A charged particle carrying charge `q=10 muC` moves with velocity `v_1=10^6 ,s^-1` at angle `45^@` with x-axis in the xy plane and experience a force `F_1=5sqrt2 mN` along the negative z-axis. When the same particle moves with velocity `v_2=10^6 ms^-1` along the z-axis, it experiences a force `F_2` in y-direction.
Find the magnitude of the force `F_2`.

Options

A. `10^-2N`
B. `10^-3N`
C. `10^-4N`
D. `10^-5N`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We have a charged particle with:
- Charge \( q = 10 \, \mu C = 10 \times 10^{-6} \, C \)
- Velocity \( v_1 = 10^6 \, m/s \) at an angle of \( 45^\circ \) with the x-axis in the xy-plane.
- Force \( F_1 = 5\sqrt{2} \, mN = 5\sqrt{2} \times 10^{-3} \, N \) acting along the negative z-axis.

### Step 2: Convert Velocity into Components
The velocity \( v_1 \) can be broken down into its x and y components:
- \( v_{1x} = v_1 \cos(45^\circ) = 10^6 \cdot \frac{1}{\sqrt{2}} = 5 \times 10^5 \, m/s \)
- \( v_{1y} = v_1 \sin(45^\circ) = 10^6 \cdot \frac{1}{\sqrt{2}} = 5 \times 10^5 \, m/s \)

Thus, we can write:
\[ \mathbf{v_1} = (5 \times 10^5 \hat{i} + 5 \times 10^5 \hat{j}) \, m/s \]

### Step 3: Use the Lorentz Force Equation
The force experienced by a charged particle in a magnetic field is given by:
\[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \]
Given that \( F_1 \) acts in the negative z-direction, we can express this as:
\[ \mathbf{F_1} = 5\sqrt{2} \times 10^{-3} \hat{k} \, N \]

### Step 4: Find the Magnetic Field \( \mathbf{B} \)
From the equation \( \mathbf{F_1} = q (\mathbf{v_1} \times \mathbf{B}) \), we can rearrange it to find \( \mathbf{B} \):
\[ \mathbf{B} = \frac{\mathbf{F_1}}{q \cdot \mathbf{v_1}} \]

### Step 5: Calculate the Magnitude of the Magnetic Field
Using the magnitude of \( \mathbf{F_1} \):
\[ |\mathbf{F_1}| = 5\sqrt{2} \times 10^{-3} \, N \]
Substituting the values:
\[ \mathbf{B} = \frac{5\sqrt{2} \times 10^{-3}}{10 \times 10^{-6} \cdot 10^6} = \frac{5\sqrt{2} \times 10^{-3}}{10 \times 10^{-6} \cdot 10^6} = 10^{-3} \, T \hat{i} \]

### Step 6: Analyze the Second Situation
Now, when the particle moves with velocity \( v_2 = 10^6 \, m/s \) along the z-axis, we have:
\[ \mathbf{v_2} = 10^6 \hat{k} \, m/s \]

### Step 7: Calculate the Force \( F_2 \)
Using the Lorentz force equation again:
\[ \mathbf{F_2} = q (\mathbf{v_2} \times \mathbf{B}) \]
Substituting the values:
\[ \mathbf{F_2} = 10 \times 10^{-6} \cdot (10^6 \hat{k} \times 10^{-3} \hat{i}) \]

### Step 8: Perform the Cross Product
Using the right-hand rule:
\[ \hat{k} \times \hat{i} = -\hat{j} \]
Thus:
\[ \mathbf{F_2} = 10 \times 10^{-6} \cdot 10^{-3} (-\hat{j}) = -10^{-8} \hat{j} \, N \]

### Step 9: Find the Magnitude of \( F_2 \)
The magnitude of \( F_2 \) is:
\[ |F_2| = 10^{-8} \, N \]

### Final Answer
The magnitude of the force \( F_2 \) is \( 10^{-8} \, N \).

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