Exercise 27.9 Part A A group of particles is traveling in a magnetic field of un
ID: 1429264 • Letter: E
Question
Exercise 27.9 Part A A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.30 km/s in the +x-direction experiences a force of 2.06x10-16 N in the +y direction, and an electron moving at 4.40 km/s in the -z-direction experiences a force of 8.40x10-16 N in the +y-direction What is the magnitude of the magnetic field? Submit My Answers Give Up Incorrect; Try Again; 9 attempts remaining; no points deducted Part B What is the direction of the magnetic field? (in the xz-plane) from the-z-direction Submit My Answers Give UpExplanation / Answer
The z component of the magnetic field is
2.06x10^(-16)N/(1.602x10(-19)C x 1.30 x 10^3 m/s) = 0.9891 T
The x component of the magnetic field is
8.4 x 10^(-16)N / (1.602x10^(-19)C x 4.40 x 10^3 m/s) = 1.191 T,
but we can't know whether this is positive or negative,
because you didn't say whether the electron is deflected
in the +y direction or the -y direction.
If I assume that the force on the electron also acts in the +y direction,
then the x component of the magnetic field is also positive,
since (-)(-k) x (i) = +j.
I conclude that the magnetic field has magnitude
sqrt(1.191^2 + 0.9891^2) T = 1.54 T
and its direction is in the x-z plane at
arctan (0.9891/1.191) = 39.70 degrees away from the + x-axis
and 50.3 degrees away from the + z-axis.
(b) In doing this part, I will again assume that the electron moving in the -z direction
was deflected in the +y direction.
F = qv x B
= (-1.602 x 10^(-19)C)(-3.50 x 10^3 m/s j) x (1.191 i + 0.9891k) T
The magnitude of this force is
sqrt(4.6) x 10^(-16) N = 4.6 x 10^(-16) N
and its direction is in the x-z plane,
perpendicular to the magnetic field,
so 39.70 degrees away from the negative z axis
and 50.3 degrees away from the positive x axis.
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