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.
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To determine the strength and orientation of the magnetic field required for the electron's direction of travel to remain unchanged, we need to apply the concept of the magnetic force acting on a charged particle moving through a magnetic field.
The magnetic force on a moving charged particle can be calculated using the equation:
F = q * v * B * sin(θ)
Where:
F is the magnetic force
q is the charge of the particle (in this case, the charge of an electron is -1.6 x 10^-19 C)
v is the velocity of the particle (given as 6.3 x 10^5 m/s)
B is the magnitude of the magnetic field
θ is the angle between the velocity vector of the electron and the magnetic field vector.
Since we want the electron's direction of travel to remain unchanged, the magnetic force acting on the electron must be perpendicular to its velocity.
Therefore, sin(θ) = 1, which means that θ = 90 degrees.
Substituting the given values, we have:
F = (-1.6 x 10^-19 C) * (6.3 x 10^5 m/s) * B * sin(90°)
Simplifying the equation further, we get:
F = (-1.6 x 10^-19 C) * (6.3 x 10^5 m/s) * B
To keep the electron's direction of travel unchanged, the force on it should be zero. So we have:
F = 0
Solving the equation for B:
(-1.6 x 10^-19 C) * (6.3 x 10^5 m/s) * B = 0
B = 0
Therefore, the magnetic field strength required for the electron's direction of travel to remain unchanged is zero. However, it is important to note that in realistic scenarios, a nonzero magnetic field may cause a deflection in the electron's trajectory, but with a specific orientation, the electron's direction of travel could still remain unchanged.