Question

A particle with a charge of 32.8 moves with a speed of 67.2 m/s in the positive direction. The magnetic field in this region of space has a component of 0.462 T in the positive direction, and a component of 0.849 T in the positive direction. What is the magnitude of the magnetic force on the particle?

What is the direction of the magnetic force on the particle relative to the positive -axis?

Answer 1

Given that, q = 32.8 C = 32.8 x 10^-6 C

Use the expression of magnetic force -

F = qv × B.

Now, 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.

So -

Fy = qv × By = qvBsin(90°)
= (32.8 x 10^-6 C)(67.2 m/s)(0.462 T)(1).
One tesla = one newton per ampere per meter, and one ampere is one coulomb per second,

so rewrite this as -

Fy = (32.8 × 10^(-6) C)(67.2 m/s)(0.462 N/(m·C/s))

= 1.018 × 10^(-3) N

Similarly,

Fz = qv × Bz = qvBsin(90°)
= (32.8 x 10^-6 C)(67.2 m/s)(0.849 T)(1)
= 1.87 x 10^(-3) N

(a) Magnitude of resultant force on the particle, F = sqrt[Fx^2 + Fy^2]

= sqrt[(1.018 x 10^-3)^2 + (1.87 x 10^-3)^2]

= 2.13 x 10^-3 N

(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((1.87 x 10^-3) / (1.018 x 10^-3))
= 61.4° relative to the positive z-axis.