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?
- 6.0x10-1T
-
6.0x10T
- 6.0x10-3T
- 6.0x10-4T
The correct answer is: B
Explanation
F = BQVB = \(\frac{F}{QV}\)
B = \(\frac{1.8 \times 10^{-8}}{1.0 \times 10^{-8} \times 3.0 \times 10^{-2}}\)
There is an explanation video available .