A circular wire loop of radius R is placed in the X-Y plane with its centre at the origin. The loop carries a current I in the clockwise direction when viewed from a point on the positive Z-axis. A uniform magnetic field `vecB=B_(0)((hati+sqrt(3)hatj)/2)` exists in the region `x gt (R sqrt(3))/2`. The magnitued of the magnetic force on the loop due to the magnetic field is `(B_(0)IR)/n`. The value of n is______________.

Explanation

To solve the problem, we need to calculate the magnetic force acting on a circular wire loop placed in a magnetic field. The loop carries a current in a specific direction, and we will use the relevant formulas to find the required value of \( n \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a circular wire loop of radius \( R \) centered at the origin in the X-Y plane.
- The current \( I \) flows in a clockwise direction when viewed from the positive Z-axis.
- A uniform magnetic field \( \vec{B} = B_0 \left( \hat{i} + \sqrt{3} \hat{j} \right)/2 \) exists in the region \( x > (R \sqrt{3})/2 \).

2. **Determine the Direction of the Magnetic Field**:
- The magnetic field can be expressed in terms of its components: \( B_x = \frac{B_0}{2} \) and \( B_y = \frac{B_0 \sqrt{3}}{2} \).
- The angle of the magnetic field with respect to the X-axis can be calculated using \( \tan \theta = \frac{B_y}{B_x} = \sqrt{3} \), which gives \( \theta = 60^\circ \).

3. **Calculate the Length Vector**:
- The length vector \( \vec{L} \) for the loop can be expressed as \( \vec{L} = -R \hat{j} \) since the current is flowing clockwise.

4. **Using the Force Formula**:
- The magnetic force \( \vec{F} \) on the loop can be calculated using the formula:
\[ \vec{F} = I \vec{L} \times \vec{B} \]
- Substituting the values, we have:
\[ \vec{F} = I (-R \hat{j}) \times \left( \frac{B_0}{2} \hat{i} + \frac{B_0 \sqrt{3}}{2} \hat{j} \right) \]

5. **Calculating the Cross Product**:
- The cross product can be computed as follows:
\[ \vec{F} = I (-R) \left( \hat{j} \times \left( \frac{B_0}{2} \hat{i} + \frac{B_0 \sqrt{3}}{2} \hat{j} \right) \right) \]
- This simplifies to:
\[ \vec{F} = I (-R) \left( \frac{B_0}{2} (\hat{j} \times \hat{i}) + \frac{B_0 \sqrt{3}}{2} (\hat{j} \times \hat{j}) \right) \]
- Since \( \hat{j} \times \hat{j} = 0 \) and \( \hat{j} \times \hat{i} = -\hat{k} \):
\[ \vec{F} = I (-R) \left( \frac{B_0}{2} (-\hat{k}) \right) = \frac{I R B_0}{2} \hat{k} \]

6. **Magnitude of the Force**:
- The magnitude of the force is:
\[ |\vec{F}| = \frac{I R B_0}{2} \]

7. **Relating to Given Expression**:
- We know from the problem statement that the magnitude of the magnetic force is given by:
\[ |\vec{F}| = \frac{B_0 I R}{n} \]
- Equating the two expressions:
\[ \frac{I R B_0}{2} = \frac{B_0 I R}{n} \]
- From this, we can deduce that:
\[ n = 2 \]

### Final Answer:
The value of \( n \) is \( 2 \).