1. An electron is traveling horizontally due north. It enters a uniform 150-mT m
ID: 1588254 • Letter: 1
Question
1. An electron is traveling horizontally due north. It enters a uniform 150-mT magnetic field which is known to be perpendicular to the velocity of the electron. The magnetic field creates a horizontal force towards the west of 4.8 × 10 N^-13 on the electron. Determine the direction of the magnetic field and the speed of the electron.
2. A horizontal copper wire of diameter 2.05mm (12 gauge wire) carries a current of 11.5 A towards the east. If the earth’s magnetic field at this location is horizontally north and has magnitude 52 mico-T, what is the magnetic force per meter of wire? Is it fair to neglect this force when wiring a building?
Explanation / Answer
1.)
B = 150 * 10^-3 T
F = 4.8 * 10^-13 N
q = 1.6 * 10^-19 C
Direction of the Magnetic Field - Perpendicular to the velocity of the electron - Out of the Page.
Force = q* VxB
F = q* VB
4.8 * 10^-13 = 1.6 * 10^-19 * V * 150 * 10^-3
V = 2.0 * 10^7 m/s
Speed of the electron, V = 2.0 * 10^7 m/s
please post seperate questions in seperate post only.
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