A charged particle (electron or proton) is introduced at the origin `(x = 0, y = 0, z = 0`) with a given initial velocity `vecv`. A uniform electric field `vecE` and a uniform magnetic field `vecB` exist everywhere. The velocity `vecv`, electric field `vecE` and magnetic field `vecB` are given in columns 1,2 and 3 respectively. The quantities `E_(0), B_(0)` are positive in magnitude.
`{:("Column-1","Column-2","Column-3"),("Electron with" vecv=2(E_(0))/(B_(0))hatx, (i)vecE=E_(0)vecz,(P) vecB=-B_(0)hatx),((II)"Electron with" vecv=(E_(0))/(B_(0))haty,(ii)vecE=-E_(0)hatx,(Q)vecB=B_(0)hatx),((VI)"Proton with" vecv=0,(iii)vecE=-E_(0)hatx,(R)vecB=B_(0)haty),((VI)"Proton with" vecv=2(E_(0))/(B_(0))hatx,(iv)vecE=E_(0)hatx,(S)vecB=B_(0)hat2):}`
In which case will the particle move in a straight line with constant velocity ?

Options

A. (IV) (i) (S)
B. (III) (ii) (R)
C. (II) (iii) (S)
D. (III) (iii) (P

Answer

To determine in which case the charged particle (electron or proton) moves in a straight line with constant velocity, we need to analyze the forces acting on the particle due to the electric field \(\vec{E}\) and the magnetic field \(\vec{B}\).

### Step-by-Step Solution:

1. **Understanding the Forces**: The net force on a charged particle in an electric field \(\vec{E}\) and a magnetic field \(\vec{B}\) is given by the Lorentz force equation:
\[ \vec{F} = q\vec{E} + q(\vec{v} \times \vec{B}) \]
For the particle to move in a straight line with constant velocity, the net force \(\vec{F}\) must be zero:
\[ q\vec{E} + q(\vec{v} \times \vec{B}) = 0 \]
This implies:
\[ \vec{E} = -(\vec{v} \times \vec{B}) \]

2. **Analyzing Each Case**: We will analyze each case provided in the columns to check if the forces balance out.

- **Case (i)**:
- \(\vec{v} = 2\frac{E_0}{B_0}\hat{x}\)
- \(\vec{E} = E_0\hat{z}\)
- \(\vec{B} = -B_0\hat{x}\)
- Here, \(\vec{v} \times \vec{B} = (2\frac{E_0}{B_0}\hat{x}) \times (-B_0\hat{x}) = 0\). Thus, \(\vec{E} \neq 0\) and the forces do not balance.

- **Case (ii)**:
- \(\vec{v} = \frac{E_0}{B_0}\hat{y}\)
- \(\vec{E} = -E_0\hat{x}\)
- \(\vec{B} = B_0\hat{x}\)
- Here, \(\vec{v} \times \vec{B} = (\frac{E_0}{B_0}\hat{y}) \times (B_0\hat{x}) = \frac{E_0}{B_0}B_0\hat{z} = E_0\hat{z}\). Thus, \(\vec{E} = -E_0\hat{x}\) does not balance with \(\vec{v} \times \vec{B}\).

- **Case (iii)**:
- \(\vec{v} = 0\)
- \(\vec{E} = -E_0\hat{x}\)
- \(\vec{B} = B_0\hat{y}\)
- Here, since \(\vec{v} = 0\), the magnetic force is zero, and the electric force is non-zero. The particle does not move with constant velocity.

- **Case (iv)**:
- \(\vec{v} = 2\frac{E_0}{B_0}\hat{x}\)
- \(\vec{E} = E_0\hat{x}\)
- \(\vec{B} = B_0\hat{z}\)
- Here, \(\vec{v} \times \vec{B} = (2\frac{E_0}{B_0}\hat{x}) \times (B_0\hat{z}) = 2E_0\hat{y}\). Thus, \(\vec{E} \neq -\vec{v} \times \vec{B}\).

3. **Conclusion**: After analyzing all cases, we find that none of the cases result in a balance of forces that would allow the particle to move in a straight line with constant velocity.

### Final Answer:
None of the given cases allow the particle to move in a straight line with constant velocity.