In an experiment on the shuttle, an electron with a speed of 6.3 x 105 m/s [N] and a mass of 9.11 x 10-31 kg is shot through an external magnetic field. Determine the strength and orientation of the magnetic field required so that the electron's direction of travel remains unchanged. Neglect any effects due to the Earth's magnetic field.
Tutors are volunteers and are not here 24/7. I'm sure when a physics tutor comes on line, he or she will respond. No tutor will do your homework for you, however; we check student's work and can guide you to resources to help you understand what is required of you. You show no evidence here of having attempted to solve this problem, nor are you asking questions about it. How do you expect a tutor to help you? Be specific.
To determine the strength and orientation of the magnetic field required for the electron's direction of travel to remain unchanged, we can use the equation for the Lorentz force:
F = q*v*B
where F is the magnetic force, q is the charge of the electron, v is its velocity, and B is the magnetic field strength.
Since we want the electron's direction of travel to remain unchanged, the magnetic force acting on it should be perpendicular to its velocity. This implies that the magnetic field should also be perpendicular to the electron's velocity.
The Lorentz force can be rewritten as:
F = m*a = m*v^2 / R
where m is the mass of the electron, a is the acceleration, and R is the radius of curvature in a magnetic field.
Since the direction of travel remains unchanged, the acceleration of the electron must be zero. Therefore, the magnetic force must be balanced by a centripetal force.
Setting the magnetic force equal to the centripetal force, we have:
qvB = m*v^2 / R
Where q is the charge of the electron:
q = -e
(e is the elementary charge, -1.6 x 10^-19 C)
Now we can solve for the magnetic field strength B:
B = (m*v) / (-e*R)
Substituting the given values:
m = 9.11 x 10^-31 kg
v = 6.3 x 10^5 m/s
e = -1.6 x 10^-19 C
We also need to find the radius of curvature R. The centripetal force acting on the electron in a magnetic field is provided by the magnetic force:
qvB = m*v^2 / R
Rearranging the equation, we can solve for R:
R = (m*v) / (q*B)
Substituting the given values:
m = 9.11 x 10^-31 kg
v = 6.3 x 10^5 m/s
q = -1.6 x 10^-19 C
Now we can calculate the required magnetic field strength B:
B = (m*v) / (-e*R)
Substituting the calculated value of R into the equation for B:
B = (9.11 x 10^-31 kg * 6.3 x 10^5 m/s) / (-1.6 x 10^-19 C * (9.11 x 10^-31 kg * 6.3 x 10^5 m/s) / (q*(-e)))
After simplifying:
B ≈ 2.27 x 10^-2 T
Therefore, the required strength of the magnetic field to keep the electron's direction of travel unchanged is approximately 2.27 x 10^-2 Tesla. The orientation of the magnetic field should be perpendicular to the direction of the electron's velocity.
To determine the strength and orientation of the magnetic field required for the electron's direction of travel to remain unchanged, we need to use the right-hand rule for charged particles moving in a magnetic field.
Here are the steps to find the values:
1. Identify the given information:
- Velocity of the electron (v) = 6.3 x 10^5 m/s (north direction)
- Mass of the electron (m) = 9.11 x 10^-31 kg
2. Recall the formula for the force experienced by a charged particle moving in a magnetic field:
F = qvBsinθ
Where:
- F is the force on the electron perpendicular to its velocity
- q is the charge of the electron (which is -1.6 x 10^-19 C)
- v is the velocity of the electron
- B is the magnetic field strength
- θ is the angle between the velocity vector and the magnetic field vector
3. Since we want the electron's direction of travel to remain unchanged, the force acting on it must be zero. This means that the angle θ between the velocity vector and the magnetic field vector should be 90 degrees (or π/2 radians).
4. Set up the equation with all the known values and solve for B:
F = qvBsin(θ)
Since sin(π/2) = 1, the equation simplifies to:
0 = qvB
Solving for B:
B = 0 / (qv)
B = 0
This means that the magnetic field strength required for the electron's direction of travel to remain unchanged is zero.
5. Regarding the orientation of the magnetic field, since the magnetic field strength is zero, the direction of the magnetic field is irrelevant in this case.
Therefore, the strength of the magnetic field required is zero, and its orientation can be in any direction.