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
Angular frequency of rotation of particle, also called the `cyclotron frequency' is

Options

A. `8 xx 10^(5) rad s^(-1)`
B. `12.5 xx 10^(4) rad s^(-1)`
C. `6.2 xx 10^(6) rad s^(-1)`
D. `4 xx 10^(7) rad s^(-1)`

Explanation

To solve the problem step by step, we need to find the angular frequency (cyclotron frequency) of a charged particle moving in a magnetic field. Here’s how we can do it:

### Step 1: Understand the Force on the Charged Particle
The force experienced by a charged particle in a magnetic field is given by:
\[ \vec{F} = q (\vec{v} \times \vec{B}) \]
where \( q \) is the charge of the particle, \( \vec{v} \) is the velocity vector, and \( \vec{B} \) is the magnetic field vector.

### Step 2: Given Values
From the problem, we have:
- 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 force, \( F = 1.6 \) N

### Step 3: Calculate the Charge \( q \)
Using the formula for the magnitude of the force:
\[ F = q |\vec{v} \times \vec{B}| \]
we need to calculate \( |\vec{v} \times \vec{B}| \).

### Step 4: Calculate the Cross Product \( \vec{v} \times \vec{B} \)
First, we write the vectors:
\[ \vec{v} = (8 \hat{i} - 6 \hat{j} + 4 \hat{k}) \times 10^6 \]
\[ \vec{B} = -0.4 \hat{k} \]

Now, we compute the cross product:
\[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 8 \times 10^6 & -6 \times 10^6 & 4 \times 10^6 \\ 0 & 0 & -0.4 \end{vmatrix} \]

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

### Step 5: Calculate the Magnitude of \( \vec{v} \times \vec{B} \)
\[ |\vec{v} \times \vec{B}| = \sqrt{(2.4 \times 10^6)^2 + (3.2 \times 10^6)^2} \]
Calculating this:
\[ = \sqrt{(5.76 \times 10^{12}) + (10.24 \times 10^{12})} = \sqrt{16 \times 10^{12}} = 4 \times 10^6 \]

### Step 6: Substitute Back to Find \( q \)
Now, substituting back into the force equation:
\[ F = q |\vec{v} \times \vec{B}| \]
\[ 1.6 = q (4 \times 10^6) \]
Solving for \( q \):
\[ q = \frac{1.6}{4 \times 10^6} = 0.4 \times 10^{-6} \text{ C} = 4 \times 10^{-7} \text{ C} \]

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

### Final Answer
The angular frequency of rotation of the particle is:
\[ \omega = 4 \times 10^{7} \text{ rad/s} \]

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