Question

A particle with a charge of 35 micro-coloumb moves with a speed of 70 m/s in the positive x direction. The magnetic field in this region of space has a component of 0.40 T in the positive y direction, and a component of 0.86 T in the positive z direction.

What is the magnitude of the magnetic force on the particle?

What is the direction of the magnetic force on the particle? (Find the angle measured from the positive z-axis toward the negative y-axis in the yz-plane.) ?

Answer 1

Answer:

 F = 2.317 10⁻³ N ,   θ =  65º

Explanation:

The magnetic force is

          F = q v x B

The bold are vectors.  In the case of having several addresses it is better to work with the determinant

           $F= q \left[\begin{array}{ccc}i&j&k\\70&0&0\\0&0.40&0.86\end{array}\right]$

           F = q (i ^ (0 -0) + j ^ (0-70 0.86) + k ^ (70 0.40 -0))

           F = 35 10⁻⁶ (-60.2 j ^ + 28 k ^)

           F = -2.1 10⁻³ j ^ + 9.8 10⁻⁴ k ^

For the magnitude let's use the Pythagorean theorem

           F = √ ((2.1 10⁻³)² + (9.8 10⁻⁴)²)

           F = 2.317 10⁻³ N

For the direction let's use trigonometry

            tan θ = $F_{y}/F_{z}$

            θ = tan⁻¹ F_{y}/F_{z}

            θ = tan⁻¹ 2.1/0.98

             θ =  65º

Measured from the positive part of the z axis to the negative part of the y axis