A particle, travelling at 5.2x10^3 ms, with a charge of 1.6uC, is moving up the page. It is at right angles to a magnetic field of 6.4 T in the direction of right. What is the magnitude and direction of the force on the particle?
To find the magnitude and direction of the force on the particle, we can use the formula for the magnetic force on a charged particle:
F = q * v * B * sin(theta)
Where:
F is the force on the particle,
q is the charge of the particle,
v is the velocity of the particle,
B is the magnetic field strength,
and theta is the angle between the velocity and the magnetic field.
Given:
q = 1.6 μC (1.6 × 10^-6 C)
v = 5.2 × 10^3 m/s
B = 6.4 T
theta = 90 degrees (since the particle is moving at right angles to the magnetic field)
Now we can substitute the given values into the formula:
F = (1.6 × 10^-6 C) * (5.2 × 10^3 m/s) * (6.4 T) * sin(90)
sin(90) = 1, so we can simplify the equation:
F = (1.6 × 10^-6 C) * (5.2 × 10^3 m/s) * (6.4 T) * 1
Calculating this equation, we get:
F = 53.248 N
So, the magnitude of the force on the particle is 53.248 N.
Now let's consider the direction. Since the particle has a positive charge and the magnetic field is directed to the right, we can determine the direction of the force using the right-hand rule:
If you point your thumb in the direction of the velocity (up the page) and your fingers in the direction of the magnetic field (to the right), your palm will face in the direction of the force. In this case, your palm would be facing towards you (out of the page).
Therefore, the direction of the force on the particle is out of the page.