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 can use the formula for the magnetic force acting on a moving charged particle.
The magnetic force experienced by a charged particle moving through a magnetic field is given by:
F = qvB sin(θ)
where F is the force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector.
In this case, the condition for the electron's direction of travel to remain unchanged is that the magnetic force acting on the electron is perpendicular to its velocity vector. Therefore, the angle θ between the velocity vector and the magnetic field vector should be 90 degrees.
Since the direction of travel remains unchanged, we can conclude that the magnetic force acting on the electron is equal to zero. This means that the equation qvB sin(θ) = 0 holds true.
Since the charge of an electron (q) and its velocity (v) are constants, the equation can only be satisfied if either B = 0, which implies no magnetic field is required, or sin(θ) = 0, which implies that the angle θ between the velocity and the magnetic field vectors is 0 degrees (parallel).
Therefore, to keep the electron's direction of travel unchanged, the strength of the magnetic field (B) should be zero or the magnetic field should be parallel to the electron's velocity.
In summary, either no magnetic field is required or a magnetic field parallel to the electron's velocity is needed for the electron's direction of travel to remain unchanged.