Question

A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction. At a particular instant, the velocity of the particle has components Vx=−1.68×10^4m/s, Vy=−2.61×10^4m/s, and Vz=5.85×10^4m/s. What is the z-component of the force on the particle at this time?

Answer 1

Answer:

The  z-component of the force is  $\= F_z = 0.00141 \ N$    

Explanation:

From the question we are told that

          The charge on the particle is $q = 7.76 *0^{-8} \ C$    

           The magnitude of the magnetic field is  $B = 0.700\r i \ T$

            The  velocity of the particle toward the x-direction is  $v_x = -1.68*10^{4}\r i \ m/s$

           The  velocity of the particle toward the y-direction is

$v_y = -2.61*10^{4}\ \r j \ m/s$

           The  velocity of the particle toward the z-direction is

$v_y = -5.85*10^{4}\ \r k \ m/s$

Generally the force on this particle is mathematically represented as

          $\= F = q (\= v X \= B )$

So  we have    

          $\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)$

         $\= F = q (v_y B(-\r k) + v_z B\r j)$      

  substituting values

       $\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)$    

      $\= F= 0.00303\ \r j +0.00141\ \r k$                  

So the z-component of the force is  $\= F_z = 0.00141 \ N$    

Note :  The  cross-multiplication template of unit vectors is  shown on the first uploaded image  ( From Wikibooks ).