Question

A particle with charge$\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{60}\mathbf{nC}$is moving in a uniform magnetic fieldrole="math" localid="1655717557369" $\mathit{B}\mathbf{=}\mathbf{-}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{}\mathbf{T}\mathbf{)}\hat{\mathit{k}}$

The magnetic force on the particle is measured to berole="math" localid="1655717706597" $\stackrel{\mathbf{\rightharpoonup }}{\mathit{F}}\mathbf{=}\mathbf{-}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{40}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{N}\mathbf{)}\hat{\mathit{i}}\mathbf{-}\mathbf{(}\mathbf{7}\mathbf{.}\mathbf{40}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{N}\mathbf{)}\hat{\mathit{j}}$ (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there
components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}\mathit{֏}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}$. What is the angle between velocity and force?

Short answer

(a) The velocity along x axis is $V_x=-106\mathrm{m}/\mathrm{s}$and along y axis is $V_y=-48.6mls$

(b) The z component is not determined, since there is no force along this component hence measurement doesn’t tell us anything about $V_z$

(c) The angle between the two is 90 degrees

Step-by-step solution

The significance of magnetic field

The magnetic field force is given by

$$\begin{gathered} {\mathit{F}}_{\mathbf{B}}\mathbf{=}\mathit{q}\left(v\times B\right) \\ \mathit{F}\mathbf{=}\mathit{q}\mathit{v}\mathit{B}\mathbf{sin}\mathit{\theta } \end{gathered}$$

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

Determination of the components of the velocity of the particle 

We know that the magnetic field force is given by

$F=q(v\times B)$

We know

$\stackrel{\rightharpoonup }{F}=q֏det\left(\begin{array}{ccc}\overline{i}& \overline{j}& \overline{k}\\ v_x& v_y& v_z\\ B_x& B_y& B_z\end{array}\right)$

role="math" localid="1655718671037" $F_x\hat{i}+F_y\hat{j}+F_z\hat{k}=q\left(v_y{B}_z-v_z{B}_y\right)\hat{i}+q\left(v_z{B}_x-v_x{B}_z\right)\hat{j}+q\left(v_x{B}_y-v_y{B}_x\right)\hat{k}$

We know that$B_y=B_x=0$

Applying this in the equation we get

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

Equating the two sides we get

$$\begin{gathered} F_x=q{v}_y{B}_z \\ {F}_y=-q{v}_x{B}_z \\ {F}_z=0 \end{gathered}$$

Plugging in the values we get

$$\begin{gathered} v_x=\frac{F_y}{-q{B}_z} \\ =\frac{7.40\times {10}^{-7}\mathrm{N}}{-\left(5.60\times {10}^{-9}\mathrm{C}\right)\left(-1.25\mathrm{T}\right)} \\ =-106\mathrm{m}/\mathrm{s} \\ {V}_y=\frac{F_x}{q{B}_z} \\ =\frac{-3.40\times {10}^{-7}\mathrm{N}}{\left(5.60\times {10}^{-9}\mathrm{C}\right)\left(-1.25\mathrm{T}\right)} \\ =-48.6\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the velocity along x axis is${\mathrm{v}}_{\mathrm{x}}=-106\mathrm{m}/\mathrm{s}$ and along y axis is${\mathrm{v}}_y=-48.6\mathrm{m}/\mathrm{s}$

Determination scalar Product

We see

$$\begin{gathered} \stackrel{\rightharpoonup }{v}֏\stackrel{\rightharpoonup }{F}=v_x{F}_x+V_y{F}_y+V_z{F}_z \\ \stackrel{\rightharpoonup }{v}֏\stackrel{\rightharpoonup }{F}=\frac{F_y}{-q{B}_z}{F}_x+\frac{F_x}{q{B}_z}{F}_y \\ \stackrel{\rightharpoonup }{v}֏\stackrel{\rightharpoonup }{F}=0 \end{gathered}$$

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