A Bainbridge mass spectrometer is shown in the figure. A charged particle with mass m, charge |q|=3.2 ×10−19C and speed v=3 ×10@6 m/s enters from the bottom of the figure and traces out the trajectory shown in the fields shown. The only electric field E=11 ×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=0.150 m above where it entered the upper region and a horizontal distance L=0.357 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=0.2 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).
Help please with formula!
To find the answers to these questions, we will need to use a combination of the Lorentz force equation and the equations governing the motion of a charged particle in a magnetic field.
(a) To find the magnitude of the magnetic field B in the first segment of the trajectory, we can use the Lorentz force equation:
F = q * (v x B)
Since the particle is moving in a straight line, the magnetic force must be balanced by the electric force:
F = q * E
Setting these two equations equal to each other, we have:
q * (v x B) = q * E
The magnitude of the cross product of v and B is given by:
|v x B| = v * B * sin(theta)
where theta is the angle between v and B. Since the trajectory is a straight line, theta is 90 degrees, so we can simplify the equation to:
|v x B| = v * B
Now we can solve for B:
v * B = E
B = E / v
Substituting the given values, we have:
B = (11 × 10^3 V/m) / (3 × 10^6 m/s)
B = 3.67 × 10^-3 T
Therefore, the magnitude of the magnetic field B in Tesla in the first segment of the trajectory is 3.67 × 10^-3 T.
(b) To find the radius r of the trajectory, we can use the equation for the centripetal force:
F = m * (v^2 / r)
In this case, the centripetal force is provided by the magnetic force:
F = q * v * B
Setting these two equations equal to each other and solving for r, we have:
m * (v^2 / r) = q * v * B
Simplifying, we find:
r = (m * v) / (q * B)
Substituting the given values, we have:
r = (m * (3 × 10^6 m/s)) / ((3.2 × 10^-19 C) * (3.67 × 10^-3 T))
r = 2.67 × 10^3 m
Therefore, the radius r of the trajectory in meters is 2.67 × 10^3 m.
(c) To determine the mass of the particle, we can rearrange the equation for the radius r:
m = (r * q * B) / v
Substituting the given values, we have:
m = ((2.67 × 10^3 m) * (3.2 × 10^-19 C) * (0.2 T)) / (3 × 10^6 m/s)
m = 1.07 × 10^-22 kg
Therefore, the mass of the particle is approximately 1.07 × 10^-22 kg.