A non-uniform magnetic field `vecB=B_(0)(1+ y/d)(-hatk)` is present in the region of space between `y=0` and `y=d`. A particle of mass `m` and positive charge `q` has velocity `vecv=(3qB_(0)d)/mhati` at origin `O`. Find the angle made by velocity of the particle withe the positive `x`-axis when it leaves the field. (Ignore any interaction other than magnetic field)

Options

A. `90^(@)`
B. `30^(@)`
C. `60^(@)`
D. `45^(@)`

Explanation

To solve the problem, we need to analyze the motion of a charged particle in a non-uniform magnetic field. The magnetic field is given by \(\vec{B} = B_0(1 + \frac{y}{d})(-\hat{k})\) and the particle has an initial velocity \(\vec{v} = \frac{3qB_0d}{m}\hat{i}\) at the origin. We will find the angle made by the velocity of the particle with the positive x-axis when it leaves the field.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Particle**:
The magnetic force acting on a charged particle moving in a magnetic field is given by:
\[ \vec{F} = q \vec{v} \times \vec{B} \]
Here, \(\vec{v} = \frac{3qB_0d}{m}\hat{i}\) and \(\vec{B} = B_0(1 + \frac{y}{d})(-\hat{k})\).

2. **Calculate the Cross Product**:
We need to calculate \(\vec{v} \times \vec{B}\):
\[ \vec{F} = q \left(\frac{3qB_0d}{m}\hat{i}\right) \times \left(B_0(1 + \frac{y}{d})(-\hat{k})\right) \]
Using the right-hand rule for the cross product:
\[ \hat{i} \times (-\hat{k}) = \hat{j} \]
Therefore, the force becomes:
\[ \vec{F} = -qB_0(1 + \frac{y}{d})\frac{3qB_0d}{m}\hat{j} \]
Simplifying this gives:
\[ \vec{F} = -\frac{3q^2B_0^2d}{m}(1 + \frac{y}{d})\hat{j} \]

3. **Determine the Acceleration**:
The acceleration \(\vec{a}\) can be found using Newton's second law:
\[ \vec{F} = m\vec{a} \implies \vec{a} = \frac{\vec{F}}{m} = -\frac{3q^2B_0^2d}{m^2}(1 + \frac{y}{d})\hat{j} \]

4. **Find the Change in Velocity**:
The velocity in the y-direction changes due to the acceleration. The time spent in the magnetic field can be determined by the distance traveled in the y-direction, which is \(d\). The time \(t\) can be calculated using:
\[ t = \frac{d}{v_x} = \frac{d}{\frac{3qB_0d}{m}} = \frac{m}{3qB_0} \]

5. **Calculate the Change in Velocity in the y-direction**:
The change in velocity in the y-direction is given by:
\[ \Delta v_y = a_y \cdot t = -\frac{3q^2B_0^2d}{m^2}(1 + \frac{d}{d}) \cdot \frac{m}{3qB_0} \]
Simplifying this gives:
\[ \Delta v_y = -\frac{3q^2B_0^2d}{m^2} \cdot \frac{m}{3qB_0} \cdot 2 = -\frac{2qB_0d}{m} \]

6. **Final Velocity Components**:
The final velocity in the y-direction is:
\[ v_y = 0 + \Delta v_y = -\frac{2qB_0d}{m} \]
The final velocity in the x-direction remains unchanged:
\[ v_x = \frac{3qB_0d}{m} \]

7. **Calculate the Angle**:
The angle \(\theta\) with respect to the x-axis is given by:
\[ \tan(\theta) = \frac{v_y}{v_x} = \frac{-\frac{2qB_0d}{m}}{\frac{3qB_0d}{m}} = -\frac{2}{3} \]
Therefore, the angle \(\theta\) is:
\[ \theta = \tan^{-1}\left(-\frac{2}{3}\right) \]

### Final Result:
The angle made by the velocity of the particle with the positive x-axis when it leaves the field is:
\[ \theta \approx -33.69^\circ \]
This indicates that the particle is moving downwards relative to the positive x-axis.