A current I flows in a rectangularly shaped wire whose centre lies at `(x_0,0,0)` and whose vertices are located at the points A `(x_0+d,-a,-b),B(x_0-d,a,-b),C(x_0-d,a,+b) and D(x_0+d,-a,b)` respectively. Assume that a,b,d << `x_0`. Find the magnitude of magnetic dipole moment vector of the rectangular wire frame in `J//T`. (Given, b=10 m,I=0.01A, d=4m,a=3m).

Options

A. `2 J T^(-1)`
B. `4 J T^(-1)`
C. `3 J T^(-1)`
D. `9 J T^(-1)`

Explanation

To find the magnitude of the magnetic dipole moment vector of the rectangular wire frame, we can follow these steps:

### Step 1: Identify the Coordinates of the Vertices
The vertices of the rectangular wire frame are given as:
- A: \((x_0 + d, -a, -b)\)
- B: \((x_0 - d, a, -b)\)
- C: \((x_0 - d, a, b)\)
- D: \((x_0 + d, -a, b)\)

### Step 2: Calculate the Position Vectors
The position vectors of points A, B, C, and D can be expressed as:
- \(\vec{A} = (x_0 + d) \hat{i} - a \hat{j} - b \hat{k}\)
- \(\vec{B} = (x_0 - d) \hat{i} + a \hat{j} - b \hat{k}\)
- \(\vec{C} = (x_0 - d) \hat{i} + a \hat{j} + b \hat{k}\)
- \(\vec{D} = (x_0 + d) \hat{i} - a \hat{j} + b \hat{k}\)

### Step 3: Calculate the Area Vector
The area vector \(\vec{A}\) can be calculated using the cross product of two adjacent sides of the rectangle. We can take the vectors \(\vec{AB}\) and \(\vec{AD}\):
- \(\vec{AB} = \vec{B} - \vec{A}\)
- \(\vec{AD} = \vec{D} - \vec{A}\)

Calculating \(\vec{AB}\):
\[ \vec{AB} = [(x_0 - d) - (x_0 + d)] \hat{i} + [a - (-a)] \hat{j} + [-b - (-b)] \hat{k} = -2d \hat{i} + 2a \hat{j} + 0 \hat{k} \]

Calculating \(\vec{AD}\):
\[ \vec{AD} = [(x_0 + d) - (x_0 + d)] \hat{i} + [-a - (-a)] \hat{j} + [b - (-b)] \hat{k} = 0 \hat{i} + 0 \hat{j} + 2b \hat{k} \]

### Step 4: Compute the Area Vector
Now, we calculate the area vector \(\vec{A}\):
\[ \vec{A} = \vec{AB} \times \vec{AD} \]
Calculating the cross product:
\[ \vec{A} = (-2d \hat{i} + 2a \hat{j}) \times (2b \hat{k}) \]
Using the determinant method for cross product:
\[ \vec{A} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2d & 2a & 0 \\ 0 & 0 & 2b \end{vmatrix} \]
Calculating this determinant gives:
\[ \vec{A} = (2a)(2b) \hat{i} - (0)(-2d) \hat{j} + (0)(2a) \hat{k} = 4ab \hat{i} \]

### Step 5: Calculate the Magnitude of the Area Vector
The magnitude of the area vector is:
\[ |\vec{A}| = 4ab \]

### Step 6: Calculate the Magnetic Dipole Moment
The magnetic dipole moment \(\vec{M}\) is given by:
\[ \vec{M} = I \cdot \vec{A} \]
Where \(I\) is the current flowing through the wire. Therefore, the magnitude of the magnetic dipole moment is:
\[ |\vec{M}| = I \cdot |\vec{A}| \]
Substituting the values:
\[ |\vec{M}| = 0.01 \, \text{A} \cdot 4ab \]

### Step 7: Substitute Given Values
Given \(a = 3 \, \text{m}\), \(b = 10 \, \text{m}\):
\[ |\vec{M}| = 0.01 \cdot 4 \cdot 3 \cdot 10 = 0.01 \cdot 120 = 1.2 \, \text{J/T} \]

### Final Result
The magnitude of the magnetic dipole moment vector of the rectangular wire frame is:
\[ \boxed{1.2 \, \text{J/T}} \]