A charged particle of unit mass and unit charge moves with velocity `vecv=(8hati+6hatj)ms^-1` in magnetic field of `vecB=2hatkT`. Choose the correct alternative (s).

Options

A. (a) the path of the particle may be `x^(2)+y^(2)-4x-21=0`
B. (b) the path of the particle may be `x^(2)+y^(2)=25`
C. (c) the path of the particle may be `y^(2)+z^(2)=25`
D. (d) the time period of the particle will be 3.14 s

Answer

To solve the problem, we need to analyze the motion of a charged particle moving in a magnetic field. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Given Information
- Velocity of the particle: \(\vec{v} = 8\hat{i} + 6\hat{j} \, \text{m/s}\)
- Magnetic field: \(\vec{B} = 2\hat{k} \, \text{T}\)
- Mass of the particle: \(m = 1 \, \text{kg}\) (unit mass)
- Charge of the particle: \(q = 1 \, \text{C}\) (unit charge)

### Step 2: Determine the Magnitude of the Velocity
To find the magnitude of the velocity \(|\vec{v}|\), we use the formula:
\[ |\vec{v}| = \sqrt{(v_x)^2 + (v_y)^2} \]
Substituting the values:
\[ |\vec{v}| = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \, \text{m/s} \]

### Step 3: Check the Perpendicularity of Velocity and Magnetic Field
The velocity vector \(\vec{v}\) is in the xy-plane, and the magnetic field \(\vec{B}\) is along the z-axis. Since they are perpendicular, we can conclude that the motion of the charged particle will be circular.

### Step 4: Calculate the Radius of the Circular Path
The radius \(r\) of the circular path can be calculated using the formula:
\[ r = \frac{mv}{Bq} \]
Substituting the known values:
\[ r = \frac{1 \cdot 10}{2 \cdot 1} = \frac{10}{2} = 5 \, \text{m} \]

### Step 5: Calculate the Time Period of the Circular Motion
The time period \(T\) of the motion is given by:
\[ T = \frac{2\pi m}{Bq} \]
Substituting the values:
\[ T = \frac{2\pi \cdot 1}{2 \cdot 1} = \frac{2\pi}{2} = \pi \, \text{s} \approx 3.14 \, \text{s} \]

### Step 6: Determine the Path of the Particle
The path of the particle is circular in the xy-plane with a radius of 5 m. The general equation of a circle in the xy-plane is:
\[ x^2 + y^2 = r^2 \]
Substituting \(r = 5\):
\[ x^2 + y^2 = 25 \]

### Conclusion
Based on the calculations:
- The radius of the circular path is \(5 \, \text{m}\).
- The time period of the motion is approximately \(3.14 \, \text{s}\).
- The equation of the path is \(x^2 + y^2 = 25\).