Question

An electron has an initial velocity of (19.0 j + 18.0 k) km/s and a constant acceleration of (3.00 ✕ 1012 m/s2)i in a region in which uniform electric and magnetic fields are present. If B = (400 µT)i, find the electric field E.

Answer 1

Answer:

$E=(-17.08 i +7.2 j -7.6 k )N/C$

Explanation:

$v= (19.0j+18.k)km/s$

$a=3.0x10^{13}m/s^2$

$\beta= 400x10{-6}T$

Electron information needed to solve the question:

$m_e=9.11x10^{-31}kg$

$q=-1.6x10{-19}C$

$F=F_E+F_B=q*(E+Vx \beta)$

$F=m*a$

$m*a=q*(E+Vx\beta)$

$E=\frac{m*a}{q}-(Vx\beta )$

$E=\frac{9.11x10{-31}kg*3.0x10^{12}m/s^2}{-1.6x10{-19}C}-[(19.0x10^3mj+18.0x10^3m)xi(400x10^{-6}T)]$

$E=-i17.08N/C-[7.6(-k)+7.2(j)]N/C$

$E=(-17.08 i +7.2 j -7.6 k )N/C$

Answer 2

Newton's second law and the electric and magnetic force allow us to find the result for the existing electric field is:

            E = (-17.1 i^ - 7.2 j^ + 7.6 k ^) N / C

Given parameters

• The electron charge q = - 1.6 10⁻¹⁹ C

• The velocity of the electron v = (19.0 j ^ + 18.0 k ^) 10³ m / s

• The magnetic field is B = 400 10⁻⁶  i ^  T

• The acceleration a = 3.00 10¹² i ^ m / s²

To find

• The electric field

Newton's second law indicates that the net force is equal to the product of the mass and the acceleration of the body

          ∑ F = m a

where the bold letters indicate vectors, m is the mass and the acceleration of the body.

When we are in an electric field the force is given by

       $F_e$  = q E

Where E is the electric field and q is the charge, we can see that the force has the direction of the electric field.

If we are in an area with a magnetic field the force is given by a vector product

         $F_m = q\ v x B$

Where  v is the velocity, B the magnetic field, this vector product can be resolved in the form of a determinant.

        $F_m = q \ \left[\begin{array}{ccc}i&j&k\\v_x&v_y&v_z\\B_x&B_y&B_z\end{array}\right]$  

Let's solve this problem for parts.

1st part. We look for the magnitude of the magnetic force.

We solve the determinant

       $Fm = q \ \left[\begin{array}{ccc}i&j&k\\0&v_y&v_z\\B_x&0&0\end{array}\right]$  

       $F_m = q \ ( v_z B_x j \ - B_x v_y k)$  

2nd part. We resolved the  Newton's second law for each axis.

x-axis

         $F_e_x + F_m_x = m a_x$  

         $F_e_x + 0 = m a_x$

         q Eₓ = m aₓ

         Eₓ = $\frac{m}{q} \ a_x$

     

Let's calculate

          Eₓ = $- \frac{9.1 \ 10^{-31}}{1.6 \ 10^{-19} } \ 3 \ 10^1^2$  

          Eₓ = - 17.1 N / C

y-axis

There is no acceleration on this axis.

          $F_e_y + F_m_y = 0$

We substitute

         $q E_y = - q v_k B_x$  

Let's calculate

         $E_y = - 18.0 \ 10^3 \ 400 \ 10^{-6}\\E_y = - 7.2 \ N/C$

z-axis

In this direction there is no acceleration.

        $F_e_z +F_m_z = 0\\q E_z = q B_x v_y$  

         

Let's calculate.

       $E_z = 400 \ 10^{-6} \ 19.0 \ 10^3 \\E_z = 7.60 \ N/C$

       

3rd part. We construct the vector of the electric field.

        $E = E_x i + E_y j + E_z k$  

        E = (-17.1 i ^ - 7.2 j ^ + 7.6 k ^) N / C

In conclusion using Newton's second law and the electric and magnetic force we can find the result for the existing electric field is:

            E = (-17.1 i ^ - 7.2 j ^ + 7.6 k ^) N / C

Learn more here:  brainly.com/question/13791875