91Ó°ÊÓ

$$\begin{gathered} \stackrel{â‡\P Ä}{B}=20\hat{i}-50\hat{j}-30\hat{k}mT=\left(20\hat{i}-50\hat{j}-30\hat{k}\right)×{10}^{-3}T \\ \stackrel{â‡\P Ä}{v}=20\hat{i}-30\hat{j}+50\hat{k}m/s \end{gathered}$$

We are given the vectors for$\stackrel{\mathbf{â‡\P Ä}}{\mathbf{v}}$and$\stackrel{\mathbf{â‡\P Ä}}{\mathbf{B}}$. We can find the cross product for them. From this, we can find the angle between them. We have the formula for the radius of the helical path so we can find the required answer.

Formula:

localid="1663951121578" $\begin{array}{rcl}\stackrel{â‡\P Ä}{v}×\stackrel{â‡\P Ä}{B}& =& \left|v\right|\left|B\right|\sin \mathrm{Ï•}\\ qvB& =& \frac{m{v}^2}{r}\end{array}$

$\begin{array}{rcl}\stackrel{â‡\P Ä}{v}×\stackrel{â‡\P Ä}{B}& =& \left(20\hat{i}-30\hat{j}+50\hat{k}\right)×\left(20\hat{i}-50\hat{j}-30\hat{k}\right)×{10}^{-3}\\ & =& 3.4\hat{i}+1.6\hat{j}-0.4\hat{k}\\ \left|\stackrel{â‡\P Ä}{v}×\stackrel{â‡\P Ä}{B}\right|& =& \sqrt{3.4^2+1.6^2-0.4^2}\\ & =& 3.78\frac{m}{s}.T\\ \left|\stackrel{â‡\P Ä}{v}\right|& =& \sqrt{{20}^2+{30}^2+{50}^2}\\ & =& 62\frac{m}{s}\\ \left|\stackrel{â‡\P Ä}{B}\right|& =& \left(\sqrt{{20}^2+{50}^2+{30}^2}\right)×{10}^{-3}\\ & =& 61.6×{10}^{-3}T\\ \stackrel{â‡\P Ä}{v}×\stackrel{â‡\P Ä}{B}& =& \left|\stackrel{â‡\P Ä}{v}\right|\left|\stackrel{â‡\P Ä}{B}\right|\sin \mathrm{Ï•}\\ \sin \mathrm{Ï•}& =& \frac{\left|\stackrel{â‡\P Ä}{v}×\stackrel{â‡\P Ä}{B}\right|}{\left|\stackrel{â‡\P Ä}{v}\right|\left|\stackrel{â‡\P Ä}{B}\right|}\end{array}$

The angle can be calculated as:

$\begin{array}{rcl}\mathrm{Ï•}& =& {\sin }^{-1}\left(\frac{\left|\stackrel{â‡\P Ä}{v}×\stackrel{â‡\P Ä}{B}\right|}{\left|\stackrel{â‡\P Ä}{v}\right|\left|\stackrel{â‡\P Ä}{B}\right|}\right)\\ & =& {\sin }^{-1}\left(\frac{3.78}{61.6×61.6×{10}^{-3}}\right)\\ & =& {84}^o\end{array}$

The angle between $\stackrel{â‡\P Ä}{v}$ and$\stackrel{â‡\P Ä}{B}$ is 84^o .

The magnetic field changes the direction of motion of the particle, but it can’t change its speed, so the speed does not change with time.

From the above diagram, we can see that the angle between velocity and magnetic field does not change with time. As the particle changes its path according to the magnetic field, the angle between them remains constant.

If v is the speed of the electron, then the speed of the electron along the plane perpendicular to the magnetic field will be $v\sin \mathrm{Ï•}$, as the angle between velocity and field is given by $\mathrm{Ï•}$.

We know that:

role="math" localid="1662723909566" $\begin{array}{rcl}qvB& =& \frac{m{v}^2}{r}\\ qB& =& \frac{mv}{r}\\ r& =& \frac{mv}{qB}\\ r& =& \frac{m×v\sin \mathrm{ϕ}}{qB}\\ & =& \frac{9.1×{10}^{-31}×62×\sin {84}^o}{1.6×{10}^{-19}×61.6×{10}^{-3}}\\ & =& 5.7×{10}^{-9}m\end{array}$

The radius of the helical path is $\begin{array}{rcl}& & 5.7×{10}^{-9}m\end{array}$.