A charged particle moving in x-y plane enters a region of uniform magnetic field of intensity B⃗ = -B0k̂ at P(0, 0) with velocity 𝑣 1 = 𝑣î and leaves the magnetic field at Q(a, b) with velocity 𝑣 2 = 𝑣′î + √3𝑣′ĵ. Choose the correct statement(s):

Question

G1 Physics · Accepted Answer

Magnetic force does no work, so |v₁|=|v₂|, giving v=2v′ and an exit direction 60° above +x; thus the deviation is 60° (B). The curvature shown for B into the page requires negative charge, so the particle cannot be a positron (A). Hence A,B.