Question

A particle of mass $\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{g}}$moves atrole="math" localid="1663161943446" $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\raisebox{1ex}{$\mathbf{\text{km}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{s}}$}\right.$in an$\mathit{x}\mathit{y}$ plane, in a region with a uniform magnetic field given by role="math" localid="1663162607660" $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\widehat{\mathbf{\text{i}}}\mathbf{\text{mT}}$. At one instant, when the particle’s velocity is directed${\mathbf{37}}^{\mathbf{\text{o}}}$counterclockwise from the positive direction of the xaxis, the magnetic force on the particle is$\mathbf{0}\mathbf{.}\mathbf{48}\hat{\mathbf{k}}\mathbf{\text{N}}$.What is the particle’s charge?

Short answer

The charge on particles is$q=-0.04\text{C}$

Step-by-step solution

Identification of given data

$m=6\text{g}$

$$\begin{gathered} v=4\raisebox{1ex}{$\text{km}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right. \\ \overrightarrow{B}=5.0\hat{i}\text{mT} \\ \theta ={37}^{\text{o}} \\ \overrightarrow{F}=0.48\hat{k}\text{N} \end{gathered}$$

Significance of magnetic force

The influence of a magnetic field produced by one charge on another is what is known as the magnetic force between two moving charges.

By using the relation between the magnetic force , velocity of particle and magnetic field applied, we can find the charge on the particle.

Formula:

$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$

Determining the charge on the particles.

By using equation 28-2, we can see that the force is along positive$\hat{k}$. But the magnetic field in positive $\hat{i}.$So the particle velocity is along positive y axis. Hence, the particle is negative. Now examining the magnitude by using the 28-3, we find

$\overrightarrow{|F}\left|=\left|q\right|v\right|\overrightarrow{B}|sin\varphi$

By substituting the value, we can get

$0.48 \text{N}=\left|q\right|\left(4000 \text{m/s}\right)\left(0.0050 \text{T}\right)\text{sin}37°$

$\left|q\right|=\frac{0.48}{20\times 0.60}\text{C}$

$\left|q\right|=0.04\text{C}$

But from the discussion, we can conclude that this charge is negative.

$q=-0.04\text{C}$