A proton (mass =1.67×10−27kg and charge =1.6×10−19C is projected in a uniform magnetic field of 2 Wb/m2 with a velocity 3.4×107m/s in a direction perpendicular to the field. The acceleration of the proton should be

F = ma = qvB⇒a =qvBm=1.6×10−19×2×3.4×1071.67×10−27=6.5×1015 m/s2

F = ma = qvB⇒a =qvBm=1.6×10−19×2×3.4×1071.67×10−27=6.5×1015 m/s2

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A proton (mass $= 1.67 \times 10^{- 27} kg and charge = 1.6 \times 10^{- 19} C)$ is projected in a uniform magnetic field of 2 Wb/m² with a velocity $3.4 \times 10^{7} m / s$ in a direction perpendicular to the field. The acceleration of the proton should be

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$F = ma = qvB$

$\Rightarrow a = \frac{qvB}{m} = \frac{1 .6 \times 10^{- 19} \times 2 \times 3 .4 \times 10^{7}}{1 .67 \times 10^{- 27}}$

$= 6.5 \times 10^{15} m / s^{2}$

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