A particle carrying a charge of 1. 0 x 10-8C enters a magnetic field at 3.0x 10-2 ms -1 at right angles to the field. If the force on this particle is 1.8 x 10-8N, what is the magnitude of the field?
-
A.
6.0×10-1T -
B.
6.0x10T -
C.
6.0×10-3T -
D.
6.0×10-4T
Correct Answer: Option B
Explanation
F = BQV
B = (frac{F}{QV})
B = (frac{1.8 times 10^{-8}}{1.0 times 10^{-8} times 3.0 times 10^{-2}})