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 in a magnetic field, we can follow these steps:

### Step 1: Understand the Given Information
We have a circular loop of radius \( R \) carrying a current \( I \) in the XZ plane. The magnetic field is given as:
\[ \vec{B} = B_0 \hat{i} + 2B_0 \hat{j} + 3B_0 \hat{k} \]
where \( B_0 \) is a positive constant.

### Step 2: Calculate the Area Vector
The area vector \( \vec{A} \) of the loop is perpendicular to the plane of the loop. Since the loop is in the XZ plane, the area vector will point in the direction of the Y-axis. The area of the loop is:
\[ A = \pi R^2 \]
Thus, the area vector can be expressed as:
\[ \vec{A} = A \hat{n} = \pi R^2 (-\hat{j}) \quad \text{(assuming the area vector points in the negative Y direction)} \]

### Step 3: Calculate the Magnetic Moment
The magnetic moment \( \vec{m} \) of the loop is given by:
\[ \vec{m} = I \vec{A} = I \pi R^2 (-\hat{j}) = -I \pi R^2 \hat{j} \]

### Step 4: Calculate the Torque
The torque \( \vec{\tau} \) acting on the loop in a magnetic field is given by the cross product:
\[ \vec{\tau} = \vec{m} \times \vec{B} \]
Substituting the values of \( \vec{m} \) and \( \vec{B} \):
\[ \vec{\tau} = (-I \pi R^2 \hat{j}) \times (B_0 \hat{i} + 2B_0 \hat{j} + 3B_0 \hat{k}) \]

### Step 5: Calculate the Cross Product
Using the properties of the cross product:
\[ \vec{\tau} = -I \pi R^2 \left( \hat{j} \times (B_0 \hat{i} + 2B_0 \hat{j} + 3B_0 \hat{k}) \right) \]
Calculating the cross products:
- \( \hat{j} \times \hat{i} = -\hat{k} \)
- \( \hat{j} \times \hat{j} = 0 \)
- \( \hat{j} \times \hat{k} = \hat{i} \)

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

### Step 6: Calculate the Magnitude of the Torque
The magnitude of the torque is given by:
\[ |\vec{\tau}| = I \pi R^2 \sqrt{(3B_0)^2 + (-B_0)^2} = I \pi R^2 \sqrt{9B_0^2 + B_0^2} = I \pi R^2 \sqrt{10B_0^2} \]
This simplifies to:
\[ |\vec{\tau}| = I \pi R^2 B_0 \sqrt{10} \]

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

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