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 would the particle move in a straight line along the negative direction of y-axis (i.e., move along `haty)`

Options

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

Answer

To determine in which case the charged particle (either an electron or a proton) would move in a straight line along the negative direction of the y-axis, we need to analyze the forces acting on the particle due to the electric and magnetic fields.

### Step-by-Step Solution:

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

2. **Analyzing Each Case**:
We will analyze each case provided in the columns to see if the net force can be zero.

- **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, \(\vec{v} \times \vec{B} = (2\frac{E_0}{B_0}\hat{x}) \times (-B_0\hat{x}) = 0\).
- Therefore, \(\vec{E} + 0 \neq 0\). This case does not yield a straight line motion.

- **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, \(\vec{E} + E_0\hat{z} \neq 0\). This case does not yield a straight line motion.

- **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. The electric force is \(-E_0\hat{x}\), but since the particle is not moving, it can be considered to move in a straight line along the y-axis. This case can yield straight line motion.

- **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, \(\vec{E} + 2E_0\hat{y} \neq 0\). This case does not yield a straight line motion.

3. **Conclusion**:
The only case where the particle can move in a straight line along the negative direction of the y-axis is **Case (iii)**, where the particle is at rest (\(\vec{v} = 0\)).

### Final Answer:
The particle would move in a straight line along the negative direction of the y-axis in **Case (iii)**.