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)

The  magnitude of the Force is    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  then this magnetic field would be mathematically represented as

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

Now this kinetic energy can be mathematically represented as

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

Now making the subject of the formula

Substituting values we have

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

Now the magnetic Force can be mathematically represented as

Where q is the charge on the proton which has a general value of  

Now substituting the value

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

So the force can be denoted as

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s