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  • A magnetic field of 0.40T is applied on a proton moving with a velocity of <img alt="5 \times 10^6 \mathrm{~m} / \mathrm{s}" src="https://entrancecorner.oncodecogs.com/gif.latex?5%20%5Ctimes%2

A magnetic field of 0.40T is applied on a proton moving with a velocity of 5 \times 10^6 \mathrm{~m} / \mathrm{s}. The magnetic field act at an angle 30^{\circ} with the direction of velocity of the proton?

Option: 1

10^{14} \mathrm{~m} / \mathrm{s}^2


Option: 2

10^{19} \mathrm{~m} / \mathrm{s}


Option: 3

10^2 \mathrm{~m} / \mathrm{s}


Option: 4

None


Answers (1)

best_answer

Force acting on the proton, F = qVBSin\theta

\therefore Acceleration of the proton = \frac{\text { force acting }}{\text { mass of proton }} =\frac{q v B \sin \theta}{m} 

\therefore \text { Acceleration }=\frac{1.6 \times 10^{-19} \times 5 \times 10^6 \times 0.40 \times \sin 30^{\circ}}{1.6 \times 10^{-27}}=10^{14} \mathrm{~m} / \mathrm{s}^2

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chirag

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