As a charged particle 'q' moving with a velocity `vec(v)` enters a uniform magnetic field `vec(B)`, it experience a force `vec(F) = q(vec(v) xx vec(B)). For theta = 0^(@) or 180^(@), theta` being the angle between `vec(v) and vec(B)`, force experienced is zero and the particle passes undeflected. For `theta = 90^(@)`, the particle moves along a circular arc and the magnetic force (qvB) provides the necessary centripetal force `(mv^(2)//r)`. For other values of `theta (theta !=0^(@), 180^(@), 90^(@))`, the charged particle moves along a helical path which is the resultant motion of simultaneous circular and translational motions.
Suppose a particle that carries a charge of magnitude q and has a mass `4 xx 10^(-15)` kg is moving in a region containing a uniform magnetic field `vec(B) = -0.4 hat(k) T`. At some instant, velocity of the particle is `vec(v) = (8 hat(i) - 6 hat(j) 4 hat(k)) xx 10^(6) m s^(-1)` and force acting on it has a magnitude 1.6 N
If the coordinates of the particle at t = 0 are (2 m, 1 m, 0), coordinates at a time t = 3 T, where T is the time period of circular component of motion. will be (take `pi = 3.14`)

Options

A. (2 m, 1 m, 0.942 m)
B. (0.142 m, 130m, 0)
C. (2 m, 1 m, 1.884 m)
D. (142 m, 130 m, 628 m)

Explanation

To solve the problem step by step, we will follow these procedures:

### Step 1: Identify the Given Data
- Mass of the particle, \( m = 4 \times 10^{-15} \) kg
- Magnetic field, \( \vec{B} = -0.4 \hat{k} \) T
- Velocity of the particle, \( \vec{v} = (8 \hat{i} - 6 \hat{j} + 4 \hat{k}) \times 10^6 \) m/s
- Magnitude of the force, \( F = 1.6 \) N

### Step 2: Calculate the Charge of the Particle
Using the equation for magnetic force:
\[ F = q (\vec{v} \times \vec{B}) \]
We need to calculate \( \vec{v} \times \vec{B} \).

#### Step 2.1: Compute the Cross Product \( \vec{v} \times \vec{B} \)
Set up the determinant:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 8 \times 10^6 & -6 \times 10^6 & 4 \times 10^6 \\ 0 & 0 & -0.4 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{v} \times \vec{B} = \hat{i} \left((-6 \times 10^6)(-0.4) - (4 \times 10^6)(0)\right) - \hat{j} \left((8 \times 10^6)(-0.4) - (4 \times 10^6)(0)\right) + \hat{k} \left((8 \times 10^6)(0) - (-6 \times 10^6)(0)\right) \]
\[ = \hat{i} (2.4 \times 10^6) - \hat{j} (-3.2 \times 10^6) + 0 \]
\[ = (2.4 \hat{i} + 3.2 \hat{j}) \times 10^6 \]

#### Step 2.2: Calculate the Magnitude of \( \vec{v} \times \vec{B} \)
\[ |\vec{v} \times \vec{B}| = \sqrt{(2.4 \times 10^6)^2 + (3.2 \times 10^6)^2} \]
\[ = \sqrt{5.76 \times 10^{12} + 10.24 \times 10^{12}} = \sqrt{16 \times 10^{12}} = 4 \times 10^6 \]

#### Step 2.3: Solve for Charge \( q \)
Using \( F = q |\vec{v} \times \vec{B}| \):
\[ 1.6 = q (4 \times 10^6) \]
\[ q = \frac{1.6}{4 \times 10^6} = 0.4 \times 10^{-6} \text{ C} = 4 \times 10^{-7} \text{ C} \]

### Step 3: Calculate the Angular Frequency \( \omega \)
The formula for angular frequency is:
\[ \omega = \frac{qB}{m} \]
Substituting the values:
\[ \omega = \frac{(0.4 \times 10^{-6})(0.4)}{4 \times 10^{-15}} = \frac{0.16 \times 10^{-6}}{4 \times 10^{-15}} = 4 \times 10^7 \text{ rad/s} \]

### Step 4: Calculate the Time Period \( T \)
The time period \( T \) is given by:
\[ T = \frac{2\pi}{\omega} \]
Substituting \( \pi = 3.14 \):
\[ T = \frac{2 \times 3.14}{4 \times 10^7} = \frac{6.28}{4 \times 10^7} = 1.57 \times 10^{-7} \text{ s} \]

### Step 5: Determine the Coordinates at \( t = 3T \)
The particle completes 3 full circles in time \( 3T \). The circular motion does not change the \( x \) and \( y \) coordinates, but the \( z \) coordinate changes due to the helical motion.

#### Step 5.1: Calculate the Pitch
The pitch \( P \) is given by:
\[ P = v_z \cdot T \]
Where \( v_z = 4 \times 10^6 \) m/s:
\[ P = (4 \times 10^6) \cdot (1.57 \times 10^{-7}) = 0.628 \text{ m} \]

#### Step 5.2: Calculate the Change in \( z \) Coordinate
In \( 3T \):
\[ \Delta z = 3P = 3 \times 0.628 = 1.884 \text{ m} \]

### Step 6: Final Coordinates
Initial coordinates at \( t = 0 \) are \( (2, 1, 0) \). After \( 3T \):
\[ \text{Final coordinates} = (2, 1, 0 + 1.884) = (2, 1, 1.884) \]

### Final Answer
The coordinates of the particle at \( t = 3T \) are \( (2, 1, 1.884) \).

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