A small particle with positive charge q = +4.25 x 10^-4C and mass m = 5.00 x 10^-5 kg is moving in a region of uniform electric and magnetic fields. The magnetic field is B=4.00 T in the +z-direction. The electric field is also in the +z-direction and has magnitude E = 60.0 N/C. At time t = 0 the particle is on the y-axis at y = +1.00 m and has velocity v = 30.0 m/s in the +x-direction. Neglect gravity.
a) What are the x-, y-, and z-coordinates of the particle at t = 0.0200 s?
b) What is the speed of the particle at t = 0.0200 s?

A) The coordinates of the particle in  x,  y and z axis is written as  :

( 0.6 i - 2039 j + 0.102 k )m   ( i.e. x = 0.6,  y = 2039,  z = 0.102 )

B) The speed of the particle at t = 0.0200 s

A ) Determine the coordinates of the particle in  x, y and z axis

applying Newton's second law ;  a = ( Fe + Fm ) / m

where :  Fe = 2.55 * 10⁻² k

The magnetic force ( Fm ) = qv * B  =

∴ Fm = 0

Acceleration in each axis ( x , y and z )

Ax = 0  because there is no force

Ay =  - 1.02 * 10⁷  j/s²

Az =   5.1 * 10² km/s²

i) For the x- axis

x =  Vot  +  ½ * Ax * t²

Vo = 30 m/s

t = 0.02

Ax = 0

∴ x - coordinate of the particle = 0.6m

ii) For the y-axis

y = I + Vo*t + ½ *Ay* t²

y = - 2039 m

iii) For the z-axis

 z = zo + v₀* t + ½ * Az* t²

 z = 0.102 m .

B ) Determine the speed of the particle in all three axis

Vx = 30 m/s

Vy = v₀ + Ay *  t

     = 30 + - 1.02 * 10⁷ * 0.020

     = 2.04 * 10⁵ m/s

Vz =  Vo + Az* t

    = 1.02 * 10⁻¹

Hence we can conclude that The coordinates of the particle in  x,  y and z axis is written as  : ( 0.6 i - 2039 j + 0.102 k )m   ( i.e. x = 0.6,  y = 2039,  z = 0.102 )The speed of the particle at t = 0.0200 s

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a)   r = (0.6 i- 2039 j ^ + 0.102 k⁾ m  and b) vₓ = 30.0 m / s , v_{y} = 2.04 10⁵ m / s   c) v_{z}  = 1.02 10⁻¹m / s

Explanation:

a) To find the position of the particle at a given moment we must know the approximation of the body, use Newton's second law to find the acceleration

         Fe + Fm = m a

         a = (Fe + Fm) / m

the electric force is

         Fe = q E   k ^

         Fe = 4.25 10-4 60 k ^

         Fe = 2.55 10-2 k ^

the magnetic force is

         Fm = q v x B

         Fm = 4.25 10⁻⁴  

         fm = 4.25 10⁻⁴ (-j ^ 30 4)

         fm 0 = ^ -5,10 10⁻² j

We look for every component of acceleration

X axis

      aₓ = 0

there is no force

Axis y

      ay = -5.10 10²/5 10⁻⁵ j ^

      ay = -1.02 107 j ^ / s2

z axis

      az = 2.55 10⁻² / 5 10⁻⁵ k ^

      az = 5.1 10² k ^ m / s²

Having the acceleration in each axis we can encocoar the position using kinematics

X axis

the initial velocity is vo = 30 m / s and an initial position xo = 0

           x = vo t + ½ aₓ t₂2

           x = 30 0.02 + 0

           x = 0.6m

Axis y

acceleration is ay = -1.02 10⁷ m / s², a starting position of i = 1m

           y = I + go t + ½ ay t²

           y = 1 + 0 + ½ (-1.02 10⁷) 0.02²

           y = 1 - 2.04 10³

           y = -2039 m j ^

z axis

acceleration is aza = 5.1 10² m / s², the position and initial speed are zero

          z = zo + v₀ t + ½ az t²

          z = 0 + 0 + ½ 5.1 10² 0.02²

          z = 1.02 10⁻¹ m k ^

therefore the position of the bodies is

   r = (0.6 i- 2039 j ^ + 0.102 k⁾ m

b) x axis

 since there is no acceleration the speed remains constant

          vₓ = 30.0 m / s

Axis y

  let's use the equation v = v₀ + t

          = 0 + -1.02 10⁷ 0.02

          v_{y} = 2.04 10⁵ m / s

z axis

          v_{z} = vo + az t

          v_{z} = 0 + 5.1 10² 0.02

          v_{z}  = 1.02 10⁻¹m / s

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