Question

A proton with a kinetic energy of 4.1 ✕ 10^-16 J moves perpendicular to a magnetic field of 3.6 T. What is the radius of its circular path?

(Answer in cm)

Answer 1

Equating magnetic force to centripetal force ..
Bqv = mv²/R

BqR = mv

squaring on both sides and then multiplied by 1/2,

(1/2) × (BqR)²= (1/2)×(mv)²

½ (BqR)² /m = ½ mv² = 4.1×10^-16 J

(BqR)² = 2m × (4.1×10^-16)
BqR = √[2m × (4.1×10^-16)]

R = √[2m × (4.1×10^-16)] / (Bq) (m= proton mass =1.67×10^-27 kg, q = 1.60× 10^-19C)

R = √[2(1.67×10^-27) × (4.1×10^-16)] / (3.6×1.6×10^-19) m

R = [3.70 × 10^(-43/2)]/( 5.76 × 10^-19) m

R = 0.64 × 10^-2 × 0.32 m

R = 0.20 cm