Question

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to

Answer 1

Complete Question

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)

Answer:

The velocity is  $v = 3.79 *10^{5} \ m/s$  

Explanation:

From the question we are told that

    The  magnitude of the electric field is  $E = 144 \ kV /m = 144*10^{3} \ V/m$

     The magnetic field is  $B = 0.38 \ T$

   

The force due to the electric field is mathematically represented as

      $F_e = E * q$

and

The force due to the magnetic field is mathematically represented as

    $F_b = q * v * B * sin(\theta )$

Now given that it is perpendicular ,  $\theta = 90$

=>   $F_b = q * v * B * sin(90)$

=>   $F_b = q * v * B$

Now  given that it is not deflected it means that

        $F_ e = F_b$

=>    $q * E = q * v * B$

=>   $v = \frac{E}{B }$

 substituting values

     $v = \frac{ 144 *10^{3}}{0.38 }$

     $v = 3.79 *10^{5} \ m/s$