A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50 km/s in the +x direction experiences a force of 2.25𝑥10−16𝑁 in the +y direction and an electron moving at 4.75 km/s in the –z direction experiences a force of 8.50𝑥10−16𝑁 in the +y-direction. (a) What are the magnitude and direction of the magnetic field? (b) What are the magnitude and direction of the magnetic force on an electron moving in the –y direction at 3.20 km/s?
(a) To find the magnitude of the magnetic field, we can use the equation:
F = qvB
where F is the magnetic force, q is the charge, v is the velocity, and B is the magnetic field.
For the proton:
F = 2.25x10^-16 N
q = proton charge = 1.6x10^-19 C
v = 1.50 km/s = 1.5x10^3 m/s
Plugging in these values, we get:
2.25x10^-16 N = (1.6x10^-19 C)(1.5x10^3 m/s)B
Simplifying, we find:
B = (2.25x10^-16 N) / ((1.6x10^-19 C)(1.5x10^3 m/s))
Now, let's calculate that:
B = 1.875 T
The magnitude of the magnetic field is 1.875 Tesla.
To determine the direction of the magnetic field, we can use the right-hand rule. Placing the fingers of your right hand in the direction of the proton's velocity (+x direction) and curling them towards the force (+y direction), your thumb points in the direction of the magnetic field. In this case, the magnetic field would be in the +z direction.
(b) Now, let's find the magnetic force on an electron moving in the –y direction at 3.20 km/s.
For the electron:
v = 3.20 km/s = 3.2x10^3 m/s
Using the same equation as before, F = qvB, we can rearrange it to solve for B:
B = F / (qv)
For the electron:
F = 8.50x10^-16 N
q = electron charge = -1.6x10^-19 C
v = 3.2x10^3 m/s
Plugging in these values, we get:
B = (8.50x10^-16 N) / ((-1.6x10^-19 C)(3.2x10^3 m/s))
Simplifying, we find:
B = -1.328 T
The magnitude of the magnetic field is 1.328 Tesla.
To determine the direction of the magnetic force, we can use the right-hand rule again. Placing the fingers of your right hand in the direction of the electron's velocity (-y direction) and curling them towards the force (+y direction), your thumb points in the direction of the magnetic field. In this case, the magnetic field would be in the -x direction.
To find the magnitude and direction of the magnetic field, we need to use the formula for the magnetic force on a charged particle moving in a magnetic field:
F = qvBsinθ
Let's break down the information given step-by-step and solve for the magnitude and direction of the magnetic field:
Step 1: Given information:
- Proton: v = 1.50 km/s in the +x direction, F = 2.25 × 10^(-16) N in the +y direction
- Electron: v = 4.75 km/s in the -z direction, F = 8.50 × 10^(-16) N in the +y direction
Step 2: Solve for the magnitude of the magnetic field for the proton:
Plug in the given values into the formula and solve for B:
2.25 × 10^(-16) N = (1.6 × 10^(-19) C)(1.50 × 10^3 m/s) B sinθ
B sinθ = (2.25 × 10^(-16) N) / ((1.6 × 10^(-19) C)(1.50 × 10^3 m/s))
B sinθ = 0.7031 T
Step 3: Solve for the magnitude of the magnetic field for the electron:
Plug in the given values into the formula and solve for B:
8.50 × 10^(-16) N = (-1.6 × 10^(-19) C)(4.75 × 10^3 m/s) B sinθ
B sinθ = (8.50 × 10^(-16) N) / ((1.6 × 10^(-19) C)(4.75 × 10^3 m/s))
B sinθ = - 2.1 T
Step 4: Solve for the direction of the magnetic field:
Since both the proton and the electron experience a force in the +y direction, the directions of the magnetic field must be opposite.
Step 5: Calculate the magnitude of the magnetic field:
To calculate the magnitude of the magnetic field, we can take the average of the magnitudes obtained from the proton and the electron:
|B| = (|B_proton| + |B_electron|) / 2
|B| = (0.7031 T + 2.1 T) / 2
|B| = 1.401 T
Step 6: Determine the direction of the magnetic field:
Since the proton is moving in the +x direction and experiences a force in the +y direction, the direction of the magnetic field can be determined using the right-hand rule. The thumb points in the direction of the force, the fingers point in the direction of the moving proton, and the magnetic field vector points in the direction perpendicular to both. Thus, the direction of the magnetic field is in the +z direction.
