Question

A proton travels with a speed of 3x10^6m/s at an angle of 37 with the direction of the magnetic field 0.3t. what is the magnetic field of the proton? and what is its acceleration?

Answer 1

Speed of the proton = v = 3 x $10^{6}$ m/s

Angle between the velocity and magnetic field = 37°

magnitude of magnetic field = B = 0.3T

⇒ The magnitude of the magnetic field of the proton:-

    Using the relation F = q ( V x B) we can find the force

     F = q ( V x B ) can also be written as :-

     F = q x V x B x sin∅ , substituting B and V in this equation ,

     F = q x ( 3 x $10^{6}$ m/s) x ( 0.3T) sin(37°)

     We know the charge of the proton (q)  = 1.6 x $10^{-19}$c

      F =  (1.6 x $10^{-19}$)  x ( 3 x $10^{6}$ m/s) x ( 0.3T)  x sin(37°)

         = 1.44 x $10^{-13}$ x sin(37°)

         = 8.899 x $10^{-14}$ N

⇒ To find the acceleration of the proton :-

  Now we know the value of the Force . So, we can use the relation

      $f = m x a$   to get the acceleration

     $a = \frac{f}{m}$ = $\frac{8.899 x 10^{-19}N }{1.6 x 10^{-27}kg }$ = 5.561 x $10^8$ m /s

 ∴   the acceleration of proton  = 5.561 x $10^8$ m /s

     

to know more about magnetic field go here :-

brainly.com/question/14848188

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