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

Correct answer is B

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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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?