To solve this problem, let’s analyze the influence of the magnetic and electric fields on the charge's motion.
Given:
- Total charge [tex]\(q\)[/tex]
- Mass [tex]\(m\)[/tex]
- Initial velocity [tex]\( \vec{v}_0 = v_0 \hat{i} + v_0 \hat{j} \)[/tex]
- Magnetic field [tex]\( \vec{B}_0 = B_0 \hat{j} \)[/tex]
- Electric field [tex]\( \vec{E}_0 = E_0 \hat{j} \)[/tex]
The pitch of a helical path is the distance a charged particle moves parallel to the magnetic field direction in one complete gyration.
In this scenario:
- We have [tex]\(\vec{B}_0\)[/tex] along the [tex]\(y\)[/tex]-axis.
- The force from the magnetic field will cause the particle to follow a helical path around the magnetic field lines.
- The velocity component [tex]\(v_0 \hat{i}\)[/tex] (along [tex]\(x\)[/tex]) and [tex]\(v_0 \hat{j}\)[/tex] (along [tex]\(y\)[/tex]) need to be considered for our calculations.
For the pitch [tex]\(P\)[/tex], the distance moved along the magnetic field direction after one complete revolution, we use:
[tex]\[ P = v_{\parallel} T \][/tex]
Here:
- [tex]\( v_{\parallel} \)[/tex] is the velocity component parallel to [tex]\(\vec{B}_0\)[/tex]. Since the initial velocity in the [tex]\(y\)[/tex]-direction is [tex]\(v_0 \hat{j} \)[/tex] and the electric field is also along [tex]\(y\)[/tex]-axis, we need to consider the component along [tex]\(\hat{j}\)[/tex].
- [tex]\( T \)[/tex] is the time period of one circular revolution.
### Time Period [tex]\(T\)[/tex]:
[tex]\[ T = \frac{2\pi m}{q B_0} \][/tex]
### Parallel Velocity [tex]\(v_{\parallel}\)[/tex]:
Initially, [tex]\( v_{\parallel} = v_0 \)[/tex].
Including the effect of [tex]\(E_0\)[/tex]:
[tex]\[ v_{\parallel} = v_0 + \frac{E_0}{B_0} \][/tex]
Putting it all together:
[tex]\[ P = \left( v_0 + \frac{E_0}{B_0} \right) \times \frac{2\pi m}{q B_0} \][/tex]
From the given expressions, we can see that this matches the third given expression:
[tex]\[ \frac{2 \pi m}{q B} \left( v_0 + \frac{E_0}{B_0} \right) \][/tex]
Let's now confirm the problem's result. The correct numerical result matches:
(3) [tex]\(\frac{2 \pi m}{q B}\left(v_0+\frac{E_0}{B_0}\right)\)[/tex]
By verifying this, we conclude the correct expression for the pitch during the first revolution is indeed:
[tex]\[ \frac{2 \pi m}{q B}\left(v_0+\frac{E_0}{B_0}\right) \][/tex]
Thus, the correct answer is:
[tex]\[ (3) \frac{2 \pi m}{q B}\left(v_0+\frac{E_0}{B_0}\right) \][/tex]
