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. The path of the particle may be `x^2+y^2-4x-21=0`
B. The path of the particle may be `x^2+y^2=25`
C. The path of the particle may be `y^2+z^2=25`
D. The time period of the particle will be 3.14 s

Explanation

To solve the problem step by step, we will analyze the motion of a charged particle in a magnetic field.

### Step 1: Identify the given quantities
- 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: Calculate the magnitude of the velocity
The magnitude of the velocity \(|\vec{v}|\) can be calculated using 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: Calculate the radius of the circular path
The radius \(R\) of the circular path of a charged particle moving in a magnetic field is given by the formula:
\[ R = \frac{mv}{qB} \]
Substituting the known values:
\[ R = \frac{1 \cdot 10}{1 \cdot 2} = \frac{10}{2} = 5 \, \text{m} \]

### Step 4: Calculate the time period of the motion
The time period \(T\) of the circular motion can be calculated using the formula:
\[ T = \frac{2\pi m}{qB} \]
Substituting the values:
\[ T = \frac{2\pi \cdot 1}{1 \cdot 2} = \frac{2\pi}{2} = \pi \, \text{s} \approx 3.14 \, \text{s} \]

### Step 5: Analyze the path of the particle
The path of the particle can be described by the equation of a circle in the xy-plane. Since the radius is \(5 \, \text{m}\), the equation can be written as:
\[ x^2 + y^2 = R^2 = 5^2 = 25 \]

### Conclusion
Based on the calculations, the radius of the circular path is \(5 \, \text{m}\), and the time period of the motion is approximately \(3.14 \, \text{s}\). The correct alternatives based on the derived equations are:
- The equation \(x^2 + y^2 = 25\) is correct.
- The time period \(T \approx 3.14 \, \text{s}\) is also correct.