A charged particle moves in a uniform magnetic field `B=(2hati-3hatj)` T

Options

A. (a) if velocity of the particles is `(6hatK)m//s`, particle moves in a circle
B. (b) if velocity of the particle is `(-4hati+6hatj)m//s`, particle moves in a straight me
C. (c) if velocity of the particle is `(hati+2hatj)m//s`, particle moves in a helical path
D. (d) In all the above three cases speed of the particle remains unchanged

Answer

To solve the problem of a charged particle moving in a uniform magnetic field \( \mathbf{B} = 2\hat{i} - 3\hat{j} \) T, we will analyze the motion based on different velocity vectors provided in the question. The magnetic force acting on the charged particle is given by the equation:

\[ \mathbf{F} = q(\mathbf{v} \times \mathbf{B}) \]

where \( q \) is the charge of the particle, \( \mathbf{v} \) is the velocity of the particle, and \( \mathbf{B} \) is the magnetic field.

### Step 1: Analyze the first case
**Given:** Velocity \( \mathbf{v} = 6\hat{k} \) m/s.

- The magnetic field \( \mathbf{B} = 2\hat{i} - 3\hat{j} \) is in the XY-plane.
- The velocity \( \mathbf{v} = 6\hat{k} \) is in the Z-direction.

**Analysis:**
- Since \( \mathbf{v} \) is perpendicular to \( \mathbf{B} \), the angle \( \theta \) between \( \mathbf{v} \) and \( \mathbf{B} \) is \( 90^\circ \).
- The force \( \mathbf{F} \) will be maximum and will cause the particle to move in a circular path.

**Conclusion:** The particle moves in a circular path.

### Step 2: Analyze the second case
**Given:** Velocity \( \mathbf{v} = -4\hat{i} + 6\hat{j} \) m/s.

- We can rewrite this as \( \mathbf{v} = -2(2\hat{i} - 3\hat{j}) \).

**Analysis:**
- The velocity vector is anti-parallel to the magnetic field vector \( \mathbf{B} \).
- The angle \( \theta \) between \( \mathbf{v} \) and \( \mathbf{B} \) is \( 180^\circ \).
- The magnetic force \( \mathbf{F} \) will be zero because \( \sin(180^\circ) = 0 \).

**Conclusion:** The particle moves in a straight line.

### Step 3: Analyze the third case
**Given:** Velocity \( \mathbf{v} = \hat{i} + 2\hat{j} \) m/s.

**Analysis:**
- The angle \( \theta \) between \( \mathbf{v} \) and \( \mathbf{B} \) is neither \( 0^\circ \) nor \( 90^\circ \) nor \( 180^\circ \).
- Since the velocity is at an arbitrary angle to the magnetic field, the particle will experience a magnetic force that will cause it to move in a helical path.

**Conclusion:** The particle moves in a helical path.

### Step 4: Analyze the fourth case
**Conclusion for all cases:**
- In all three cases, the magnetic force does not do work on the charged particle because the magnetic force is always perpendicular to the velocity of the particle.
- Therefore, the speed of the particle remains unchanged in all cases.

### Final Conclusion:
All four options provided in the question are correct:
- (A) The particle moves in a circle.
- (B) The particle moves in a straight line.
- (C) The particle moves in a helical path.
- (D) The speed of the particle remains unchanged in all cases.