An electron travels at a uniform speed of 2.50 × 106 m/s toward the top of this page. It then enters a uniform magnetic field and experiences a maximum force of 5.15 × 10-8 N that points toward the right of the page.
What is the magnitude of the magnetic field?
What is the direction of the magnetic field?
F = Bvq
v=2.5e6
F=5.15e-8
q=1.6e-19
Solve for B
To find the magnitude of the magnetic field, we can use the formula for the force experienced by a charged particle moving through a magnetic field:
F = q v B sin(theta)
Where:
F is the force,
q is the charge of the particle (in this case, the charge of an electron is -1.6 x 10^-19 C),
v is the velocity of the particle (2.5 x 10^6 m/s),
B is the magnitude of the magnetic field, and
theta is the angle between the direction of the velocity and the magnetic field (90 degrees in this case, because the force is perpendicular to the velocity).
Plugging in the given values:
5.15 x 10^-8 N = (-1.6 x 10^-19 C) * (2.5 x 10^6 m/s) * B * sin(90°)
Simplifying:
B = (5.15 x 10^-8 N) / ((-1.6 x 10^-19 C) * (2.5 x 10^6 m/s) * sin(90°))
B ≈ 2.03 x 10^-3 T (tesla)
So, the magnitude of the magnetic field is approximately 2.03 x 10^-3 T.
To determine the direction of the magnetic field, we need to apply the right-hand rule. If the force on a positive charge is in the right direction, then the magnetic field is directed into the page. If the force is in the left direction, then the magnetic field is directed out of the page.
In this case, since the force on the electron is directed to the right, the magnetic field must be directed into the page.