A particle carrying a charge of 1 0 x 10-8C ent... - JAMB Physics 2012 Question
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
6.0x10-1T
B
6.0x10T
C
6.0x10-3T
D
6.0x10-4T
correct option: b
F = BQV
B = \(\frac{F}{QV}\)
B = \(\frac{1.8 \times 10^{-8}}{1.0 \times 10^{-8} \times 3.0 \times 10^{-2}}\)
B = \(\frac{F}{QV}\)
B = \(\frac{1.8 \times 10^{-8}}{1.0 \times 10^{-8} \times 3.0 \times 10^{-2}}\)
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