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$$
\begin{array}{l}{\text { A particle with mass } 1.81 \times 10^{-3} \mathrm{kg} \text { and a charge of }} \\ {1.22 \times 10^{-8} \mathrm{C} \text { has, at a given instant, a velocity } \vec{v}=(3.00 \times} \\ {10^{4} \mathrm{m} / \mathrm{s} ) \hat{\jmath} . \text { What are the magnitude and direction of the particle's }} \\ {\text { acceleration produced by a uniform magnetic field } \vec{B}=(1.63 \mathrm{T}) \hat{\imath}+} \\ {(0.980 \mathrm{T}) \hat{J} ?}\end{array}
$$

$$
\begin{array}{l}{\vec{F}=q \vec{V} \times \vec{B}} \\ {\vec{F}=m \vec{a}}\end{array} ] \rightarrow \quad m \vec{a}=q \vec{V} \times \vec{B}
$$

$$
\longrightarrow {\vec{a}}=\frac{q \vec{V} \times \vec{B}}{m}
$$

$$
\vec{a}=\frac{1.22*10^{-8} * 3 * 10^{4}(1.63)\hat{j} \times \hat{i}}{1.81 * 10^{-3}}=-0.33 \hat{k}
$$

$$
\begin{array}{l}{\text { A particle with charge } 7.80 \mu \mathrm{C} \text { is moving with veloc- }} \\ {\text { ity } \overrightarrow{\boldsymbol{v}}=-\left(3.80 \times 10^{3} \mathrm{m} / \mathrm{s}\right) \hat{\jmath} . \text { The magnetic force on the particle }} \\ {\text { is measured to be } \overrightarrow{\boldsymbol{F}}=+\left(7.60 \times 10^{-3} \mathrm{N}\right) \hat{\imath}-\left(5.20 \times 10^{-3} \mathrm{N}\right) \hat{k}} \\ {\text { (a) Calculate all the components of the magnetic field you can }}\end{array}
$$

$$
\begin{array}{l}{\text { from this information. (b) Are there components of the magnetic }} \\ {\text { field that are not determined by the measurement of the force? }} \\ {\text { Explain. (c) Calculate the scalar product } \vec{B} \cdot \vec{F} . \text { What is the angle }} \\ {\text { between } \vec{B} \text { and } \vec{F} ?}\end{array}
$$

$$
F=|q| V B \sin \phi
$$

$$
\rightarrow V=\frac{F}{| q | B \sin \phi}
$$

$$
q=-1.6 * 10^{-19} \mathrm{C}
$$

$$
V=\frac{4.6 * 10^{-15}}{1.6 * 10^{-19} * 3.5 * 10^{-3} \sin 60}=9.49 * 10^{6} \mathrm{m}  / \mathrm{s}
$$

(a) $$
F_{x}=q\left(V_ y B_{z}-V_{z} B _y\right)=q V_{y} B_ z
$$

$$
\rightarrow B_{z}=\frac{F_{x}}{q V_{y}}=\frac{7.6 * 10^{-3}}{7.8 * 10^{-6} *\left(-3.8 * 10^{3}\right)}=-0.256 \mathrm{T}
$$

$$
F y=q\left(V_{z} B _x-V \times B _z\right)=0
$$

$$
F_{z}=q\left(V_{x} B_{y}-V_{y} B_ x\right)=
$$

$$
\rightarrow B_{x}=-\frac{F_{z}}{q V y}=-0.175 \mathrm{T}
$$

(c) $$
\vec{B} \cdot \vec{F}=B x F x
$$$$
+By F_{y}+B_{z} F_{z}
$$$$
=(-0.175)\left(7.6 * 10^{-3}\right)+(-0.256)\left(-5.2*10^{-3}\right)
$$

$$
=0 \quad \vec{B} \& \vec{F} \perp\left(\theta=90^{\circ}\right)
$$

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