A particle with a charge of 2.10×10 8 C is moving with an instantaneous velocity
ID: 1356034 • Letter: A
Question
A particle with a charge of 2.10×108 C is moving with an instantaneous velocity of magnitude 40.5 km/s in the xy -plane at an angle of 53.0 ocounterclockwise from the +x axis.
What is the direction of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the -x direction?
-y
-z
-x
+y
+z
+x
What is the magnitude of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the -x direction?
F1=
What is the direction of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the +z direction?
Please, enter your answer as a counterclockwise angle from the +y direction to the direction of the force.
o=
What is the magnitude of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the +z direction?
F2=
Explanation / Answer
a)
Positive Y direction
b)
Magnetic force
F=qVBSin(o)=(2.1*10-8)(40.5*103)*2*SIn53
F=1.36*10-3 N
c)
o=53o
d)
F=qVBSin90=(2.1*10-8)*(40.5*103)*2*Sin90
F=1.7*10-3 N
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