An electron is moving parallel to the Earth’s surface at the equator in a direction 25° south of east. Its velocity is 7.3 ´ 10^4 m/s and a magnetic force of 1.8 ´ 10^-18 N is exerted on the electron. If the magnetic field points south at this location, what is the direction of the magnetic force on the electron and the magnitude of the magnetic field?
3 answers
Lemme go look in my drawer. I have a Maxwell equation T shirt somewhere.
ah, yes, F = q * V cross B
q = - 1.6*10^-19 Coulombs
Vx = 7.3*10^4 (cos 25)
Vy = -7.3*10^4(sin 25)
Bx = 0
By = -b
so
F =q times * 10^4 times determinant of
i j k
6.62 -3.09 0
0 - b 0
so cross product is
q [ 0 i + 0 j - 6.62 k ] 10^4 logical, only x component of V is 90 deg from B
I think you can take it from there :)
q = - 1.6*10^-19 Coulombs
Vx = 7.3*10^4 (cos 25)
Vy = -7.3*10^4(sin 25)
Bx = 0
By = -b
so
F =q times * 10^4 times determinant of
i j k
6.62 -3.09 0
0 - b 0
so cross product is
q [ 0 i + 0 j - 6.62 k ] 10^4 logical, only x component of V is 90 deg from B
I think you can take it from there :)
q [ 0 i + 0 j - 6.62 b k ] 10^4 logical, only x component of V is 90 deg from B
TYPO, left b out
TYPO, left b out
0 answers