Question

An alpha particle is accelerated through a potential difference of 10kV and moves along x- axis. It enters in a region of uniform magnetic field B=2x 10-3 T acting along Y-axis. Find the radius of its path.

Answer 1

Answer:

0.1

Explanation:

r=mv/qb

put values

Answer 2

10

Explanation:

r=√(2mqV) / qB

as given alpha particle q=2e=2×1.6×10^-19

V=10^4V

m=6.4×10^-27kg

B=2×10^-3T

r=√(2 × 6.4×10^-27 × 2× 1.6×10^-19 × 10^4) /1.6×10^-19 × 2×10^-3

=√(4096×10^-44) / 64 × 10^-23

=(64 × 10^-22) / (64 × 10^-23)

=10

therefore radius is 10m