An electron is moving parallel to the Earth’s surface at the equator in a direction 25° south of east. Its velocity is 7.3 ´ 10^4 m/s and a magnetic force of 1.8 ´ 10^-18 N is exerted on the electron. If the magnetic field points south at this location, what is the direction of the magnetic force on the electron and the magnitude of the magnetic field?
q [ 0 i + 0 j - 6.62 b k ] 10^4 logical, only x component of V is 90 deg from B
TYPO, left b out
To determine the direction of the magnetic force on the electron and the magnitude of the magnetic field, we can use the equation for magnetic force:
F = q * v * B * sin(θ)
where:
F is the magnetic force,
q is the charge of the electron (1.6 × 10^-19 C),
v is the velocity of the electron (7.3 × 10^4 m/s),
B is the magnitude of the magnetic field, and
θ is the angle between the velocity vector (v) and the magnetic field vector (B).
We are given the following information:
- The magnetic force exerted on the electron is 1.8 × 10^-18 N.
- The velocity of the electron is 7.3 × 10^4 m/s.
- The magnetic field points south.
Let's solve for the magnitude of the magnetic field first:
1.8 × 10^-18 = (1.6 × 10^-19 C) * (7.3 × 10^4 m/s) * B * sin(θ)
We know that sin(θ) = sin(25° south of east) = sin(180° - 25°) = sin(155°) = -0.4226 (negative because the angle is south).
Substituting the values:
1.8 × 10^-18 = (1.6 × 10^-19 C) * (7.3 × 10^4 m/s) * B * (-0.4226)
Simplifying the equation:
B = (1.8 × 10^-18) / ((1.6 × 10^-19 C) * (7.3 × 10^4 m/s) * (-0.4226))
Calculating the value of B:
B ≈ 0.092 T (Tesla)
Therefore, the magnitude of the magnetic field at this location is approximately 0.092 T.
To find the direction of the magnetic force, we can use the right-hand rule:
1. Point your thumb in the direction of the velocity vector (25° south of east).
2. Bend your fingers in the direction of the magnetic field (south).
3. Your palm will face in the direction of the magnetic force.
Using this rule, it can be determined that the magnetic force on the electron is directed downward, or towards the Earth's surface.
Lemme go look in my drawer. I have a Maxwell equation T shirt somewhere.
ah, yes, F = q * V cross B
q = - 1.6*10^-19 Coulombs
Vx = 7.3*10^4 (cos 25)
Vy = -7.3*10^4(sin 25)
Bx = 0
By = -b
so
F =q times * 10^4 times determinant of
i j k
6.62 -3.09 0
0 - b 0
so cross product is
q [ 0 i + 0 j - 6.62 k ] 10^4 logical, only x component of V is 90 deg from B
I think you can take it from there :)