Problem 76-3    Source:  Problem 13 - page 253 - HALIDAY & RESNICK -    Book: Fundamentos da Física - 8ª edição - Ed. LTC - 2009.

A Figure 76-03.1 shows a proton moving with velocity v = -200 j (m/s) in direction to a long straight wire that carries an electrical current with an intensity I = 350 mA. At the time shown, the distance between the proton and the wire is d = 2.89 cm. In terms of unit vectors, what is the magnetic force to which the proton is subjected?

Solution of the Problem 76-3   

Scalar Analysis   

Initially, let's look at how to solve the problem without using vector analysis (which will be done below). Notice that the current flowing through the wire produces a magnetic induction field at the point where the proton is. By the right hand rule, this field is coming off the page, forming an angle of 90° with the direction of the proton's velocity. Let's calculate the value of the field where the proton is located. For this calculation we will use the eq. 74-17 studied in item 2.4 of chapter 74. Below, we see the equation.

Substituting the numerical values, remembering that I = 0.35 A e d = 0.0289 m, we have

Performing the calculation we found

As we know the speed of the proton, the magnetic induction field where the proton is and the charge on the proton, which is exactly equal to the charge on the electron, ie q = 1.6 x 10^-19C, it is possible to calculate the magnetic force acting on the proton using the eq. 74-02 studied in item 2 of chapter 74 and repeated below.

Substituting numerical values and remembering that θ = 90°, we have sen 90° = 1. So

Performing the calculation we found

And to determine the direction and sense of the magnetic force on the proton, just use the right hand rule. By doing this, we conclude that the force is in the direction of the x axis and in the negative direction, that is, to the left. In other words, looking from the proton's point of view, it will deviate to the right.

Vector Analysis   

In a vector analysis, we must first determine the magnitude of the magnetic induction field vector. As for the direction and sense, we can easily see that the field points outside the page, that is, in the positive direction of the z axis, which will be represented by the unit vector k^→. So, as the module was already calculated above, so:

As we know the speed of the proton, the magnetic induction field where the proton is and the charge of the proton, which is exactly equal to the charge of the electron, ie q = 1.6 x 10^-19 C, it is possible to calculate the magnetic force acting on the proton using the eq. 74-01 studied in item 2 of chapter 74 and repeated below.

So, substituting the numerical values, we have:

And from algebra, we know that the cross product   j^→ x  k^→  =  -  i^→. Therefore, the end result is:

Notice that the vector sign i^→ means that the magnetic force has direction and sense to the negative side of the axis x.