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 2 .5 T. The force on the proton is

Moderate

A

2.5x10^-10N
B

8x10^-11 N
C

2.5 x10^-11 N
D

8 x10^-12 N

Solution

$\begin{array}{l} \frac{1}{2} {mv}^{2} = 2 \times 10^{6} eV = 2 \times 10^{6} \times e joules \\ ∴ v = {[\frac{4 \times 10^{6} \times e}{m}]}^{1 / 2} \\ F = evB = eB {[\frac{4 \times 10^{6} \times e}{m}]}^{1 / 2} \\ = (1 .6 \times 10^{- 19}) (2 \cdot 5) {[\frac{4 \times 10^{6} \times (1 .6 \times 10^{- 19})}{1 .67 \times 10^{- 27}}]}^{1 / 2} \\ = 8 \times 10^{- 12} N \end{array}$

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