91Ó°ÊÓ

$\stackrel{→}{v}=\left(-2.00\hat{i}+4.00\hat{j}-6.00\hat{k}\right)\text{m/s}$

$\stackrel{→}{B}=\left(2.00\hat{i}-4.00\hat{j}+8.00\hat{k}\right)\text{mT}$

By using the formula for magnetic force, we first find the magnetic force. After that, by taking dot product, we can find the angle between the force and velocity, and we can also find the angle between the velocity and magnetic field.

Formula:

$$\begin{gathered} \stackrel{→}{F}=q\left(\stackrel{→}{v}×\stackrel{→}{B}\right) \\ \stackrel{→}{F}·\stackrel{→}{v}=Fv\text{cos}\mathrm{θ} \\ \stackrel{→}{v}·\stackrel{→}{B}=\left|v\right|\left|B\right|\text{cos}\mathrm{θ} \end{gathered}$$

The proton is having positive charge, i.e.,$+e$, so from the equation,

$\stackrel{→}{F}=q\stackrel{→}{v}×\stackrel{→}{B}$

We can find the force as$$\begin{gathered} \stackrel{→}{F}=+e\left(-2.00\hat{i}+4.00\hat{j}-6.00\hat{k}\right)×\left(2.00\hat{i}-4.00\hat{j}+8.00\hat{k}\right) \\ \stackrel{→}{F}=+e\left[\left(-2×-4\right)\hat{k}+\left(-2×8\right)\left(-\hat{j}\right)+\left(4×2\right)\left(-\hat{k}\right)+\left(4×8\right)\hat{i}+\left(-6×2\right)\hat{j}+\left(-6×\left(-4\right)\left(-\hat{i}\right)\right)\right]×{10}^{-3} \\ \stackrel{→}{F}=+e\left[8\hat{k}+16\hat{j}-8\hat{k}+32\hat{i}-12\hat{j}-24\hat{i}\right]×{10}^{-3} \\ \stackrel{→}{F}=+e\left(8\hat{i}+4\hat{j}\right)×{10}^{-3} \end{gathered}$$

Where$e=1.6×{10}^{-19}C$

$$\begin{gathered} \stackrel{→}{F}=1.6×{10}^{-19}×\left(8\hat{i}+4\hat{j}\right)×{10}^{-3} \\ \stackrel{→}{F}=1.6×{10}^{-22}×\left(8\hat{i}+4\hat{j}\right) \\ \stackrel{→}{F}=\left(12.8\hat{i}+6.4\hat{j}\right)×{10}^{-22}N \end{gathered}$$

Hence, the magnetic force acting on the proton is$\stackrel{→}{F}=\left(12.8\hat{i}+6.4\hat{j}\right)×{10}^{-22}N$

We know the dot product of two vectors is

$\stackrel{→}{F}·\stackrel{→}{v}=Fv\text{cos}\mathrm{θ}$

$\mathrm{θ}$is the angle between $\stackrel{→}{F}$and $\stackrel{→}{B}$

$$\begin{gathered} \stackrel{→}{F}.\stackrel{→}{v}=\left(\left(12.8\hat{i}+6.4\hat{j}\right)×{10}^{-22}\right)×\left(-2.00\hat{i}+4.00\hat{j}-6.00\hat{k}\right) \\ \stackrel{→}{F}.\stackrel{→}{v}=\left(\left(12.8\hat{i}+6.4\hat{j}\right)×{10}^{-22}\right)×\left(-2.00\hat{i}+4.00\hat{j}-6.00\hat{k}\right) \\ \stackrel{→}{F}.\stackrel{→}{v}=-2.56+2.56 \\ \stackrel{→}{F}.\stackrel{→}{v}=0 \end{gathered}$$

From the above equation, we can conclude that $\stackrel{→}{F}$and $\stackrel{→}{v}$are perpendicular to each other so that

$$\begin{gathered} 0=Fv\text{cos}\mathrm{θ} \\ \text{cos}\mathrm{θ}=0 \\ \mathrm{θ}=90° \end{gathered}$$

Hence, the angle between $\stackrel{→}{v}$and $\stackrel{→}{F}$is$\mathrm{θ}=90°$.

Now, for finding the angle between$\stackrel{→}{v}$and magnetic field$\stackrel{→}{B}$we take dot product as

$\stackrel{→}{v}·\stackrel{→}{B}=\left|v\right|\left|B\right|\text{cos}\mathrm{θ}······\left(1\right)$

So first find the dot product as

$$\begin{gathered} \stackrel{→}{v}·\stackrel{→}{B}=\left(-2.00\hat{i}+4.00\hat{j}-6.00\hat{k}\right)·\left(2.00\hat{i}-4.00\hat{j}+8.00\hat{k}\right) \\ \stackrel{→}{v}·\stackrel{→}{B}=\left(-4.00-16.00-48.00\right) \\ \stackrel{→}{v}·\stackrel{→}{B}=-68 \end{gathered}$$

Now find the magnitude of the each vector.

$$\begin{gathered} \left|\stackrel{→}{v}\right|=\sqrt{4+16+36} \\ \left|\stackrel{→}{v}\right|=\sqrt{56} \\ \left|\stackrel{→}{v}\right|=7.48 \end{gathered}$$

$$\begin{gathered} \left|\stackrel{→}{B}\right|=\sqrt{4+16+64} \\ \left|\stackrel{→}{B}\right|=\sqrt{84} \\ \left|\stackrel{→}{B}\right|=9.1 \end{gathered}$$

From equation we can find the angle between them.

$$\begin{gathered} \stackrel{→}{v}·\stackrel{→}{B}=\left|v\right|\left|B\right|\cos \mathrm{θ} \\ \cos \mathrm{θ}=\frac{\stackrel{→}{v}·\stackrel{→}{B}}{\left|v\right|\left|B\right|} \end{gathered}$$

By putting the value,

$$\begin{gathered} \cos \mathrm{θ}=-\frac{68}{7.48×9.16} \\ \cos \mathrm{θ}=-\frac{68}{68.58} \\ \cos \mathrm{θ}=-0.9915 \\ \mathrm{θ}=-0.9915 \\ \mathrm{θ}=172.52 \\ \mathrm{θ}≈173° \end{gathered}$$

Hence, the angle between $\stackrel{→}{v}$and $\stackrel{→}{B}$is$\mathrm{θ}=173°$