The area shown is the boundary of a magnetic field directed in the positive z direction. An electron with a velocity along the
x-axis enters the field at coordinates (x, y) =
(−2 m, 0) and exits 0.35 µs later at point A
whose coordinates are (x, y) = (0, 2 m).
What is the magnitude of B~
? The mass
of the electron is 9.109 × 10
−31
kg and the
elemental charge is −1.602 × 10
−19
C.
Answer in units of µT
To find the magnitude of the magnetic field, we can use the formula for the magnetic force on a charged particle.
The equation for the magnetic force on a charged particle is given by:
F = q * (v x B)
where F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field.
Since we are given the velocity of the electron along the x-axis, we can assume y-component of velocity is 0.
Using the coordinates of the entry and exit points of the electron, we can calculate the displacement along the x-axis:
Δx = (0 m) - (-2 m) = 2 m
The time interval is given as 0.35 μs, which needs to be converted to seconds:
Δt = 0.35 μs = 0.35 * 10^(-6) s
Using the formula for displacement and time:
v = Δx/Δt
v = 2 m / 0.35 * 10^(-6) s
v ≈ 5.714 * 10^6 m/s
Now let's find the magnetic field (B) using the formula for magnetic force:
F = q * (v x B)
Since the electron experiences a force perpendicular to its velocity (along the y-axis), we can rearrange the formula as:
B = F / (q * v)
The force experienced by the electron can be found using Newton's second law:
F = mass * acceleration
The acceleration can be calculated by dividing the change in velocity by the time interval:
acceleration = (v - 0) / Δt
Substituting the values:
acceleration = (5.714 * 10^6 m/s - 0) / (0.35 * 10^(-6) s)
acceleration ≈ 1.633 * 10^13 m/s^2
The force can now be calculated as:
F = (9.109 * 10^(-31) kg) * (1.633 * 10^13 m/s^2)
F ≈ 1.488 * 10^(-17) N
Now substituting the values back into the equation for B:
B = F / (q * v)
B = (1.488 * 10^(-17) N) / ((-1.602 * 10^(-19) C) * (5.714 * 10^6 m/s))
B ≈ -460.0 T
Taking the magnitude of B, we get:
|B| ≈ 460.0 T = 460.0 µT
Therefore, the magnitude of the magnetic field is approximately 460.0 µT.
To find the magnitude of the magnetic field B~, we can use the equation for the force experienced by a charged particle moving through a magnetic field:
F~ = q (v~ × B~)
Where F~ is the force, q is the charge of the particle, v~ is the velocity of the particle, and B~ is the magnetic field. In this case, the charged particle is an electron with a charge of -1.602 × 10^-19 C.
First, we need to find the velocity of the electron. We know that the electron enters the field at coordinates (x, y) = (-2 m, 0) and exits later at coordinates (x, y) = (0, 2 m). The time it takes for the electron to travel from (-2 m, 0) to (0, 2 m) is given as 0.35 µs.
Using the formula for average velocity, we can calculate the velocity of the electron:
v~ = Δr~/Δt
Where Δr~ is the change in position and Δt is the change in time.
Δr~ = (0 - (-2))m î + (2 - 0)m ĵ
= 2m î + 2m ĵ
Δt = 0.35 µs = 0.35 × 10^-6 s
v~ = (2m î + 2m ĵ) / (0.35 × 10^-6 s)
= 5.71 × 10^6 m/s î + 5.71 × 10^6 m/s ĵ
Now that we have the velocity of the electron, we can rearrange the equation for the magnetic force and solve for the magnetic field B~:
B~ = F~ / (q v~)
Substituting the values:
B~ = (q (v~ × B~)) / (q v~)
= (v~ × B~) / v~
Cross product (v~ × B~) can be calculated as:
v~ × B~ = (v_x B_z - v_z B_x) î + (v_z B_x - v_x B_z) ĵ + (v_x B_y - v_y B_x) k̂
In this case, the electron's velocity is along the x-axis, so the cross product simplifies to:
v~ × B~ = -v_z B_x ĵ
Substituting this into the equation for B~:
B~ = (-v_z B_x ĵ) / v~
Comparing the x-components, we can eliminate B_x from the equation:
B~ = (-v_z ĵ) / v~
Now we need to find the z-component of the electron's velocity, v_z. Since the electron's velocity is only along the x-axis, v_z = 0.
Substituting this into the equation for B~:
B~ = (-0 ĵ) / v~
= 0
Therefore, the magnitude of the magnetic field B~ is 0 µT.
16.2
Electron enters the magnetic field perpendicular to the magnetic field, therefore, it moves with centripetal acceleration along the curvilinear path due to Lorentz force action
m•a=q•v•B •sin α.
Since α=90º, sin α=1; q=e; v=at.=>
m•a =e•v•B=e•a•t•B =>
m= e•t•B.
B=m/e•t=
=9.109•10^-31/1.602•10^-19•0.35•10^-6 =1.62•10^-5 T =16.2 μT