An electron moves through a uniform magnetic fiedl given by `B=B_xhati+(3B_x)hatj`. At a particular instant, the electron has the velocity `v=(2.0hati+4.0hatj)` m/s and the magnetic force acting on its is `(6.4xx10^-19N)hatk` Find `B_x`.

Explanation

To solve the problem, we need to find the value of \( B_x \) given the magnetic field, the velocity of the electron, and the magnetic force acting on it. We will use the formula for magnetic force, which is given by:

\[ \vec{F} = q (\vec{v} \times \vec{B}) \]

where:
- \( \vec{F} \) is the magnetic force,
- \( q \) is the charge of the electron,
- \( \vec{v} \) is the velocity vector,
- \( \vec{B} \) is the magnetic field vector.

### Step 1: Write down the given quantities
- The magnetic field is given by:
\[ \vec{B} = B_x \hat{i} + 3B_x \hat{j} \]
- The velocity of the electron is:
\[ \vec{v} = 2.0 \hat{i} + 4.0 \hat{j} \text{ m/s} \]
- The magnetic force acting on the electron is:
\[ \vec{F} = 6.4 \times 10^{-19} \hat{k} \text{ N} \]
- The charge of the electron is:
\[ q = -1.6 \times 10^{-19} \text{ C} \]

### Step 2: Calculate the cross product \( \vec{v} \times \vec{B} \)
Using the determinant method to find the cross product:

\[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 0 \\ B_x & 3B_x & 0 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{v} \times \vec{B} = \hat{i} \left( 4 \cdot 0 - 0 \cdot 3B_x \right) - \hat{j} \left( 2 \cdot 0 - 0 \cdot B_x \right) + \hat{k} \left( 2 \cdot 3B_x - 4 \cdot B_x \right) \]

This simplifies to:

\[ \vec{v} \times \vec{B} = \hat{k} (6B_x - 4B_x) = 2B_x \hat{k} \]

### Step 3: Substitute into the force equation
Now, substituting \( \vec{v} \times \vec{B} \) into the force equation:

\[ \vec{F} = q (\vec{v} \times \vec{B}) = -1.6 \times 10^{-19} (2B_x \hat{k}) \]

This gives:

\[ \vec{F} = -3.2 \times 10^{-19} B_x \hat{k} \]

### Step 4: Set the magnetic force equal to the calculated force
We know that the magnetic force is also given as:

\[ \vec{F} = 6.4 \times 10^{-19} \hat{k} \]

Setting the two expressions for force equal to each other:

\[ -3.2 \times 10^{-19} B_x = 6.4 \times 10^{-19} \]

### Step 5: Solve for \( B_x \)
Dividing both sides by \(-3.2 \times 10^{-19}\):

\[ B_x = \frac{6.4 \times 10^{-19}}{-3.2 \times 10^{-19}} = -2 \]

Thus, the value of \( B_x \) is:

\[ B_x = -2 \text{ T} \]

### Final Answer
The value of \( B_x \) is \(-2 \text{ T}\).