One component of a magnetic field has a magnitude of 0.048 T and points along th

Question

One component of a magnetic field has a magnitude of 0.048 T and points along the +x axis, while the other component has a magnitude of 0.065 T and points along the -y axis. A particle carrying a charge of +2.0 x 10 ^ (-5) C is moving along the +z axis at a speed of 4.2 X 10 ^ (3) m/s. A)Find the magnitude of the magnetic force that acts on the particle due to the component of the magnetic field pointing along the +x axis. B) find the direction of the force calculated in part a. How idd you determine the direction? Explain. C) Find the magnitude of the magnetic force that acts on the particle due to the component of the magnetic field pointing along the -y axis. D) Find the direction of the force calcuated in part c. How did you determine the direction? Explain. E) Find the magnitude of the net magnetic force that acts on the particle? F) Determine the angle that the net force makes with respect to the +x axis

Explanation / Answer

v = [0i ,0j ,2800k] m/s B= [0.036i ,-0.058j ,0k] T En = vXB = 162.4i +100.8j + 0k q = 3*10 ^ -5 C F = q vXB = q*En = [0.004872i + 0.003024j + 0k] N |F| = sqr(Fx^2+Fy^2) = 5.734E-03 N Angle positive of the x-axis: Inverse tan (Fy/Fx) = 31.827 degrees

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