A particle of mass $\mathbf{10}$g and charge$\mathbf{80}$µC moves through a uniform magnetic field, in a region where the free-fall acceleration is$\mathbf{-}\mathbf{9}\mathbf{.}\mathbf{8}\hat{\mathbf{j}}\mathbf{}\frac{\mathbf{m}}{{\mathbf{s}}^{\mathbf{2}}}$.The velocity of the particle is a constant$\mathbf{20}\hat{\mathbf{i}}$$\mathbf{-}\mathbf{9}\mathbf{.}\mathbf{8}\hat{\mathbf{j}}\mathbf{}\frac{\mathbf{\text{k}}\mathbf{m}}{\mathbf{s}}$which is perpendicular to the magnetic field. What then is the magnetic field?

Consider the mass is:

$$\begin{gathered} m=10 \\ g=0.010\text{kg} \end{gathered}$$

Consider the value of the charge:

$q=\text{80}$$\mu C$

$=\text{80}\times {\text{10}}^{-\text{6}}\text{C}$

Consider the acceleration due to gravity as:

$g=-9.8\hat{j}\frac{\text{m}}{{\text{s}}^{\text{2}}}$

Consider the value of the velocity as:

$v$$=20\hat{i}$$\frac{\text{km}}{\text{s}}$

$=20000\hat{i}$$\frac{\text{m}}{\text{s}}$

Consider the formula for the force:

$F_B=evBsin\varphi$

Consider the formula for the kinetic energy as:

$K.E=\frac{1}{2}m{v}^2$

Here, F_B is magnetic force, v is velocity, m is mass, K.E. is kinetic energy, B is magnetic field, e is charge of particle.

Since, gravity is acting in –y direction, the magnetic force should act in the +y direction. The particle is moving in +x direction, therefore, according to the right hand rule, the magnetic field should be in –z direction.

Gravitational force acting on the particle is balanced by the magnetic force.

$mg=qvB\sin \varphi$

Substitute the value and solve as:

$B=\frac{0.010\left(9.8\right)}{80\times {10}^{-6}\left(20000\right)\sin 90°}\left(-\hat{k}\right)$

$B=(-0.061T)\hat{k}$

Hence,themagnetic field is $B=(-0.061T)\hat{k}$.

Therefore, by using Right hand rule magnetic field is determined.