Enhanced EOC: Problem 24.23 MC A proton moves with a speed of 4.5x10 m/s in the
ID: 1445049 • Letter: E
Question
Enhanced EOC: Problem 24.23 MC A proton moves with a speed of 4.5x10 m/s in the directions shown in the figure. A 0.55 T magnetic field points in the positive x-direction.(Figure 1) You may want to revie (DA ages 777-783 For help with math skills, you may want to review: Vector Components Figure 1 V of 1 Part C If the particle is an electron, what is the direction of the magnetic force in figure (a)? z direction Submit Hints My Answers Give Up Review Part Correct Part D What is the magnitude of the magnetic force in figure (b)? Express your answer using three significant figures. F 3.96x10 2 N Hints My rs Give Up Review Part Subm Correct Part E What is the direction of the magnetic force in figure (b)? zy direction (45 degrees between the -z and -y axes Submit Hints My Answers Give Up Review Part Incorrect, Try Again Part F If the particle is an electron, what is the direction of the magnetic force in figure (b)? Please ChooseExplanation / Answer
a) Magnetic force on moving charge,
Fm = q ( v X B) {cross prodcut}
q = - 1.6 x 10^-19 C
v = - 4.5 x 10^7 j m/s
B = 0.55 T i
Fm = -1.6 x 10^-19 ( -4.5x10^7j X 0.55 i)
cross product of -j x i = - (-k) = k
Fm = -1.6 x 10^-19 (0.55 x 4.5 x 10^7 k )
Fm = 3.96 x 10^-12 (-k) N
direction - > -k = along -ve z direction
D) magnitude = 3.96 x 10^-12 N
E) v = 4.5 x 10^7 ( -cos45 j + sin45 k)
for proton. q = +v
F = 1.6 x 10^-19 [ (-3.18E+7 j + 3.18E+7k) x (0.55 i) ]
F = (1.6 x 10^19) [ (1.75 x 10^7 k ) + (1.75 x 10^7 j ) ]
F = 2.8 x 10^-12 j + 2.8 x 10^-12 k N
direction:
= 45 degrees between +ve y and + z axes )
magnitude = sqrt(2.8^2 + 2.8^2) x 10^-12 =3.96 x 10^-12 N
F) for electron,
q = - 1.6 x 10^-19 C
everything else is same.
so direction will be 45 deg between -y and -z axes.
magnitude = 3.96 x 10^-12 N
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