Question

A positive charge coasts upward at a constant velocity for a long time. Then at $\mathit{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$(Figure 23.120) a force acts downward on it for $\mathbf{1}\mathbf{}\mathit{n}\mathit{s}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{}\mathbf{s}\mathbf{)}$. After this force stops acting, the charge coasts upward at a smaller constant speed for 1 ns; then a force acts upward for and it resumes its original speed. The new position reached at $\mathit{t}\mathbf{=}\mathbf{3}\mathbf{}\mathit{n}\mathit{s}$is much less than a millimeter from the original position.

You stand at location A $\mathbf{30}\mathbf{}\mathit{m}$, to the right of the charge (Figure 23.120), with instruments for measuring electric and magnetic fields. What will you observe due to the motion of the positive charge, at what times? You do not need to calculate the magnitudes of the electric and magnetic fields, but you do need to specify their directions, and the times when these fields are observed.

Short answer

Hence the time taken by the radiation pulse to react at the observer position is$100\text{ns}$ and direction of the fields is out of the page.

Step-by-step solution

Identification of given data

The distance covered by the charge,$L=30\text{m}$

At ,$t=0$ a force act on charge for the $1\text{ns}$.

Determine the formulas to calculate time forradiation pulse to reach at the point A.

The expression to calculate the time for radiation pulse to reach at point$A$ is given as follows.

role="math" localid="1668588905933" $\mathit{t}\mathbf{=}\frac{\mathbf{L}}{\mathbf{c}}$ …(i)

Here, c is the speed of the light.

Determining the time for radiation pulse to reach at the point A.

Calculate the time to reach at the point A.

Substitute $3\times {10}^{8}m/s$for $\mathrm{c}$cand$30m$for L into equation (i).

$$\begin{gathered} t=\frac{30 \text{m} }{3\times {10}^{8}m/s} \\ =10\times {10}^{-8} \text{s} \\ =100\times {10}^{-9} \text{s} \\ =100\text{ns} \end{gathered}$$

Let assume observer is at position $A$. The propagation of wave, direction of magnetic field and electric field are shown below.

The direction of acceleration from $t=0\text{nsto}t=1\text{ns}$is downward and the direction of radiative field from $100\text{nsto}101\text{ns}$upward. Since the direction of the propagation is $v=E\times B$the right therefore the direction of the magnetic field is out of the page.

From role="math" localid="1668589871682" $101\text{nsto}102\text{ns}$the radiative field is zero due to zero acceleration. From $t=102\text{nsto}103\text{ns}$, there is radiative electric field in the downward direction and propagation direction is $v=E\times B$. Therefore, the direction of the magnetic field is out of the page.