Question

A particle with a charge of 37 μC moves with a speed of 77 m/s in the positive x direction. The magnetic field in this region of space has a component of 0.42 T in the positive y direction, and a component of 0.87 T in the positive z direction.

Part A:

What is the magnitude of the magnetic force on the particle? Express your answer using two significant figures.

Part B:

What is the direction of the magnetic force on the particle? (Find the angle measured from the positive zz-axis toward the negative yy-axis in the yzyz-plane.) Express your answer using two significant figures.

Answer #1

Solution) q = 37 uC = 37×10^(-6) C

V = 77 m/s ( in +X direction )

V = 77(i) m/s

By = 0.42(j) T

Bz = 0.87(k) T

(a) Magnetic force , F = ?

F = qV × B

F = q(77(i) × (0.42(j) + 0.87(k)))

F = q((77×0.42)(i×j) + (77×0.87)(i×k))

F = q(32.34(k) + 66.99(- j ))

F = (37×10^(-6))(32.34(k) - 67(j))

F = 1196.58×10^(-6)(k) - 2479×10^(-6)(j)

Magnitude of force ,

F = ((1196.58×10^(-6))^2 + (-2479×10^(-6))^2)^(1/2)

F = 2752.67×10^(-6) N

F = 2.75×10^(-3) N

F = 2.7×10^(-3) N

(b) Direction , theeta = ?

Theeta =

Tan inverse((-2479×10^(-6))÷(1196.58×10^(-6)))

Theeta = - 64.23°

Theeta = - 64°

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