Question

CHEGG 42 mT magnetic field points due west. If a proton of kinetic energy 9 x 10-12 J enters this field in an upward direction, find the magnetic force acting on the proton, in magnitude and direction. (6.98 x 10-13 N , South)

Answer 1

Answer:

The  magnitude of the Force is  $F = 697 *10^{-15}N$  and the direction is South  

Explanation:

From the question we are told that

         The magnetic field point due west and since East point toward the positive x -axis$(i)$  then this magnetic field would be mathematically represented as

                $\= B = 42(-i)mT = 42*10^{-3} (-i) T$

Now from the question we are told that the kinetic energy is

             $KE = 9*10^{-12}J$

Now this kinetic energy can be mathematically represented as

                  $KE = \frac{1}{2}mv^2$

Where m is the mass of proton which has a general value of

           $m = 1.67*10^{-27}kg$

Now making the subject of the formula

                $v = \sqrt{\frac{KE}{0.5 * m} }$

Substituting values we have

               $v = \sqrt{\frac{9*10^{-12}}{0.5 * 1.67*10^{-27}} }$

                 $= 10.37*10^7m/s$

Now from the question we are told that proton is moving upward which is in the positive z direction so the velocity of the proton would be in the positive

So the velocity would be

            $\= v = 10.37*10^{7} \r k \ m/s$

Now the magnetic Force can be mathematically represented as

          $\= F = q \= v * \= B$

Where q is the charge on the proton which has a general value of  $q =1.6*10^{-19}C$

Now substituting the value

          $\= F = 1.6*10^{-19 } * (10.37 *10^7) \r k * (42 *10^{-3})(-i)$

              $= 697*10^{-15} J$

Now according to Fleming's left hand rule the direction of the magnetic force is south toward the negative Y - direction $(-j)$

So the force can be denoted as

                 $\= F = 697*10^{-15}(-j) N$