Question

A beam of protons is directed in a straight line along the positive zz ‑direction through a region of space in which there are crossed electric and magnetic fields. If the electric field magnitude is E=450E=450 V/m in the negative yy ‑direction and the protons move at a constant speed of v=7.9×105v=7.9×105 m/s, what must the direction and magnitude of the magnetic field be in order for the beam of protons to continue undeflected along its straight-line trajectory? Select the direction of the magnetic field BB .

Answer 1

Answer:

The magnitude is $B = \frac{450}{7.9* 10^5}$

The direction is  the positive x axis

Explanation:

From the question we are told that

   The  electric field is  E = 450 V/m in the negative y ‑direction

   The speed of the proton is  $v= 7.9* 10^5\ m/s$ in the positive z direction

Generally the overall force acting on  the proton is mathematical represented as

              $F_E = q(\vec E + \vec v * \vec B)$

Now for the beam of protons to continue un-deflected along its straight-line trajectory then  $F_E =0$

So

          $0 = q( E (-y) + v(z) * \vec B)$

=>      $E\^y = v \^ z * \vec B$

Generally from unit vector cross product vector multiplication

         $\^ z \ * \ \^ x = \^ y$

So the direction of  B (magnetic field must be in the positive x -axis )

So

       $E\^y = v \^ z * B\^ x$

=>     $E\^y = vB ( \^ z * \^ x)$

=>     $E\^y = vB ( \^y)$        

=>      $E = vB$  

=>      $B = \frac{450}{7.9* 10^5}$

=>      $B = 0.0005696 \ T$