Please show all work, and legible please! Part A A group of particles is traveli

Question

Please show all work, and legible please!

Part A A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.70 km/s in the +r- direction experiences a force of 2.06x10-16 N in the +y-direction, and an electron moving at 4.60 km/s in the -z-direction experiences a force dof 8.50x10-16 N in the +y-direction What is the magnitude of the magnetic field? Submit My Answers Giv Part B What is the direction of the magnetic field? (in the x2-plane) o from the direction Submit My Answers Giv Part C What is the magnitude of the magnetic force on an electron moving in the -y-direction at 3.70 km/s? Submit My Answers Giv

Explanation / Answer

A)

We use eqn

F=q(v x B)=qvBsin

For proton B is in the –ve z axis. Hence

Fy= qvB-z

Plugging values,

2.06*10^-16 = (1.6*10^-19)*(1.70*10^3)*(B-z)

B-z = 0.7573 T ------------(1)

For electron B is in the +ve x axis. Hence

Fy= qv-zB+x

Plugging values,

8.50*10^-16 = (1.6*10^-19)*(4.60*10^3)*(B+x)

B+x = 1.155 T ------------(2)

B= sqrt(B+x^2+ B-z^2)=sqrt(1.155^2+0.7573^2)= 1.38 T

B)

=tan^-1[(B+x)/( B-z)] = tan^-1(1.155/0.7573) = 56.75deg …from –z axis

C)

F+z= qv-yB+x = (1.6*10^-19)*(3.70*10^3)*(1.155)

F+z= 6.638*10^-16 N

F+x= qv-yB-z = (1.6*10^-19)*(3.70*10^3)*(0.7573)

F+x= 4.48*10^-16 N

F=sqrt(F+x^2+ F+x^2) = sqrt[(6.638*10^-16)^2+(4.48*10^-16)^2] = 8.00*10^-16 N

D)

=tan^-1[(B+z)/( B+x)] = tan^-1(6.638/4.48) = 56.deg …from +x axis

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