

Pages:

5
Author:

Jabin F.

Motion of a charge particle in a magnetic field Lorentz force: Fm = q(v x B) Fm = qVBsinθ θ = angle b/w v and B n̂ → unit vector along dirn (v x B) Limiting Cases Case 1: v = 0 (charge is at rest) Fm = 0 Case 2: If charge particle is moving with a constant velocity (v) (a) If v is parallel or antiparallel to B θ = 0° or θ = 180° Fm = 0 (b) v is ⊥ to B (θ = 90° or 270°) Fm = 2υβζϊηθή ⇒ Fm = 2υβm dirn of magnetic force can be found by right hand grip rule. B pointing into the page B pointing out of the page Ques. 17 Page-40 v(0) = 0, E^z, B^x Cycloid, E x B dirn z x x = +y direction Ques. 6 Page-38 B = 1 tesla z^ E = 1 v/m y^ v(0) = 300 m/s = Vx + Vy for straight line u(0) = E/B plane -> x-y plane dirn of motion of speed = 1 m/s x dirn (option-d) Ques. 20 Page-40 B = B^j E = -E^k if speed v(0) = E/B (-k x j) (undeflected) = (E/B)c^ True for Both type of charges Ques. 16 P-40 Cycloid -> x-y plane Fe = qE (ad rect) Fm = 0 (parallel) Ques. 3 P-38 Vi = 2VL Pitch P = ViT P = Vi(2πR/VL) μ = 2π(Vi/VL) μ = 2π(√2) = π(√2) R = 4π Ques. 5 P-38 B = Boz^ v = (3e^x + 2e^z) Vi = 2m/s VL = 3m/s d = Vit t = d/Vi t = 1 sec Helical path, t = 1 sec Vi₁ = v cos θ (undeflected) VI = v sin θ (circularly) (Helical motion) Pitch of helical y B V F z y B VL F x z y ViT VI x z B Pitch of helical path in which translation path travelled by charge particle