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  • A  magnetic field of 0.40 T is applied on a proton moving with a velocity of .

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

Option: 1

10^{14}m/s^2


Option: 2

10^{19}m/s^2


Option: 3

10^{2}m/s


Option: 4

None


Answers (1)

best_answer

Force acting on the proton, F=qvB sin \theta

\therefore Acceleration of the proton 

=\frac{Force \;acting} {mass \;of\; proton}

=\frac{qv B sin\theta}{m}

\therefore Acceleration =\frac{1.6 \times 10^{-19}\times 5 \times10^6\times 0.40\times sin30\degree}{1.6\times 10^{-27}}

=10^{14} m/s^2

Posted by

avinash.dongre

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