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, we will follow these steps:

### Step 1: Identify the Given Values
- Charge of the particle: \( q \) (unknown)
- 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
- Initial coordinates: \( (x_0, y_0, z_0) = (2 \text{ m}, 1 \text{ m}, 0) \)
- Time: \( t = 3T \) where \( T \) is the time period of circular motion.

### Step 2: Calculate the Cross Product \( \vec{v} \times \vec{B} \)
Using the formula for the magnetic force:
\[ \vec{F} = q (\vec{v} \times \vec{B}) \]
First, we need to compute \( \vec{v} \times \vec{B} \):
\[ \vec{v} = (8 \hat{i} - 6 \hat{j} + 4 \hat{k}) \times 10^6 \]
\[ \vec{B} = -0.4 \hat{k} \]
The cross product is calculated using 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 this determinant:
\[ = \hat{i} \left((-6 \times 10^6)(-0.4) - (4 \times 10^6)(0)\right) - \hat{j} \left((8 \times 10^6)(-0.4) - (4 \times 10^6)(0)\right) + \hat{k} \left((8 \times 10^6)(0) - (-6 \times 10^6)(0)\right) \]
\[ = \hat{i} (2.4 \times 10^6) + \hat{j} (3.2 \times 10^6) + 0 \]
Thus,
\[ \vec{v} \times \vec{B} = (2.4 \hat{i} + 3.2 \hat{j}) \times 10^6 \]

### Step 3: Calculate the Magnitude of \( \vec{v} \times \vec{B} \)
The magnitude is given by:
\[ |\vec{v} \times \vec{B}| = \sqrt{(2.4 \times 10^6)^2 + (3.2 \times 10^6)^2} \]
Calculating this:
\[ = \sqrt{(5.76 \times 10^{12}) + (10.24 \times 10^{12})} = \sqrt{16 \times 10^{12}} = 4 \times 10^6 \]

### Step 4: Solve for Charge \( q \)
Using the force equation:
\[ F = q |\vec{v} \times \vec{B}| \]
Substituting the known values:
\[ 1.6 = q (4 \times 10^6) \]
Thus,
\[ q = \frac{1.6}{4 \times 10^6} = 0.4 \times 10^{-6} \text{ C} \]

### Step 5: Calculate Angular Frequency \( \omega \)
Using the formula:
\[ \omega = \frac{qB}{m} \]
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 6: Calculate the Time Period \( T \)
The time period \( T \) is given by:
\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{4 \times 10^7} \approx \frac{6.28}{4 \times 10^7} \approx 1.57 \times 10^{-8} \text{ s} \]

### Step 7: Calculate the Distance Travelled in the Z-Direction
The velocity in the z-direction is \( 4 \times 10^6 \) m/s. The distance travelled in the z-direction in time \( t = 3T \):
\[ \text{Distance} = v_z \times t = 4 \times 10^6 \times (3 \times 1.57 \times 10^{-8}) \approx 0.1884 \text{ m} \]

### Step 8: Find the New Coordinates
The new coordinates after time \( t = 3T \):
\[ (x, y, z) = (2, 1, 0 + 0.1884) = (2, 1, 0.1884) \]

### Final Answer
The coordinates of the particle at \( t = 3T \) are:
\[ (2 \text{ m}, 1 \text{ m}, 0.1884 \text{ m}) \]