A circular loop of radius R carrying current I is kept in XZ plane. A uniform and constant magnetic field `vecB=(B_(0)hati+2B_(0)hatj+3B_(0)hatk)` exists in the region (`B_(0)-` a positive constant). Then the magnitude of the torque acting on the loop will be

Options

A. `2IpiR^(2)B_(0)`
B. `sqrt5IpiR^(2)B_(0)`
C. `sqrt(10)IpiR^(2)B_(0)`
D. `6IpiR^(2)B_(0)`

Answer

To find the magnitude of the torque acting on a circular loop carrying current \( I \) in a uniform magnetic field \( \vec{B} \), we can follow these steps:

### Step 1: Identify the magnetic moment \( \vec{m} \)
The magnetic moment \( \vec{m} \) of a circular loop is given by the formula:
\[ \vec{m} = n \cdot I \cdot \vec{A} \]
where:
- \( n \) is the number of turns (for a single loop, \( n = 1 \)),
- \( I \) is the current,
- \( \vec{A} \) is the area vector of the loop.

For a circular loop of radius \( R \), the area \( A \) is:
\[ A = \pi R^2 \]
The direction of the area vector \( \vec{A} \) is perpendicular to the plane of the loop. Since the loop is in the XZ plane, the area vector will point in the negative Y direction:
\[ \vec{A} = \pi R^2 (-\hat{j}) \]
Thus, the magnetic moment becomes:
\[ \vec{m} = I \cdot \pi R^2 (-\hat{j}) = -\pi R^2 I \hat{j} \]

### Step 2: Write the magnetic field \( \vec{B} \)
The magnetic field is given as:
\[ \vec{B} = B_0 \hat{i} + 2B_0 \hat{j} + 3B_0 \hat{k} \]

### Step 3: Calculate the torque \( \vec{\tau} \)
The torque \( \vec{\tau} \) acting on the loop is given by the cross product of the magnetic moment and the magnetic field:
\[ \vec{\tau} = \vec{m} \times \vec{B} \]
Substituting the values of \( \vec{m} \) and \( \vec{B} \):
\[ \vec{\tau} = (-\pi R^2 I \hat{j}) \times (B_0 \hat{i} + 2B_0 \hat{j} + 3B_0 \hat{k}) \]

### Step 4: Perform the cross product
Using the distributive property of the cross product:
\[ \vec{\tau} = -\pi R^2 I \left( \hat{j} \times (B_0 \hat{i}) + \hat{j} \times (2B_0 \hat{j}) + \hat{j} \times (3B_0 \hat{k}) \right) \]
Calculating each term:
1. \( \hat{j} \times \hat{i} = -\hat{k} \)
2. \( \hat{j} \times \hat{j} = 0 \)
3. \( \hat{j} \times \hat{k} = \hat{i} \)

Thus, we have:
\[ \vec{\tau} = -\pi R^2 I \left( -B_0 \hat{k} + 0 + 3B_0 \hat{i} \right) \]
This simplifies to:
\[ \vec{\tau} = \pi R^2 I (3B_0 \hat{i} - B_0 \hat{k}) \]

### Step 5: Calculate the magnitude of the torque
The magnitude of the torque vector is given by:
\[ |\vec{\tau}| = \sqrt{(3B_0)^2 + (-B_0)^2} \cdot \pi R^2 I \]
Calculating the magnitude:
\[ |\vec{\tau}| = \sqrt{9B_0^2 + B_0^2} \cdot \pi R^2 I = \sqrt{10B_0^2} \cdot \pi R^2 I = \sqrt{10} B_0 \pi R^2 I \]

### Final Answer
The magnitude of the torque acting on the loop is:
\[ |\vec{\tau}| = \sqrt{10} B_0 \pi R^2 I \]