First slide

Force on a charged particle moving in a magnetic field

Question

A 2 MeV proton is moving perpendicular to a uniform magnetic field of 2.5 T. The force on the proton is

Moderate

A

$2.5 \times 10^{- 10} N$
B

$8 \times 10^{- 11} N$
C

$2.5 \times 10^{- 11} N$
D

$8 \times 10^{- 12} N$

Solution

$\begin{array}{l} \frac{1}{2} m v^{2} = 2 \times 1.6 \times 10^{- 19} \times 10^{6} \\ \frac{1}{2} 1.6 \times 10^{- 27} v^{2} = 2 \times 1.6 \times 10^{- 13} \\ v^{2} = 4 \times 10^{14} \\ v = 2 \times 10^{7} \\ F = 1.6 \times 10^{- 19} \times 2 \times 10^{7} \times 2.5 \\ = 8 \times 10^{- 12} N \end{array}$

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