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

Question

A particle with a charge of 35 mu 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. What is the magnitude of the magnetic force on the particle? Express your answer using two significant figures. 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.) Express your answer using two significant figures.

Explanation / Answer

Use F = qv × B.
Let By and Bz be the magnetic field components in the y-direction and z-direction, respectively.
Let Fy and Fz be the forces caused by By and Bz, respectively.
Since the velocity is normal to both axes, the angle between the vectors when the cross product is computed is 90°, and sin(90°) = 1.

Fy = qv × By = qvBsin(90°)
= (35 µC)(73 m/s)(0.42 T)(1).
One tesla = one newton per ampere per meter, and one ampere is one coulomb per second, so rewrite this as
Fy = (35 × 10^(-6) C)(73 m/s)(0.41 N/(m·C/s)) = 1.047 × 10^(-3) N, or 1.047 mN.

Similarly,
Fz = qv × Bz = qvBsin(90°)
= (35 µC)(73 m/s)(0.86 T)(1)
= 2.197 × 10^(-3) N, or 2.197 mN.

a) The magnitude of the resultant is ((1.047)² + (2.197)²) = 2.434 × 10^(-3) N, or 2.434 mN.

b) Fz is the shorter leg of a right triangle; Fy is the longer leg. (The resultant vector is the hypotenuse.)
The angle from the z-axis to the resultant vector is tan¹(Fz/Fy)
= tan¹(2.197 mN / 1.047 mN)
= 64°51" (approximately).

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