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

The magnitude of the magnetic force on the particle is approximately 2.4 × 10^-3 N, and the direction of the magnetic force is in the negative y direction.

Step-by-step explanation:

The magnitude of the magnetic force on a charged particle can be calculated using the equation F = qvBsin(θ), where F is the force, q is the charge, v is the velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field.

In this case, the charge of the particle is 35 micro-Coulomb (35 × 10-6 C), the velocity is 70 m/s, and the magnetic field has components of 0.40 T in the y direction and 0.86 T in the z direction.

To find the magnitude of the force, we can use vector addition to find the total magnetic field. The vector addition of the y and z components of the magnetic field gives a total magnetic field of approximately 0.97 T.

Plugging in the values into the equation, we get:

F = (35 × 10-6 C)(70 m/s)(0.97 T)sin(θ)

F = 2.4 × 10-3 N

The magnitude of the magnetic force on the particle is approximately 2.4 × 10-3 N.

The direction of the magnetic force can be found by considering the right-hand rule. If the particle has a positive charge and is moving in the positive x direction, the force will be perpendicular to both the velocity and the magnetic field. It will point in the negative y direction (opposite to the positive y direction).

The angle θ can be found using trigonometry. It is the angle between the positive z axis and the negative y axis in the yz-plane. Since the force is in the negative y direction, θ is 90 degrees.

Therefore, the direction of the magnetic force on the particle is in the negative y direction.

Answer 2

F = 2.317 10⁻³ N , θ = 65º

Step-by-step 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