Question

A particle of charge q > 0 is moving at speed v in the -direction through a region of uniform magnetic field . The magnetic force on the particle is $\overrightarrow{\mathbf{F}}\mathbf{=}{\mathbf{F}}_{\mathbf{0}}\mathbf{(}\mathbf{3}\widehat{\mathbf{i}}\mathbf{+}\mathbf{4}\widehat{\mathbf{j}}\mathbf{)}$, where is a positive constant. (a) Determine the components, , and, or at least as many of the three components as is possible from the information given. (b) If it is given in addition that the magnetic field has magnitude$\mathbf{6}{\mathit{F}}_{\mathbf{0}}\mathbf{/}\mathit{q}\mathit{v}$ , determine as much as you can about the remaining components of .

Short answer

(a) The component of ${\mathrm{B}}_{\mathrm{y}}=-\frac{3{\mathrm{F}}_0}{\mathrm{qv}},{\mathrm{B}}_{\mathrm{x}}=\frac{4{\mathrm{F}}_0}{\mathrm{qv}}$ .

(b) The remaining component of$\overrightarrow{B}\text{is}\pm \frac{\sqrt{11}{F}_0}{qv}$ .

Step-by-step solution

Definition of Magnetic field

The term magnetic field may be defined as the area around the magnet behaves like a magnet.

Step 2: Identification of the given data

The magnetic force of the particle is,$\overrightarrow{\mathrm{F}}={\mathrm{F}}_0(3\widehat{\mathrm{i}}+4\widehat{\mathrm{j}})$ .

The magnetic field is,$\mathrm{B}=\frac{6{\mathrm{F}}_0}{\mathrm{qv}}$ .

Determine the Component of Magnetic field

The force can be calculated as

$\overrightarrow{\mathrm{F}}=\mathrm{q}\overrightarrow{\mathrm{V}}\times \overrightarrow{\mathrm{B}}$

Put all the given values

$\begin{array}{r}\overrightarrow{\mathrm{F}}=\mathrm{q}\left(\mathrm{v}\widehat{\mathrm{k}}\right)\times \left({\mathrm{B}}_{\mathrm{x}}\widehat{\mathrm{i}}+{\mathrm{B}}_{\mathrm{y}}\widehat{\mathrm{j}}+{\mathrm{B}}_{\mathrm{z}}\widehat{\mathrm{k}}\right)\\ \overrightarrow{\mathrm{F}}=-{\mathrm{qvB}}_{\mathrm{y}}\widehat{\mathrm{i}}+{\mathrm{qvB}}_{\mathrm{x}}\widehat{\mathrm{j}}\\ {\mathrm{F}}_0(3\widehat{\mathrm{i}}+4\widehat{\mathrm{j}})=-{\mathrm{qvB}}_{\mathrm{y}}\widehat{\mathrm{i}}+{\mathrm{qvB}}_{\mathrm{x}}\widehat{\mathrm{j}}\\ \mathrm{Compare}\mathrm{both}\mathrm{sides}\mathrm{get}\\ 3{\mathrm{F}}_0=-{\mathrm{qvB}}_{\mathrm{y}}\\ {\mathrm{B}}_{\mathrm{y}}=-\frac{3{\mathrm{F}}_0}{\mathrm{qv}}\\ 4{\mathrm{F}}_0={\mathrm{qvB}}_{\mathrm{x}}\\ {\mathrm{B}}_{\mathrm{x}}=\frac{4{\mathrm{F}}_0}{\mathrm{qv}}\end{array}$

Hence, the component of ${\mathrm{B}}_{\mathrm{y}}=-\frac{3{\mathrm{F}}_0}{\mathrm{qv}},{\mathrm{B}}_{\mathrm{x}}=\frac{4{\mathrm{F}}_0}{\mathrm{qv}}$.

Determine the remaining Component of Magnetic field

The magnitude of the magnetic field is calculated as,

$\begin{array}{r}\mathrm{B}=\frac{6{\mathrm{F}}_0}{\mathrm{qV}}\\ \sqrt{{\mathrm{B}}_{\mathrm{x}}^2+{\mathrm{B}}_{\mathrm{y}}^2+{\mathrm{B}}_{\mathrm{z}}^2}=\frac{6{\mathrm{F}}_0}{\mathrm{qV}}\\ \mathrm{Substitute}\mathrm{all}\mathrm{the}\mathrm{value}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}.\\ \sqrt{{\left(\frac{4{\mathrm{F}}_0}{\mathrm{qV}}\right)}^2+{\left(-\frac{3{\mathrm{F}}_0}{\mathrm{qV}}\right)}^2+{\mathrm{B}}_{\mathrm{z}}^2}=\frac{6{\mathrm{F}}_0}{\mathrm{qv}}\\ 6=\sqrt{9+16+{\left(\frac{\mathrm{qv}}{{\mathrm{F}}_0}\right)}^2{\mathrm{B}}_{\mathrm{z}}^2}\\ 36=25+{\left(\frac{\mathrm{qv}}{{\mathrm{F}}_0}\right)}^2{\mathrm{B}}_{\mathrm{z}}^2\end{array}$

$\begin{array}{r}11={\left(\frac{\mathrm{qv}}{{\mathrm{F}}_0}\right)}^2{\mathrm{B}}_{\mathrm{z}}^2\\ {\mathrm{B}}_{\mathrm{z}}=\pm \frac{\sqrt{11}{\mathrm{F}}_0}{\mathrm{qv}}\end{array}$

Hence, the remaining component of$\overrightarrow{\mathrm{B}}\mathrm{is}\pm \frac{\sqrt{11}{\mathrm{F}}_0}{\mathrm{qv}}$ .