A particle accelerator moves an alpha particle with a speed of 2 \times 10\(^5\)ms\(^{−1}\) perpendicular to a 5 T magnetic field. Calculate the force on this particle. [Q = 1.6 \times 10\(^{−19}\) C]
A
0.9 x 10\(^{-13}\)N
B
4.5 x 10\(^{-14}\)N
C
3.2 x 10\(^{-13}\)N
D
1.2 x 10\(^{-13}\)N
Verified Correct Answer: C.
Physical Principles & Numerical Solution
An alpha particle has charge q = 2e = 2 \times 1.6 \times 10\(^{−19}\) = 3.2 \times 10\(^{−19}\) C
The magnetic force is given by F = q v B (since the velocity is perpendicular to the field).
F = (3.2 \times 10\(^{−19}\)) \times (2 \times 10\(^5\)) \times 5 = 3.2 \times 10\(^{−13}\)N
Source: Myschool Past Questions Archive (Record #78269)