Magnetic field exist in the space and given as `vecB=-(B_(0))/(l^(2))x^(2)hatk`, where `B_(0)`and `l` positive constants. A particle having positive charge `'q'` and mass `'m'` is pojected wit speed `'v_(0)'` along positive `x` axis from the origin. What is the maimum distance of the charged particle from the `y`-axis before it turns back due to the magnetic field. (Ignore any interaction other than magnetic field)

Options

A. `((ml^(2)v_(0))/(3qB_(0)))^(1//3)`
B. `((3ml^(2)v_(0))/(2qB_(0)))^(1//3)`
C. `((3ml^(2)v_(0))/(qB_(0)))^(1//3)`
D. `((ml^(2)v_(0))/(qB_(0)))^(1//3)`

Explanation

To solve the problem, we need to analyze the motion of a charged particle in a magnetic field. The magnetic field is given as:

\[ \vec{B} = -\frac{B_0}{l^2} x^2 \hat{k} \]

where \(B_0\) and \(l\) are positive constants. The particle has a positive charge \(q\) and mass \(m\), and it is projected with an initial speed \(v_0\) along the positive x-axis from the origin.

### Step 1: Determine the Force on the Charged Particle

The magnetic force acting on a charged particle moving in a magnetic field is given by the Lorentz force equation:

\[ \vec{F} = q \vec{v} \times \vec{B} \]

Here, the velocity vector \(\vec{v}\) is along the x-axis, which can be represented as:

\[ \vec{v} = v_0 \hat{i} \]

Substituting \(\vec{v}\) and \(\vec{B}\) into the Lorentz force equation:

\[ \vec{F} = q (v_0 \hat{i}) \times \left(-\frac{B_0}{l^2} x^2 \hat{k}\right) \]

### Step 2: Calculate the Cross Product

Using the right-hand rule for the cross product, we have:

\[ \hat{i} \times \hat{k} = \hat{j} \]

Thus, the force becomes:

\[ \vec{F} = -\frac{q B_0}{l^2} x^2 v_0 \hat{j} \]

This indicates that the force acts in the negative y-direction.

### Step 3: Write the Equation of Motion

The particle will experience a downward force due to the magnetic field, which will cause it to accelerate in the y-direction. The equation of motion in the y-direction can be written as:

\[ m \frac{d^2 y}{dt^2} = -\frac{q B_0}{l^2} x^2 v_0 \]

### Step 4: Analyze the Motion

Since the particle is moving along the x-axis, we can express \(x\) as a function of time:

\[ x = v_0 t \]

Substituting this into the force equation gives:

\[ m \frac{d^2 y}{dt^2} = -\frac{q B_0}{l^2} (v_0 t)^2 v_0 \]

This simplifies to:

\[ m \frac{d^2 y}{dt^2} = -\frac{q B_0 v_0^3}{l^2} t^2 \]

### Step 5: Integrate to Find the Position

Integrating once with respect to time:

\[ \frac{dy}{dt} = -\frac{q B_0 v_0^3}{3m l^2} t^3 + C_1 \]

Integrating again gives:

\[ y = -\frac{q B_0 v_0^3}{12m l^2} t^4 + C_1 t + C_2 \]

### Step 6: Determine Constants of Integration

At \(t = 0\), the particle starts at the origin, hence:

\[ y(0) = 0 \implies C_2 = 0 \]

The initial velocity in the y-direction is zero, so:

\[ \frac{dy}{dt}(0) = 0 \implies C_1 = 0 \]

Thus, the equation simplifies to:

\[ y = -\frac{q B_0 v_0^3}{12m l^2} t^4 \]

### Step 7: Find Maximum Distance from the y-axis

The maximum distance from the y-axis occurs when the particle turns back, which is when \(y\) reaches its maximum value. This will happen when the particle stops moving in the y-direction, which will occur when the force acting on it becomes zero.

To find the maximum distance, we need to determine the time at which the particle turns back. This can be done by setting the acceleration to zero, which corresponds to the time when the particle reaches its maximum y-displacement.

### Final Answer

The maximum distance from the y-axis can be calculated by substituting the time corresponding to the maximum y-displacement back into the equation for \(x\):

\[ x_{\text{max}} = v_0 t_{\text{max}} \]

This will yield the maximum distance of the charged particle from the y-axis before it turns back due to the magnetic field.