Question

A proton of charge e and mass m enters a uniform magnetic field $\overline{B}=B\widehat{i}$ with an initial velocity$\overline{v}={{v}_{x}}\widehat{i}+{{v}_{y}}\widehat{j}$ . Find an expression in unit vector notation for its velocity at time .

Answer 1

Hint: In the above question it is given that the magnetic field B is acting along the x-direction in 3d space. We first need to determine the direction of magnetic force on the proton. Further according we will obtain the velocity vector of the particle and express it into unit vectors.

Formula used:
$F=q(v\times B)$
$F=ma$
$v=u+at$

Complete step-by-step answer:
Let us say a particle has charge ‘q’ with velocity ‘v’ and enters a region of magnetic field ‘B’. Then the force on the particle is given by,
$F=q(v\times B)$
In the above question the magnetic field in space is $\overline{B}=B\widehat{i}$and the velocity of the particle is $\overline{v}={{v}_{x}}\widehat{i}+{{v}_{y}}\widehat{j}$ . Hence from the above expression, the magnetic force on the proton of charge ‘e’ is,
$\begin{align}
  & F=e(v\times B) \\
 & \Rightarrow F=e(({{v}_{x}}\widehat{i}+{{v}_{y}}\widehat{j})\times B\widehat{i}) \\
 & \therefore F=-e{{v}_{y}}B\widehat{k}.....(1) \\
\end{align}$
From the Newton’s second law if the particle experiences a force F of mass ‘m’ then the acceleration ‘a’ of the particle is given by,
$F=ma.....(2)$
Therefore from equation 1 and 2 we get,
$\begin{align}
  & F=ma=-e{{v}_{y}}B\widehat{k} \\
 & \therefore a=-\dfrac{e{{v}_{y}}B\widehat{k}}{m} \\
\end{align}$
The particle enters the field at initial velocity of $u={{v}_{x}}\widehat{i}+{{v}_{y}}\widehat{j}$. From Newton’s first kinematic equation, the velocity ‘v’ of the particle at time t is given by,
$\begin{align}
  & v=u+at \\
 & \Rightarrow v={{v}_{x}}\widehat{i}+{{v}_{y}}\widehat{j}+-\left( \dfrac{e{{v}_{y}}B\widehat{k}}{m} \right)t \\
 & \therefore v(t)={{v}_{x}}\widehat{i}+{{v}_{y}}\widehat{j}-\dfrac{e{{v}_{y}}Bt}{m}\widehat{k} \\
\end{align}$
The unit vector is defined as the ratio of the vector to that of the magnitude. Hence the velocity V(m)of the proton expressed in the form of its unit vector is,
$V(m)=\dfrac{{{v}_{x}}\widehat{i}+{{v}_{y}}\widehat{j}-\dfrac{e{{v}_{y}}Bt}{m}\widehat{k}}{\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}+{{\left( \dfrac{e{{v}_{y}}Bt}{m} \right)}^{2}}}}$

Note: It is to be noted that the unit vector is a vector having direction but magnitude equal to unity. The proton experiences acceleration in the z-direction. Hence the velocity of the proton is a function of time in the z-direction and independent of time in the x and y direction.