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 magnetic field `vecB`.

Options

A. `(10^-3T)(hati+hatj)`
B. `(2xx10^-3T)hati`
C. `(10^-3T)hati`
D. `(2xx10^-3T)(hati+hatj)`

Explanation

To solve the problem, we need to find the magnetic field \(\vec{B}\) given the conditions of a charged particle moving in a magnetic field. We will break down the solution step by step.

### Step 1: Understand the Given Information
We have a charged particle with:
- Charge \( q = 10 \, \mu C = 10 \times 10^{-6} \, C = 10^{-5} \, C \)
- Velocity \( v_1 = 10^6 \, m/s \) at an angle \( 45^\circ \) 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: Break Down the Velocity Vector
The velocity vector \( \vec{v_1} \) can be expressed in terms of its components:
\[ \vec{v_1} = v_{1x} \hat{i} + v_{1y} \hat{j} \]
Where:
\[ v_{1x} = v_1 \cos(45^\circ) = 10^6 \cdot \frac{1}{\sqrt{2}} \hat{i} = \frac{10^6}{\sqrt{2}} \hat{i} \]
\[ v_{1y} = v_1 \sin(45^\circ) = 10^6 \cdot \frac{1}{\sqrt{2}} \hat{j} = \frac{10^6}{\sqrt{2}} \hat{j} \]
Thus,
\[ \vec{v_1} = \frac{10^6}{\sqrt{2}} \hat{i} + \frac{10^6}{\sqrt{2}} \hat{j} \]

### Step 3: Write the Magnetic Force Equation
The magnetic force \( \vec{F_1} \) is given by:
\[ \vec{F_1} = q \vec{v_1} \times \vec{B} \]
Given that \( \vec{F_1} = -5\sqrt{2} \times 10^{-3} \hat{k} \).

### Step 4: Set Up the Cross Product
Let the magnetic field be:
\[ \vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k} \]
Now, we compute the cross product \( \vec{v_1} \times \vec{B} \):
\[ \vec{v_1} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{10^6}{\sqrt{2}} & \frac{10^6}{\sqrt{2}} & 0 \\ B_x & B_y & B_z \end{vmatrix} \]

Calculating the determinant:
\[ \vec{v_1} \times \vec{B} = \left( \frac{10^6}{\sqrt{2}} B_z - 0 \right) \hat{i} - \left( 0 - \frac{10^6}{\sqrt{2}} B_x \right) \hat{j} + \left( \frac{10^6}{\sqrt{2}} B_y - \frac{10^6}{\sqrt{2}} B_x \right) \hat{k} \]
This simplifies to:
\[ \vec{v_1} \times \vec{B} = \frac{10^6}{\sqrt{2}} B_z \hat{i} + \frac{10^6}{\sqrt{2}} B_x \hat{j} + \left( \frac{10^6}{\sqrt{2}} B_y - \frac{10^6}{\sqrt{2}} B_x \right) \hat{k} \]

### Step 5: Equate Forces
Now substituting in the force equation:
\[ q \left( \frac{10^6}{\sqrt{2}} B_z \hat{i} + \frac{10^6}{\sqrt{2}} B_x \hat{j} + \left( \frac{10^6}{\sqrt{2}} B_y - \frac{10^6}{\sqrt{2}} B_x \right) \hat{k} \right) = -5\sqrt{2} \times 10^{-3} \hat{k} \]

### Step 6: Solve for Components
From the \( \hat{k} \) component:
\[ q \left( \frac{10^6}{\sqrt{2}} B_y - \frac{10^6}{\sqrt{2}} B_x \right) = -5\sqrt{2} \times 10^{-3} \]
Substituting \( q \):
\[ 10^{-5} \left( \frac{10^6}{\sqrt{2}} B_y - \frac{10^6}{\sqrt{2}} B_x \right) = -5\sqrt{2} \times 10^{-3} \]
This leads to:
\[ \frac{10^1}{\sqrt{2}} (B_y - B_x) = -5\sqrt{2} \]
Thus,
\[ B_y - B_x = -10 \]
(Equation 1)

### Step 7: Analyze the Second Condition
When the particle moves with \( v_2 = 10^6 \, m/s \) along the z-axis, the force \( \vec{F_2} \) acts in the y-direction:
\[ \vec{F_2} = q \vec{v_2} \times \vec{B} \]
Where \( \vec{v_2} = 10^6 \hat{k} \).

Calculating the cross product:
\[ \vec{F_2} = q \left( 10^6 \hat{k} \times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k}) \right) = q \left( 10^6 (B_y \hat{i} - B_x \hat{j}) \right) \]
Setting \( \vec{F_2} = F_2 \hat{j} \):
\[ 10^{-5} \cdot 10^6 (B_y \hat{i} - B_x \hat{j}) = F_2 \hat{j} \]
From the \( \hat{j} \) component:
\[ -10^{-5} \cdot 10^6 B_x = F_2 \]
And from the \( \hat{i} \) component:
\[ 10^{-5} \cdot 10^6 B_y = 0 \Rightarrow B_y = 0 \]

### Step 8: Solve for \( B_x \)
Substituting \( B_y = 0 \) into Equation 1:
\[ 0 - B_x = -10 \Rightarrow B_x = 10 \]

### Step 9: Compile the Results
Thus, we have:
- \( B_x = 10 \, T \)
- \( B_y = 0 \)
- \( B_z = 0 \)

The magnetic field vector is:
\[ \vec{B} = 10 \hat{i} + 0 \hat{j} + 0 \hat{k} = 10 \hat{i} \, T \]

### Final Answer
The magnetic field \( \vec{B} \) is:
\[ \vec{B} = 10 \, T \, \hat{i} \]