A group of particles is travelling in a magnetic field of unknown magnitude and

Question

A group of particles is travelling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.20 km/s in the + x-direction experiences a force of 2.06 times 10^- 16 N in the + y-direction, and an electron moving at 4.60 km/s in the - z-direction experiences a force of 8.50 times 10^- 16 N in the + y-direction. What is the magnitude of the magnetic field? B = __________ T What is the direction of the magnetic field? (in the xz-plane) theta = __________ degree from the -z-direction What is the magnitude of the magnetic force on an electron moving in the -y-direction at 3.70 km/s ? F = __________ N

Explanation / Answer

The z component of the magnetic field is

2.06x10^(-16)N/(1.602x10(-19)C x 1.20 x 10^3 m/s) = 1.0716 T

The x component of the magnetic field is

8.5 x 10^(-16)N / (1.602x10^(-19)C x 4.60 x 10^3 m/s) = 1.1534 T,

a)The magnetic field has magnitude

sqrt(1.1534^2 + 1.0716^2) T = 1.5744 T

b)and its direction is in the x-z plane at

arctan (1.0716/1.1534) = 42.89 degrees away from the + x-axis

and (90-42.89) = 47.11 degrees away from the + z-axis.

(180 - 47.11) = 132.89 degrees away from the negative z axis

c)F = qv x B

= (-1.602 x 10^(-19)C)[(-3.7 x 10^3 m/s j) x (1.1534 i + 1.0716 k)] T

= (-6.8367 k + 6.3518 i) x 10^(-16) N

The magnitude of this force is

Sqrt[(6.8367^2 + 6.3518^2 x 10^(-16)] N = 6.8367 x 10^(-16) N

d)and its direction is in the x-z plane,

perpendicular to the magnetic field,

so 42.89 degrees away from the negative z axis

and 47.11 degrees away from the positive x axis.

(180 - 47.11) = 132.89 degrees away from the negative x axis

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