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.  

grav force prop 1/r squared 2r all squared is 4rsquared 1/4rsquared quartered

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards