Question

Laboratory scientists have created the electric and magnetic fields shown in FIGURE EX31.3. These fields are also seen by scientists who zoom past in a rocket traveling in the x-direction at $1.0x{10}^{6}m/s$ . According to the rocket scientists, what angle does the electric field make with the axis of the rocket?

Short answer

Angle according to the rocket scientists is $\theta =16.3°$above the x-axis.

Step-by-step solution

Given information

We have given,

Magnetic field $B=0.5T$(along positive z-axis).

Electric field has two components,

$$\begin{gathered} E_x=\frac{1x{10}^6}{\sqrt{2}}V/m \\ {E}_y=\frac{1x{10}^6}{\sqrt{2}}V/m \end{gathered}$$ (along positive x and y axis)

Velocity of the rocket $V_r=1.0x{10}^{6}m/s$.

We need to find the angle made by electric field with the axis of the rocket.

Simplify.

In the frame of rocket scientist the electric field due to movement of rocket in the magnetic field will also be considered.

So, new electric field $E\text{'}$is:

$E\text{'}=E_{initial}+V_{r}xB$

role="math" localid="1649864938158" $$\begin{gathered} E\text{'}=1.0x{10}^6\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)+1.0x{10}^6\hat{i}x0.5\hat{k}V/m \\ E\text{'}=1.0x{10}^6\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)-0.5x{10}^6\hat{j}V/m \\ E\text{'}=\frac{1.0x{10}^6}{\sqrt{2}}\hat{i}+\left(\frac{1.0x{10}^6}{\sqrt{2}}-0.5x{10}^6\right)\hat{j}V/m \\ E\text{'}=\frac{1.0x{10}^6}{\sqrt{2}}\hat{i}+\left(\frac{0.293x{10}^6}{\sqrt{2}}\right)\hat{j}V/m \end{gathered}$$

For find the angle with the x-axis:

$$\begin{gathered} \tan \theta =\frac{E_y}{E_x} \\ \tan \theta =\frac{0.293x{10}^6}{1.0x{10}^6} \\ \tan \theta =0.293 \\ \theta ={\tan }^{-1}(0.293) \\ \theta =16.3^{\circ } \end{gathered}$$

This angle is above positive x-axis.