An electron moves through a uniform magnetic fields given by `vecB=B_(x)hati+(3.0B_(x))hatj`. At aparticular instant, the electron has velocity `vecv=(2.0hati++4.0hatj)m//s` and the magnetic force acting on it is `(6.4xx10^(-19)N)hatk`. The value of `B_(x)` (in T) will

Options

A. `-3.0`
B. 3
C. 2
D. `-2.0`

Answer

To find the value of \( B_x \) in the given magnetic field, we can use the formula for the magnetic force acting on a charged particle moving in a magnetic field, which is given by:

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

### Step 1: Identify the given values
- Charge of the electron, \( q = -1.6 \times 10^{-19} \, \text{C} \)
- Velocity of the electron, \( \vec{v} = 2.0 \hat{i} + 4.0 \hat{j} \, \text{m/s} \)
- Magnetic force, \( \vec{F} = 6.4 \times 10^{-19} \hat{k} \, \text{N} \)
- Magnetic field, \( \vec{B} = B_x \hat{i} + (3.0 B_x) \hat{j} \)

### Step 2: Set up the cross product
Using the determinant method for the cross product \( \vec{v} \times \vec{B} \):

\[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2.0 & 4.0 & 0 \\ B_x & 3.0 B_x & 0 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant, we have:

\[ \vec{v} \times \vec{B} = \hat{i} \begin{vmatrix} 4.0 & 0 \\ 3.0 B_x & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 2.0 & 0 \\ B_x & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 2.0 & 4.0 \\ B_x & 3.0 B_x \end{vmatrix} \]

Calculating each of these:

1. For \( \hat{i} \):
\[ 4.0 \cdot 0 - 0 \cdot 3.0 B_x = 0 \]

2. For \( \hat{j} \):
\[ 2.0 \cdot 0 - 0 \cdot B_x = 0 \]

3. For \( \hat{k} \):
\[ 2.0 \cdot (3.0 B_x) - 4.0 \cdot B_x = 6.0 B_x - 4.0 B_x = 2.0 B_x \]

Thus, we have:
\[ \vec{v} \times \vec{B} = 0 \hat{i} - 0 \hat{j} + 2.0 B_x \hat{k} = 2.0 B_x \hat{k} \]

### Step 4: Substitute into the force equation
Now, substituting into the force equation:
\[ \vec{F} = q (\vec{v} \times \vec{B}) = -1.6 \times 10^{-19} (2.0 B_x \hat{k}) \]

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

### Step 5: Set the forces equal
Setting the magnetic force equal to the calculated force:
\[ 6.4 \times 10^{-19} \hat{k} = -3.2 \times 10^{-19} B_x \hat{k} \]

### Step 6: Solve for \( B_x \)
Dividing both sides by \( -3.2 \times 10^{-19} \):
\[ B_x = \frac{6.4 \times 10^{-19}}{-3.2 \times 10^{-19}} = -2 \]

### Conclusion
Thus, the value of \( B_x \) is:
\[ B_x = -2 \, \text{T} \]