An electron that has a velocity with x component 2.4 x 106 m/s and y component 3.6 x 106 m/s moves through a uniform magnetic field with x component 0.027 t and ycomponent -0.15 t. (a) find the magnitude of the magnetic force on the electron. (b) repeat your calculation for a proton having the same velocity.

i know that the equation is fb = qv × b, i'm just having difficulty computing the cross product directly? also, i know that part b is the same answer as part a just in the opposite direction

(a) 7.315 x 10^(-14) N

(b) - 7.315 x 10^(-14) N

Explanation:

As you referred at the final remark, the electron and proton undergo a magnetic force of same magnitude but opposite direction. Using the definition of magnetic force,  a cross product must be done. One technique is either calculate the magnitude of the velocity and magnetic field and multiplying by sin (90°), but it is necessary to assure both vectors are perpendicular between each other ( which is not the case) or do directly the cross product dealing with a determinant (which is the most convenient approach), thus,

(a) The electron has a velocity defined as:  

In respect to the magnetic field;

The magnetic force can be written as;

Bear in mind  

thus,

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e,

(b) Considering the proton charge has the same magnitude as electron does, but the sign is positive, thus

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e,

Final remarks: The cross product was performed in R3 due to the geometrical conditions of the problem.  

in order to follow the instructions and complete the statements, i need to use data from table c of your student guide. can you think of a way to let me see your student guide for a while ? i can't answer the question without the data in table c.