A particle with a charge of 2.10×10 8 C is moving with an instantaneous velocity
ID: 1356046 • 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) -z
b) when B is towards -x axis
F1 = q*v*B*sin(theta)
= 2.1*10^-8*40.5*10^3*2*sin(180 -53)
= 1.36*10^-3 N
c) when B is towards +z axis
F2 = q*v*B*sin(theta)
= 2.1*10^-8*40.5*10^3*2*sin(90)
= 1.7*10^-3 N
direction : 53 degrees with +y axis
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