An electron moves with a speed of 2.00 x 10\(^7\) ms\(^{-1}\) in an orbit in a uniform magnetic field of 1.20 x 10\(^{-3}\) T. Calculate the radius of the orbit. [Mass of an electron = 9.11 x 10\(^{-3}\) kg; charge on an electron = 1.61 x 10\(^{-19}\)C]
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Correct Answer: Option n
Explanation:
\(\frac{Mv^2}{r}\) = qvB
r = \(\frac{MV^2}{qVB} = \frac{9.11 \times 10^{31} \times (2.00 \times 10^7)^2}{1.61 \times 10^{-19} \times 2.00 \times 10^7 \times 1.2 \times 10^{-3}}\)
= 9.43 x 10\(^{-2}\)
\(\frac{Mv^2}{r}\) = qvB
r = \(\frac{MV^2}{qVB} = \frac{9.11 \times 10^{31} \times (2.00 \times 10^7)^2}{1.61 \times 10^{-19} \times 2.00 \times 10^7 \times 1.2 \times 10^{-3}}\)
= 9.43 x 10\(^{-2}\)