A charge of `4.0 muC` is moving with a velocity of `4.0 times 10^6ms^(–1)` along the positive y-axis under a magnetic field `vec B` of strength `(2 hat k)` T. The force acting on the charge is `x hat i N`. The value of x is __.

Answer

To solve the problem, we will use the formula for the magnetic force acting on a charge moving in a magnetic field, which is given by:

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

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

### Step 1: Convert the charge to coulombs
The charge is given as \(4.0 \, \mu C\). We need to convert this to coulombs:

\[ q = 4.0 \, \mu C = 4.0 \times 10^{-6} \, C \]

**Hint:** Remember that \(1 \, \mu C = 10^{-6} \, C\).

### Step 2: Write the velocity vector
The velocity is given as \(4.0 \times 10^6 \, m/s\) along the positive y-axis. Thus, we can express the velocity vector as:

\[ \vec{v} = 4.0 \times 10^6 \, \hat{j} \, m/s \]

**Hint:** The unit vector \(\hat{j}\) indicates the direction along the y-axis.

### Step 3: Write the magnetic field vector
The magnetic field is given as \(2 \hat{k} \, T\). Therefore, we can express the magnetic field vector as:

\[ \vec{B} = 2 \hat{k} \, T \]

**Hint:** The unit vector \(\hat{k}\) indicates the direction along the z-axis.

### Step 4: Calculate the cross product \(\vec{v} \times \vec{B}\)
Now we need to calculate the cross product of \(\vec{v}\) and \(\vec{B}\):

\[ \vec{F} = q(\vec{v} \times \vec{B}) = 4.0 \times 10^{-6} \, C \cdot (\hat{j} \times \hat{k}) \]

Using the right-hand rule, we find:

\[ \hat{j} \times \hat{k} = \hat{i} \]

Thus, we have:

\[ \vec{v} \times \vec{B} = (4.0 \times 10^6) \cdot 2 \hat{i} = 8.0 \times 10^6 \hat{i} \]

**Hint:** The right-hand rule helps determine the direction of the cross product.

### Step 5: Calculate the force \(\vec{F}\)
Now substituting back into the force equation:

\[ \vec{F} = q(\vec{v} \times \vec{B}) = 4.0 \times 10^{-6} \cdot (8.0 \times 10^6 \hat{i}) \]

Calculating this gives:

\[ \vec{F} = 32 \hat{i} \, N \]

### Conclusion
The value of \(x\) in the force \(\vec{F} = x \hat{i} \, N\) is:

\[ \boxed{32} \]