Question

A particle with charge 7.80 $\mathit{\mu }$C is moving with velocity $\mathit{v}\mathbf{⃗}\mathbf{=}\mathbf{-}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{80}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathit{m}\mathbf{/}\mathit{s}\mathbf{)}\hat{\mathbf{j}}$. The magnetic force on the particle is measured to be $\mathbf{(}\mathbf{7}\mathbf{.}\mathbf{60}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{N}\mathbf{)}\hat{\mathbf{i}}\mathbf{-}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{N}\mathbf{)}\hat{\mathbf{k}}$.(a) Calculate all the components of the magnetic field you canfrom this information. (b) Are there components of the magneticfield that are not determined by the measurement of the force?Explain. (c) Calculate the scalar product$\overrightarrow{\mathbf{B}}\overrightarrow{\mathbf{F}}$ what is the angle between Bands F?

Short answer

1. Magnetic field along x axis is$B_x=-0.175T$ and along z axis is$B_z=-0.256T$
2. The y component is not determined, since there is no force along this component hence measurement doesn’t tell us anything about$B_y$

c. The angle between the two is 90 degrees

Step-by-step solution

The significance of the magnetic field

The magnetic field force is given by

$$\begin{gathered} {\mathit{F}}_{\mathbf{B}}\mathbf{=}\mathit{q}\mathbf{(}\mathit{v}\mathbf{\times }\mathit{B}\mathbf{)} \\ \mathit{F}\mathbf{=}\mathit{q}\mathit{v}\mathit{B}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\theta } \end{gathered}$$

Where q is the charge of the particle, V is the velocity and B is the magnetic field

Determination of the components of the magnetic field

We know that the magnetic field force is given by

$F=q(v\times B)$

We know

$\begin{array}{r}\overrightarrow{F}=q^{-}\det \left(\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ v_x& v_y& v_z\\ B_x& B_y& B_z\end{array}\right)\\ F_x\widehat{i}+F_y\widehat{j}+F_z\widehat{k}=q\left(v_y{B}_z-v_z{B}_y\right)\widehat{i}+q\left(v_z{B}_x-v_x{B}_z\right)\widehat{j}+q\left(v_x{B}_y-v_y{B}_x\right)\widehat{k}\end{array}$

We know$V_x=0$ that$V_z=0$

$F_x\hat{i}+F_y\hat{j}+F_z\hat{k}=q{v}_y{B}_z\hat{i}-q{v}_y{B}_x\hat{k}$

Comparing we get

$\begin{array}{r}{F}_x=q\left(v_y{B}_z\right)\\ F_y=0\\ F_z=-q{v}_y{B}_x\end{array}$

Substitute all the value in the above equation.

Solve for the magnetic field components

$\begin{array}{r}{B}_z=\frac{F_x}{q{v}_y}\\ =\frac{7.60\times {10}^{-3}\mathrm{N}}{7.80\times {10}^{-6}\mathrm{C}\times \left(-3.80\times {10}^3\mathrm{m}/\mathrm{s}\right)}\\ =-0.256\mathrm{T}\\ B_x=\frac{F_z}{-q{v}_y}\\ =\frac{-5.20\times {10}^{-3}\mathrm{N}}{7.80\times {10}^{-6}\mathrm{C}\times \left(-3.80\times {10}^3\mathrm{m}/\mathrm{s}\right)}\\ =-0.175\mathrm{T}\end{array}$

Hence, Magnetic field along x axis is$B_x=-0.175T$ and along z axis is $B_z=-0.256T$.

Determination of the scalar product

We see

$\begin{array}{r}\overrightarrow{B}\overrightarrow{F}=B_x{F}_x+B_y{F}_y+B_z{F}_z\\ \overrightarrow{B}-\overrightarrow{F}=(-0.175\mathrm{T})\left(7.60\times {10}^{-3}\mathrm{N}\right)+(-0.256\mathrm{T})\left(-5.20\times {10}^{-3}\mathrm{N}\right)\\ \overrightarrow{B}\overrightarrow{F}=0\end{array}$

This means that the two vectors are perpendicular to each other, so the angle is 90 degrees.