A charged particle with mass m, charge |q|=6.4 ×10−19C and speed v=3 ×106 m/s enters from the bottom of the figure and traces out the trajectory shown in the fields shown. The only electric field E=5 ×103 V/m is in the region where the trajectory of the charge is a straight line.
(a) When the particle is moving through the first (straight-line) segment of its trajectory, what is the magnitude of the magnetic field B in Tesla?
(b) The charge hits the left wall of the spectrometer at a vertical distance h=0.114 m above where it entered the upper region and a horizontal distance L=0.272 m to the left of where it entered the upper region (see sketch). What is the radius r of the trajectory in m?
(c) The mass of the particle can be determined using the radius r, the charge q, the speed v, and the magnetic field B0. Using a value of B0=0.6 T, evaluate the mass of the particle in kg. (Note that the magnitude of the field in the curved section, B0, is NOT the same as the magnitude in the straight section, B, found in part a).
To solve this problem, we need to use the Lorentz force equation, which describes the motion of a charged particle in both electric and magnetic fields. The Lorentz force equation is given by:
F = q(E + v x B)
where F is the total force on the particle, q is the charge of the particle, E is the electric field, v is the velocity of the particle, and B is the magnetic field.
(a) Let's focus on the first straight-line segment of the trajectory. Since the particle is moving in a straight line and the Lorentz force only depends on the perpendicular component of the magnetic field, we can determine the magnitude of the magnetic field B using the equation:
F = qvB
From the given information, we have:
q = 6.4 × 10^(-19) C (charge of the particle)
v = 3 × 10^6 m/s (speed of the particle)
Substituting these values into the equation, we can solve for B as follows:
F = qvB
B = F / (qv)
The only force acting on the particle in the straight-line segment is the magnetic force. Thus, we can set F equal to the net magnetic force acting on the particle:
F = qvB
Since we are given the electric field E = 5 × 10^3 V/m, and the Lorentz force only depends on the perpendicular component of the magnetic field, we can rewrite the equation as follows:
F = qvB = qvE
Now we can substitute the known values and calculate B:
B = F / (qv) = (qvE) / (qv) = E
Therefore, the magnitude of the magnetic field B in Tesla is B = 5 × 10^3 T.
(b) To find the radius r of the trajectory, we need to use the equation of motion for a charged particle in a magnetic field:
F = qvB = mv^2 / r
Where r is the radius of the trajectory. Solving for r, we get:
r = mv / (qB)
We are given the speed v = 3 × 10^6 m/s and the magnetic field B = 5 × 10^3 T. We need to find the mass of the particle, m, to calculate the radius. But we can't directly calculate the mass until we have the radius. So, we need to solve for the radius iteratively.
1. Let's assume a value for m (let's say m = 1 kg).
2. Calculate the radius using the formula r = mv / (qB).
3. Check if the calculated radius matches the given horizontal distance L = 0.272 m. If it does, then our assumption for mass is correct. If not, update the value of mass and repeat steps 2 and 3 until the radius matches the given horizontal distance L.
(c) Once we have the correct value of the radius r, we can determine the mass of the particle using the equation:
r = mv / (qB0)
We are given the radius r, the charge q = 6.4 × 10^(-19) C, the speed v = 3 × 10^6 m/s, and the magnetic field B0 = 0.6 T. We can rearrange the equation to solve for the mass:
m = rqB0 / v
Substituting the known values, we can calculate the mass of the particle in kg.