A uniform magnetic field B is acting from south to north and is of magnitude $1.5\ wb/m^{2}$. If a proton having mass $= 1.7 imes 10^{- 27}kg$ and charge $= 1.6 imes 10^{- 19}C$ moves in this field vertically downward with energy $5\ MeV$, then the force acting on it will be

Question

A uniform magnetic field B is acting from south to north and is of magnitude $1.5\ wb/m^{2}$. If a proton having mass $= 1.7 	imes 10^{- 27}kg$ and charge $= 1.6 	imes 10^{- 19}C$ moves in this field vertically downward with energy $5\ MeV$, then the force acting on it will be

Dr. Abubacker · Accepted Answer

(b) $F = qvB$ and $K = \frac{1}{2}mv^{2} \Rightarrow F = qB\sqrt{\frac{2k}{m}}$
$= 1.6 	imes 10^{- 19} 	imes 1.5\sqrt{\frac{2 	imes 5 	imes 10^{6} 	imes 1.6 	imes 10^{- 19}}{1.7 	imes 10^{- 27}}}$
$= 7.344 	imes 10^{- 12}N$

Q. A uniform magnetic field B is acting from south to north and is of magnitude $1.5\ wb/m^{2}$. If a proton having mass $= 1.7 \times 10^{- 27}kg$ and charge $= 1.6 \times 10^{- 19}C$ moves in this field vertically downward with energy $5\ MeV$, then the force acting on it will be

Explanation

Reviewed by Dr. Abubacker

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