A uniform 5.9 T magnetic field points north.
If an electron moves vertically downward (toward the ground) with a speed of 0.85 × 107
m/s through this field.
The charge on a proton is 1.60×10−19
.
a) What is the magnitude of the force acting
on it?
Answer in units of N
F=qvBsinα
sinα = sin180= 0 => F=0
To calculate the magnitude of the force acting on the electron, we can use the formula for the magnetic force on a moving charge:
F = qvB
Where:
F is the force on the electron,
q is the charge on the electron (-1.6 x 10^-19 C),
v is the velocity of the electron (0.85 x 10^7 m/s), and
B is the magnetic field strength (5.9 T).
Plugging in the given values:
F = (-1.6 x 10^-19 C) × (0.85 x 10^7 m/s) × (5.9 T)
F = -12.88 x 10^-12 N
The magnitude of the force acting on the electron is 12.88 x 10^-12 N.
To find the magnitude of the force acting on the electron, we can use the equation for the magnetic force on a charged particle:
F = q * v * B * sin(theta)
Where:
- F is the magnitude of the force
- q is the charge of the particle
- v is the velocity of the particle
- B is the magnetic field strength
- theta is the angle between the velocity and the magnetic field
In this case, the charge of the electron is -1.6 * 10^-19 C (since it's negative), the velocity is 0.85 * 10^7 m/s, and the magnetic field strength is 5.9 T.
Since the electron is moving vertically downward and the magnetic field is pointing north, the angle theta between the velocity and the magnetic field is 90 degrees. Therefore, sin(theta) is equal to 1.
Plugging in these values into the formula, we get:
F = (-1.6 * 10^-19 C) * (0.85 * 10^7 m/s) * (5.9 T) * 1
Simplifying the expression, we find:
F ≈ -8.068 * 10^-13 N
The negative sign indicates that the force is in the opposite direction of the velocity. However, since we are asked to find the magnitude of the force, we take the absolute value, giving us:
|F| ≈ 8.068 * 10^-13 N
Therefore, the magnitude of the force acting on the electron is approximately 8.068 * 10^-13 N.