An electron travels at a speed of 3.0×10 4 m/sthrough a uniform magnetic field w

Question

An electron travels at a speed of 3.0×104 m/sthrough a uniform magnetic field whose magnitude is 1.2×103 T .

a)What is the magnitude of the magnetic force on the electron if its velocity and the magnetic field are perpendicular? ___N

b)What is the magnitude of the magnetic force on the electron if its velocity and the magnetic field make an angle of 45 ? ___N

c) What is the magnitude of the magnetic force on the electron if its velocity and the magnetic field are parallel?____N

d)What is the magnitude of the magnetic force on the electron if its velocity and the magnetic field are exactly opposite?____N

Explanation / Answer

a)Are perpendicular=> F =QvBsinF = (1.60 ^-19 C)( 3.0×104m/s)(1.2 10^-3T) sin 90F = 5.76*10^-18 N
b) Make an angle of 45o=> F =QvBsinF = ( 1.60 ^-19 C)(3.0×104 m/s)(1.2 10^-3T) sin 45F = 4.07*10^-18 N
c) Are parallel=> F = 0
d. Are exactly opposite=>F=0

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