A charged particle with charge to mass ratio `((q)/(m)) = (10)^(3)/(19) Ckg^(-1)` enters a uniform magnetic field `vec(B) = 20 hat(i) + 30 hat(j) + 50 hat(k) T` at time t = 0 with velocity `vec(V) = (20 hat(i) + 50 hat(j) + 30 hat(k)) m//s`. Assume that magnetic field exists in large space.
The frequency (in Hz) of the revolution of the particle in cycles per second will be

Options

A. `(10^3)/(pisqrt(38))`
B. `(10^4)/(pisqrt(38))`
C. `(10^4)/(pisqrt(19))`
D. `(10^4)/(2pisqrt(19))`

Explanation

To find the frequency of revolution of a charged particle in a magnetic field, we can use the following steps:

### Step 1: Identify the Given Values
- Charge to mass ratio: \(\frac{q}{m} = \frac{10^3}{19} \, \text{C/kg}\)
- Magnetic field: \(\vec{B} = 20\hat{i} + 30\hat{j} + 50\hat{k} \, \text{T}\)
- Velocity: \(\vec{V} = 20\hat{i} + 50\hat{j} + 30\hat{k} \, \text{m/s}\)

### Step 2: Calculate the Magnitude of the Magnetic Field
The magnitude of the magnetic field \(|\vec{B}|\) can be calculated using the formula:
\[ |\vec{B}| = \sqrt{B_x^2 + B_y^2 + B_z^2} \]
Substituting the values:
\[ |\vec{B}| = \sqrt{20^2 + 30^2 + 50^2} = \sqrt{400 + 900 + 2500} = \sqrt{3800} \]

### Step 3: Calculate the Frequency of Revolution
The frequency \(f\) of the charged particle in a magnetic field is given by the formula:
\[ f = \frac{q}{2\pi m} |\vec{B}| \]
Using the charge to mass ratio \(\frac{q}{m}\):
\[ f = \frac{q/m}{2\pi} |\vec{B}| \]
Substituting the values:
\[ f = \frac{10^3/19}{2\pi} \sqrt{3800} \]

### Step 4: Simplify the Expression
First, calculate \(\sqrt{3800}\):
\[ \sqrt{3800} = \sqrt{100 \times 38} = 10\sqrt{38} \]
Now substitute this back into the frequency equation:
\[ f = \frac{10^3}{19 \cdot 2\pi} \cdot 10\sqrt{38} \]
\[ f = \frac{10^4 \sqrt{38}}{19 \cdot 2\pi} \]

### Step 5: Final Calculation
Now, we can plug in the value of \(\pi \approx 3.14\):
\[ f = \frac{10^4 \sqrt{38}}{38\pi} \]
This gives us the final expression for the frequency of the revolution of the particle.

### Final Answer
The frequency \(f\) of the revolution of the particle in cycles per second is:
\[ f \approx \frac{10^4 \sqrt{38}}{38 \cdot 3.14} \]