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An electron is moving horizontally to the right with speed 2 x 106 m/s. What is the magnetic field due to this moving electron at the indicated locations in the figure? Each location is d = 5 cm from the electron, and the angle =35º. Give both magnitude and direction of the magnetic field at each location. P6 A Ps--- -----P2 E E E E E E
An electron is moving horizontally to the right with speed 2 x 106 m/s. What is the magnetic field due to this moving electron at the indicated locations in the figure? Each location is d = 5 cm from the electron, and the angle =35º. Give both magnitude and direction of the magnetic field at each location. P6 A Ps--- -----P2 E E E E E E
Final answer:To calculate themagnetic fielddue to a moving electron at different locations, we can use the formula F = qvBsinθ, where F is the magnetic force, q is the charge of theelectron, v is its velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field.Explanation:When acharged particlemoves in a magnetic field, it experiences a magnetic force. The magnitude of this force can be calculated using the formula F = qvBsinθ, where q is the charge of the particle, v is itsvelocity, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector. In this case, the electron is moving horizontally to the right, so the angle θ is 0°. We can calculate the magnitude of themagnetic fieldat each location using this formula.Location P6: For thislocation, the angle θ is 35º. So, the magnitude of the magnetic field is given by B = (2 x 10-6 T)(Sin 35º).Location P2: For this location, the angle θ is 0º. So, the magnitude of the magnetic field is given by B = (2 x 10-6 T)(Sin 0º).Learn more aboutMagnetic fieldhere:brainly.com/question/23610548#SPJ11...