One component of a magnetic field has a magnitude of 0.080 T and points along the +x axis, while the other component has a magnitude of 0.034 T and points along the -y axis. A particle carrying a charge of +1.00 10-5 C is moving along the +z axis at a speed of 3.60 103 m/s.
(a) What is the magnitude of the net magnetic force that acts on the particle?
N
(b) What is the angle that the net force makes with respect to the +x axis?
°
To find the magnitude of the net magnetic force that acts on the particle, we can use the formula:
F = q * v * B * sin(theta)
Where:
F is the magnetic force
q is the charge of the particle
v is the velocity of the particle
B is the magnetic field
theta is the angle between the velocity vector and the magnetic field vector.
(a) Substituting the given values into the formula:
q = +1.00 * 10^(-5) C (charge)
v = 3.60 * 10^3 m/s (velocity)
B = √(Bx² + By²) (magnitude of the magnetic field)
Given:
Bx = 0.080 T (component of the magnetic field along the +x axis)
By = -0.034 T (component of the magnetic field along the -y axis)
B = √(0.080² + (-0.034)²) = √(0.0064 + 0.001156) = √0.007556 = 0.0869 T
Now, let's calculate the angle theta:
cos(theta) = (By)/(B)
cos(theta) = (-0.034)/(0.0869) = -0.3916
theta = arccos(-0.3916)
theta ≈ 114.8°
Now, substituting all the values into the formula:
F = q * v * B * sin(theta)
F = (+1.00 * 10^(-5)) * (3.60 * 10^3) * (0.0869) * sin(114.8°)
F ≈ 0.00145 N
Therefore, the magnitude of the net magnetic force that acts on the particle is approximately 0.00145 N.
(b) The angle that the net force makes with respect to the +x axis is equal to the complement of theta. Therefore:
angle = 90° - theta
angle = 90° - 114.8°
angle ≈ -24.8°
The net force makes an angle of approximately -24.8° with respect to the +x axis.
To determine the net magnetic force acting on the particle, we need to consider the individual magnetic forces due to each component of the magnetic field.
(a) The magnetic force on a charged particle moving through a magnetic field is determined by the equation:
F = q * v * B * sin(theta)
Where:
F is the magnetic force,
q is the charge of the particle,
v is the velocity of the particle,
B is the magnetic field, and
theta is the angle between the velocity vector and the magnetic field vector.
First, let's find the total magnetic force along the x-axis. Since the particle is moving along the z-axis, the angle between the velocity vector and the x-axis is 90 degrees. Therefore, the magnetic force due to the x-component of the magnetic field is:
F_x = q * v * B_x * sin(90°)
B_x = 0.080 T (given)
Substituting the values:
F_x = (1.00 x 10^(-5) C) * (3.60 x 10^3 m/s) * (0.080 T) * sin(90°)
F_x = 0 N (since sin(90°) = 1)
Next, let's find the total magnetic force along the y-axis. The angle between the velocity vector and the y-axis is also 90 degrees. Therefore, the magnetic force due to the y-component of the magnetic field is:
F_y = q * v * B_y * sin(90°)
B_y = -0.034 T (given, as it points along the negative y-axis)
Substituting the values:
F_y = (1.00 x 10^(-5) C) * (3.60 x 10^3 m/s) * (-0.034 T) * sin(90°)
F_y = -1.22 x 10^(-7) N (negative sign indicates it points in the opposite direction)
The net magnetic force is the vector sum of the forces in the x and y directions:
F_net = sqrt(F_x^2 + F_y^2)
Substituting the values:
F_net = sqrt((0 N)^2 + (-1.22 x 10^(-7) N)^2)
F_net = 1.22 x 10^(-7) N
Therefore, the magnitude of the net magnetic force that acts on the particle is 1.22 x 10^(-7) N.
(b) Since the net magnetic force is calculated based on the x and y components, we need to find the angle it makes with the x-axis. We can use trigonometry to calculate this angle:
tan(theta) = F_y / F_x
Substituting the values:
tan(theta) = (-1.22 x 10^(-7) N) / (0 N)
Since F_x is 0, the angle theta is either 90 degrees or 270 degrees.
Therefore, the angle that the net force makes with respect to the +x axis is either 90 degrees or 270 degrees.