Question

Calculate the force of a particle with a net charge of 170 coulombs traveling at a speed of 135 meters/second perpendicular to the magnetic field of 5.0 × 10-5 tesla. The magnetic field runs from south to north (as shown below), and the particle is moving from east to west: ↑B ← +q

Answer 1

Answer:

F=1.14N j

Explanation:

The magnitude of the magnetic force over a charge in a constant magnetic field is given by the formula:

$|\vec{F}|=|q\vec{v} \ X\ \vec{B}|=qvsin\theta$  (|)

In this case v and B vectors are perpendicular between them. Furthermore the direction of the magnetic force is:

-i X k = +j

Finally, by replacing in (1) we obtain:

$\vec{F}=(170C)(135\frac{m}{s})(5.0*10^{-5}T)=1.14N\ \hat{j}$

hope this helps!

Answer 2

Answer:

The force on the particle, F = 1.15 N

Explanation:

Charge, q = 170 Coulombs

speed of the particle, v = 135 m/s

Magnetic field, B = 5 * 10⁻⁵ T

The force is perpendicular to the magnetic field, θ = 90°

The force of the particle is given by the formula,

F = qvBsinθ

F = 170 * 135 * 5 * 10⁻⁵ * sin90°

F =1.15 N