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

Question

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.60 km/s in the +x-direction experiences a force of 2.10×1016 N in the +y-direction, and an electron moving at 4.40 km/s in the z-direction experiences a force of 8.30×1016 N in the +y-direction. What is the magnitude of the magnetic field? What is the direction of the magnetic field? (in the xz-plane) = What is the magnitude of the magnetic force on an electron moving in the y-direction at 3.30 km/s ? What is the direction of this the magnetic force? (in the xz-plane) =

Explanation / Answer

for proton

charge q = 1.6*10^-19 C

speed v = 1.6*10^3 i

Fb = 2.1*10^-16 j

B = Bi + Byj + Bzk

Fb = q*( v X B )

2.1*10^-16 j = 1.6*10^-19*( 1.6*10^3 i x (Bxi + Byj + Bzk) )

2.1*10^-16 j = 1.6*10^-19*1.6*10^3*By k - 1.6*10^-19*1.6*10^3*Bz j

2.1*10^-16 = -1.6*10^-19*1.6*10^3*Bz

Bz = -0.82 T

for electron

charge q = -1.6*10^-19 C

speed v = -4.4*10^3 k

Fb = 8.3*10^-16 j

B = Bxi + Byj + Bzk

Fb = q*( v X B )

8.3*10^-16 j = -1.6*10^-19*( -4.4*10^3 k x (Bxi + Byj + Bzk) )

8.3*10^-16 j = 1.6*10^-19*4.4*10^3*Bx j - 1.6*10^-19*1.6*10^3*By i

8.3*10^-16 = 1.6*10^-19*4.4*10^3*Bx

Bx = 1.18 T

magnetic field B = sqrt(B^2+Bz^2)

magnitude of magnetic field B = sqrt(1.18^2 +0.82^2) = 1.44 T

direction theta = tan^-1(Bz/Bx) = 125 degrees with +x axis or 55 degrees with -x axis

===========================

Fb = -1.6*10^-19*( -3.3*10^3 j x (1.18 i - 0.82 k ) )

Fb = -1.6*10^-19*3.3*10^3*1.18 k - 1.6*10^-19*3.3*10^3*0.82 i

Fb = -4.33*10^-16 i - 6.23*10^-16 k

magnitude = sqrt((4.33*10^-16)^2 + (6.23*10^-16)^2) = 7.58*10^-16 N

direction = arctan(Fz/Fx) = 55.2 degrees

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