Question

A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It enters a magnetic field of ( ) mT. What is the acceleration of the particle?

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

The acceleration would be  $a = 0.003* 6 =0.018\ m/s^2$    

Explanation:

The objective of this solution is to obtain the acceleration of the particle

       Now looking at Newton law which is mathematically represented as

                       $F = ma$

  Where F is the force experience by a particle

              m is the mass of the particle

              a is the acceleration of the particle

  And also this force is equivalent to magnetic force in a magnetic field which is mathematically represented as

                    $F = qvB$

Where q is the charge of the particle

             v is the velocity of the  charge

             B is the magnetic field the charge is under it influence

  Now equating this two formulas

                   $ma = qvB$

 Making a the subject we have

                  $a = \frac{qvB}{m}$

In the question the direction of the is in the positive x-axis which is $i$ hence the direction would be in the $i$ direction

      So substituting  $(2.0i +3.0j+4.0k)mT = (2.0i +3.0j+4.0k)*10^{-3}T$ for B

                    $a = \frac{q}{m} * v (2.0i +3.0j +4.0k)*10^{-3}$

      Substituting       $3.0 Km /s = 3.0*10^{3}\ m/s$  for v  and $-6.0 \muC = -6.0*10^{-6} C$ for q

                     $a = \frac{-6.0*10^{-6}}{2.0*10^{-3}} * 3.0*10^{3} *(2i+3j+4k) *10^{-3}$

                       $a = 0.003 * 3i(2i+3j+4k)$

                      $a = 0.003 *((3*2)i \ \cdot i \ +(3*3) i \ \cdot \ j \ + (3*4)i \ \cdot \ k)$

According to vector multiplication

                                             $i \cdot i = j \cdot j = k\cdot k = 1\\\\and \ i\cdot j = i\cdot k = 0$

     So

               $a = 0.003* 6 =0.018\ m/s^2$