A particle (q = 5.0 nC, m = 3.0 μg) moves in a region where the magnetic field has components Bx = 2.0 mT, By = 3.0 mT, and Bz = −4.0 mT. At an instant when the speed of the particle is 5.0 km/s and the direction of its velocity is 120° relative to the magnetic field, what is the magnitude of the acceleration of the particle? A)33 m/s² B)17 m/s² C)39 m/s² D)25 m/s² E)45 m/s²

Question

Quizplus · Accepted Answer

The acceleration of a charged particle moving in a magnetic field is given by the formula a = (q/m) * (v x B), where q is the charge of the particle, m is its mass, v is its velocity, and B is the magnetic field.

First, we need to convert the charge from nanocoulombs to coulombs and the mass from micrograms to kilograms:
q = 5.0 x 10^-9 C
m = 3.0 x 10^-9 kg

Next, we need to find the vector cross product of the velocity and magnetic field, which is given by:
v x B = (v sin(theta)) * (−Bx, Bz, By)

where theta is the angle between v and B (which is 120 degrees or 2π/3 radians), and the negative signs in the vector indicate that this is a cross product in a left-handed coordinate system.

Plugging in the given values, we get:
v sin(theta) = 5.0 km/s * sin(2π/3) = 4.33 km/s = 4.33 x 10^3 m/s
v x B = (4.33 x 10^3 m/s) * (−2.0 x 10^-3 T, −4.0 x 10^-3 T, 3.0 x 10^-3 T) = (−8.663, 6.494, 12.99) m/s^2

Finally, we can find the magnitude of the acceleration by dividing the vector magnitude of v x B by the mass of the particle and multiplying by the charge:
a = (q/m) * |v x B| = (5.0 x 10^-9 C / 3.0 x 10^-9 kg) * √(8.663^2 + 6.494^2 + 12.99^2) m/s^2
a = 39 m/s^2, which is closest to choice C.

A Particle (Q = 5