Step 7: Calculate the magnetic force on an electron moving in the -y direction at 3.20 km/s:
Using the formula F = qvBsinθ, we can calculate the magnetic force on the electron:
F = (-1.6 × 10^(-19) C)(3.20 × 10^3 m/s)(1.401 T)(sin90°)
F = 4.5152 × 10^(-14) N
The magnitude of the magnetic force on the electron moving in the -y direction at 3.20 km/s is 4.5152 × 10^(-14) N.
Since the electron is moving in the -y direction, the direction of the magnetic force is in the +x direction, as determined by the right-hand rule.
To find the magnitude and direction of the magnetic field, we can use the formula for the magnetic force experienced by a charged particle:
F = qvBsinθ
where F is the force, q is the charge of the particle, v is the velocity of the particle, B is the magnitude of the magnetic field, and θ is the angle between the velocity and the magnetic field.
(a) Let's start by finding the magnitude of the magnetic field for the proton. From the given information, we have:
F_p = 2.25 × 10^(-16) N
v_p = 1.50 km/s = 1.50 × 10^3 m/s
q_p = charge of a proton = 1.6 × 10^(-19) C
The angle θ for the proton can be determined from the fact that the force is directed in the +y direction. Since the force experienced by a positive charge is perpendicular to both its velocity and the magnetic field direction, the angle between the velocity and the magnetic field is 90 degrees. So we can write:
F_p = q_p * v_p * B * sin(90°)
Simplifying this equation:
B = F_p / (q_p * v_p)
Substituting the values:
B = (2.25 × 10^(-16) N) / ((1.6 × 10^(-19) C) * (1.50 × 10^3 m/s))
Calculating this expression will give you the magnitude of the magnetic field for the proton.
To determine the direction of the magnetic field, we can use the right-hand rule. If the force on a positive charge is in the +y direction, this means the magnetic field is directed towards the +z direction (outward from the plane). So the direction of the magnetic field for the proton is +z.
Similarly, we can find the magnitude and direction of the magnetic field for the electron.
F_e = 8.50 × 10^(-16) N
v_e = 4.75 km/s = 4.75 × 10^3 m/s
q_e = charge of an electron = -1.6 × 10^(-19) C
Since the force on the electron is also in the +y direction, we can use the same equation:
F_e = q_e * v_e * B * sin(90°)
Solving for B:
B = F_e / (q_e * v_e)
Substituting the values:
B = (8.50 × 10^(-16) N) / ((-1.6 × 10^(-19) C) * (4.75 × 10^3 m/s))
Calculating this expression will give you the magnitude of the magnetic field for the electron.
Again, using the right-hand rule, since the force on the electron is in the +y direction, the magnetic field direction for the electron is opposite to that of the proton i.e., -z direction.
(b) To find the magnitude and direction of the magnetic force on an electron moving in the -y direction at 3.20 km/s, we can use the same formula for the magnetic force:
F = qvBsinθ
Given information:
v_e' = -3.20 km/s = -3.20 × 10^3 m/s
q_e = charge of an electron = -1.6 × 10^(-19) C
B = magnitude of the magnetic field for the electron
The angle θ can be determined from the fact that the force is directed in the +y direction. Since the force experienced by a negative charge (-e) is opposite in direction to the force experienced by a positive charge (+e), the angle between the velocity and the magnetic field is 180 degrees. So we can write:
F_e' = q_e * v_e' * B * sin(180°)
Simplifying this equation:
F_e' = -q_e * v_e' * B
Solving for B:
B = -F_e' / (q_e * v_e')
Substituting the given values:
B = - (8.50 × 10^(-16) N) / ((-1.6 × 10^(-19) C) * (-3.20 × 10^3 m/s))
Calculating this expression will give you the magnitude of the magnetic field for the -y moving electron.
Using the right-hand rule, since the force on the electron is in the +y direction, the magnetic field direction for the -y moving electron is opposite to that of the previous cases i.e., +z direction.