Given:
Mass of the particle, m=5 mg=5×10−6 kg
Charge, q=5π×10−6 C
Velocity, v=(3i^+2k^)×10−2 m/s
Magnetic field, B=0.1k^ Wb/m2
The velocity component parallel to the magnetic field is v∥=2×10−2 m/s.
The time period of one revolution is given by:
T=qB2πm
Substituting the given values:
T=5π×10−6×0.12π×5×10−6=0.12=20 s
The time taken to complete 5 revolutions is:
t=5T=5×20=100 s
The distance moved along the k^ direction (pitch for 5 revolutions) is:
α=v∥×t
α=2×10−2×100=2 m
Answer: 2