Question

A charged particle of mass 6.00 mg moves with a speed of 4.00 km/s in a direction that makes an angle of 37o above the positive x-axis. At the instant it enters a magnetic field of 5.00 mT which is in the positive x direction (+i), it experiences an acceleration of 8.00 m/s2 in the positive z direction (+k). What is the charge of the particle?

Answer 1

Answer:

Explanation:

mass of charged particle m = 6 x 10⁻³ kg .

speed of particle   v  = 4 x 10³ m /s

speed of particle perpendicular to magnetic field = v sin37

= 4 x 10³ sin37

= 2.41 x 10³ m / s

Force on charged particle

= B q v , B is magnetic field , q is charge on particle and v is velocity perpendicular to B

Force = ma

= 6 x 10⁻³ x 8

= 48 x 10⁻³

Force = Bqv

48 x 10⁻³ = 5 x 10⁻³ q x  2.41 x 10³

q = 48 x 10⁻³  / (5 x 2.41)

= 3.98 x 10⁻³C.

Answer 2

Answer:

Explanation:

mass, m = 6 mg = 6 x 10^-6 kg

velocity, v = 4 km/s = 4000 m/s

Angle from X axis, θ = 37°

Magnetic field, B = 5 mT = 0.005 T along X axis

Acceleration, a = 8 m/s² along Z axis

Let q is the charge on the charge particle.

Write all the parameters in vector form

$\overrightarrow{v}=4000(Cos 37\widehat{i}+Sin37\widehat{j})=3194.5\widehat{i}+2407.3\widehat{j}$

$\overrightarrow{B}=5\widehat{i}$

$\overrightarrow{a}=8\widehat{k}$

the magnetic force is equal to the force experienced by the charge particle

$m\overrightarrow{a}=q\left ( \overrightarrow{v} \times \overrightarrow{B} \right )$

$6\times 10^{-6}\times 8\widehat{k} =q\left (3194.5\widehat{i}+2407.3\widehat{j} \right )\times 5\widehat{i}$

48 x 10^-6 = -  q x 12036.5

q = - 4 x 10^-9 C