Question

A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direction, and a magnetic field of magnitude 1.25 T points in the positive z direction.

Answer 1

The given question is incomplete. The complete question is as follows.

A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direction, and a magnetic field of magnitude 1.25 T points in the positive z direction.

A) If the net force acting on the particle is $6.21 \times 10^{-3}$ N in the positive x direction, find the components of the particle's velocity. Assume the particle's velocity is in the x-y plane.

Enter your answers numerically separated by commas.

Explanation:

The given data is as follows.

           Q = $6.50 \times 10^{-6} C$

           E = 1300 N/C in the +x direction

           B = 1.02 T in the +z direction

and,    $F_{net} = 6.25 \times 10^{-3} N$ in the +x direction

Also,       $F_{net} = F_{E} - F_{b}$

                         = qE - qvB

Now, we will calculate the value of v as follows.

             v = $(\frac{1}{B}) \times (E - \frac{F_{net}}{q})$

                 = $(\frac{1}{1.02 T}) \times (1300 - \frac{6.25 \times 10^{-3}}{6.50 \times 10^{-6}})$

                v = 458.507 m/s

Using the value for velocity, we need to know which direction it's going.

You know +x direction for E, +z direction for B and +x for $F_{net}$.

Using the right hand rule where:

your right thumb goes toward the $F_{net}$, then your index finger points to B (z direction) Then curl your middle, ring, and pink 90 angle. This shows where v is going which is -y direction.

Thus, we can conclude that $v_{x}, v_{y}, v_{z}$ = 0, -(458.507), 0.