Question

Part A Review What is the magnitude of the magnetic force on the particle? A particle with a charge of 33 μC moves with a speed of 77 m/s in the positive z direction. The magnetic field in this region of space has a component of 0.45 T in the positive y direction, and a component of 0.83 T in the positive z direction Express your answer using two significant figures. mN Submit Previous Answers Request Answer XIncorrect; Try Again; 5 attempts remaining Part B 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.

Answer 1

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°)
= (33µC)(77 m/s)(0.45 T)(1).

One tesla = one newton per ampere per meter, and one ampere is one coulomb per second, so rewrite this as
Fy = (33 × 10^(-6) C)(77 m/s)(0.45 N/(m·C/s)) = 1.14 × 10^(-3) N, or 1.0 mN.

Similarly,
Fz = qv × Bz = qvBsin(90°)
= (33 µC)(77m/s)(0.83 T)(1)
= 2.1 × 10^(-3) N, or 2.1 mN.

a) The magnitude of the resultant is √((1.14)² + (2.1)²) = 2.4 × 10^(-3) N, or 2.3 mN.

b) Fz is the shorter leg of a right triangle; Fy is the longer leg. (The resultant vector is the hypotenuse.)
The angle θ from the z-axis to the resultant vector is tanˉ¹(Fz/Fy)
= tanˉ¹(2.1 mN / 1.14mN)
= 62° (approximately).

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