A particle with a charge of 35 C moves with a speed of 73 m/s in the positive x

Question

A particle with a charge of 35 C moves with a speed of 73 m/s in the positive x direction. The magnetic field in this region of space has a component of 0.41 T in the positive y direction, and a component of 0.86 T in the positive z direction. 1.) What is the magnitude of the magnetic force on the partice? F=____mN 2.) What is the direction of the magnetic force on the particle? (Find the angle measured from the positive z-axis toward the negative y-axis in the yz-plane.) =____^degress. Make sure to express both answers using 2 significant figures.

Explanation / Answer

in a magnetic field, force on a moving charge particle=F=q*(cross product of velocity and magnetic flux density)

=35*10^(-6)*(cross product of (73 i) and (0.41 j + 0.86 k))

=35*10^(-6)*(-62.78 j +29.93 k)

[note: cross product of i and j is k and cross product of i and k is -j]

magnitude of the force=35*10^(-6)*sqrt(62.78^2+29.93^2)=2.434*10^(-3) N

answer: F=2.43 mN

part b:

if angle with z axis is theta,

then cos(theta)=dot product of ((-62.78 j + 22.93 k ) and k )/(magnitude of (-62.78 j +22.93 k) * magnitude of k)

=22.93/66.8364=0.343

==>theta=69.94 degrees

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