Question

A proton of charge +eand mass menters a uniform magnetic field $\mathbf{B}\mathbf{}\mathbf{⃗}\mathbf{=}\mathbf{B}\hat{\mathbf{i}}$ with an initial velocity$\mathbf{v}\mathbf{⃗}\mathbf{=}{\mathbf{v}}_{\mathbf{0}}\mathbf{x}\hat{\mathbf{i}}\mathbf{+}{\mathbf{v}}_{\mathbf{0}}\mathbf{y}\hat{\mathbf{j}}$.Find an expression in unit-vector notation for its velocity at any later time t.

Short answer

The expression for velocity $\overrightarrow{v}$at any later time t is$v⃗\left(t\right)=v_{0x}\hat{i}+v_{0y}cos\left(\omega t\right)\hat{j}-v_{0y}sin\left(\omega t\right)\hat{k}.$.

Step-by-step solution

Identification of given data

1. The charge of the proton is q=+e.
2. The uniform magnetic field is $\overrightarrow{B}=B\hat{i}$

3. The initial velocity is role="math" localid="1662913225328" $v⃗=v_{0x}\hat{i}+v_{0y}\hat{j}.$

Understanding the concept

By substituting the formula for force in the equation of the motion for the proton and solving it, we can find the expression for velocity $\overrightarrow{v}$at any later time t.

Formula:

1. The equation of the motion for the proton is given by$F⃗=qv⃗\times B⃗$
2. The force is$F⃗=m_{p}a⃗$

Determining the expression for velocity v→ at any later time t

The equation of the motion for the proton is given by

$$\begin{gathered} F⃗=qv⃗\times B⃗ \\ F⃗=q(v_x\hat{i}+v_y\hat{j}+v_z\hat{k})\times B\hat{i} \end{gathered}$$

$F⃗=qB(v_z\hat{j}-v_y\hat{k})$ …(i)

But, we know that,$F⃗=m_{p}a⃗$, where m_p is the mass of the proton.

localid="1662914279886" $F⃗=m_p\left(\right(\frac{d{v}_x}{dt}\hat{i}+\frac{d{v}_y}{dt}\hat{j}+\frac{d{v}_z}{dt}\hat{k})$

And charge q=+e

Substituting in equation (i) we get,

$m_p\left(\right(\frac{d{v}_x}{dt}\hat{i}+\frac{d{v}_y}{dt}\hat{j}+\frac{d{v}_z}{dt}\hat{k})=eB\left(v_z\hat{j}-v_y\hat{k}\right)$

Therefore, we have

$\left(\right(\frac{d{v}_x}{dt}\hat{i}+\frac{d{v}_y}{dt}\hat{j}+\frac{d{v}_z}{dt}\hat{k})=\frac{eB}{m_p}\left(v_z\hat{j}-v_y\hat{k}\right)$

But,$\frac{eB}{m_p}=\omega$, thus,

$(\frac{d{v}_x}{dt}\hat{i}+\frac{d{v}_y}{dt}\hat{j}+\frac{d{v}_z}{dt}\hat{k})=\omega \left(v_z\hat{j}-v_y\hat{k}\right)$

Therefore,$\frac{d{v}_x}{dt}=0$,$\frac{d{v}_y}{dt}=\omega v_z$, and $\frac{d{v}_z}{dt}=-\omega v_y$.

We can solve these equations to get

$v_x=v_{ox}$$,v_y={v_0}_y\cos \left(\omega t\right)$and$V_z=-V_{0y}\sin \left(\omega t\right)$

Thus, the velocity is

$v⃗\left(t\right)=v_{0}x\hat{i}+v_{0y}cos\left(\omega t\right)\hat{j}-v_{oy}sin\left(\omega t\right)\hat{k}$