A charge of 1.20 X 10-8 C is moving at 25.0 m/sec across a magnetic field of 0.060 T. What is the magnetic force acting on the charge?
a. 0.000180 N
b. 1.8 X 10-8 N
c. 180 N
d. 0
e. 0.018 N
a
To calculate the magnetic force acting on a charge moving across a magnetic field, you can use the formula:
F = q * v * B * sin(theta)
Where:
F is the magnetic force,
q is the charge (in coulombs),
v is the velocity of the charge (in meters per second),
B is the magnetic field strength (in teslas),
theta is the angle between the velocity vector and the magnetic field vector.
In this case, the charge is 1.20 x 10^-8 C, the velocity is 25.0 m/s, and the magnetic field is 0.060 T. Since the question does not provide an angle theta, we assume that the charge moves perpendicular to the magnetic field (sin(theta) = 1).
Substituting the values into the formula:
F = (1.20 x 10^-8 C) * (25.0 m/s) * (0.060 T) * sin(90 degrees)
Since sin(90 degrees) = 1, the equation simplifies to:
F = (1.20 x 10^-8 C) * (25.0 m/s) * (0.060 T) * 1
F = 0.018 N
Therefore, the magnetic force acting on the charge is 0.018 N.
Hence, the correct answer is e. 0.018 N.
To find the magnetic force acting on a charge moving across a magnetic field, you can use the formula:
F = q * v * B * sin(θ)
Where:
F is the magnetic force
q is the charge
v is the velocity of the charge
B is the magnetic field strength
θ is the angle between the velocity and the magnetic field
In this case, the charge is 1.20 x 10^-8 C, the velocity is 25.0 m/s, and the magnetic field is 0.060 T. However, the angle (θ) is not given. We can assume that the angle is 90 degrees since the charge is moving "across" the magnetic field.
Plugging these values into the formula, we get:
F = (1.20 x 10^-8 C) * (25.0 m/s) * (0.060 T) * sin(90°)
F = (1.20 x 10^-8 C) * (25.0 m/s) * (0.060 T) * 1
F = 1.8 x 10^-8 N
So, the magnetic force acting on the charge is 1.8 x 10^-8 N.
The correct answer is b. 1.8 x 10^-8 N.