Aparticle 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

q = 2,95 10-6 C

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 =

      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

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

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explanation:

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