Question

A particle with negative charge q and mass m = 2.65×10−15 kg is traveling through a region containing a uniform magnetic field B⃗ =(− 0.130T) k^. At a particular instant of time the velocity of the particle is v⃗ =(1.20 × 106 m/s) (−3i^ + 4j^+12 k^) and the force F⃗ on the particle has a magnitude of 2.30 N . (a) Determine the charge q. (b) Determine the acceleration a⃗ of the particle. (c) Find the x, y and z-component of the acceleration

Answer 1

q = 2,95 10-6 C

Step-by-step explanation:

The magnetic force on a particle is described by the equation

F = q v x B

Where bold indicate vectors

Let's make the vector product

vxB =
`\left[\begin{array}{ccc}i&j&k\\-3&4&12\\0&0&-0.13\end{array}\right]`

v x B = 1.20 106 [i ^ (4 0.130) - j ^ (3 0.130)]

vx B = 1.20 106 [0.52 i ^ - 0.39j ^]

As they give us the force module, let's use Pythagoras' theorem,

|v xB | =1.20 10⁶ √( 0.52² + 0.39²)

|v x B| = 1.20 10⁶ 0.65

v xB = 0.78 10⁶

Let's replace and calculate

2.30 = q 0.78 10⁶

q = 2.3 / 0.78 106

q = 2,95 10-6 C