A rigid circular loop of radius `r` and mass `m` lies in the `XY-` plane of a flat table and has a current `I` flowing in it. At this particular place. The Earth's magnetic field `vec(B)=B_(x)i+B_(z)k`. The value of `I` so that the loop starts tilting is `:`

Options

A. `(mg)/(pirsqrt(B_(x)^(2)+B_(2)^(z))`
B. `(mg)/(pirB_(z))`
C. `(mg)/(pirB_(x))`
D. `(mg)/(pirsqrt(B_(x)B_(z)))`

Answer

To find the value of the current \( I \) such that the loop starts tilting, we need to analyze the torques acting on the loop due to both gravity and the magnetic field.

### Step-by-Step Solution:

1. **Identify the Forces and Torques:**
- The gravitational force acting on the loop is \( mg \), where \( m \) is the mass of the loop and \( g \) is the acceleration due to gravity.
- The torque due to the gravitational force about the pivot point (where the loop will tilt) is given by:
\[ \tau_{\text{gravity}} = mg \cdot R \]
where \( R \) is the radius of the loop.

2. **Determine the Magnetic Moment:**
- The magnetic moment \( \vec{m} \) of the loop is given by:
\[ \vec{m} = I \cdot \text{Area} \cdot \hat{n} \]
- The area \( A \) of the circular loop is \( \pi r^2 \), and since the loop lies in the XY-plane, the direction of \( \hat{n} \) (the area vector) is along the Z-axis:
\[ \vec{m} = I \cdot \pi r^2 \hat{k} \]

3. **Calculate the Torque Due to the Magnetic Field:**
- The torque \( \tau_{\text{mag}} \) on the loop in a magnetic field \( \vec{B} = B_x \hat{i} + B_z \hat{k} \) is given by:
\[ \tau_{\text{mag}} = \vec{m} \times \vec{B} \]
- Substituting \( \vec{m} \) and \( \vec{B} \):
\[ \tau_{\text{mag}} = (I \cdot \pi r^2 \hat{k}) \times (B_x \hat{i} + B_z \hat{k}) \]
- Using the properties of the cross product:
\[ \tau_{\text{mag}} = I \cdot \pi r^2 \left( \hat{k} \times B_x \hat{i} + \hat{k} \times B_z \hat{k} \right) \]
- Since \( \hat{k} \times \hat{k} = 0 \) and \( \hat{k} \times \hat{i} = -\hat{j} \):
\[ \tau_{\text{mag}} = -I \cdot \pi r^2 B_x \hat{j} \]

4. **Set Up the Condition for Tilting:**
- The loop will start tilting when the torque due to the magnetic field is greater than the torque due to gravity:
\[ |\tau_{\text{mag}}| > |\tau_{\text{gravity}}| \]
- This gives us:
\[ I \cdot \pi r^2 B_x > mgR \]

5. **Solve for Current \( I \):**
- Rearranging the inequality:
\[ I > \frac{mgR}{\pi r^2 B_x} \]
- Since \( R \) is the radius \( r \) of the loop, we can simplify:
\[ I > \frac{mg}{\pi r B_x} \]

### Final Answer:
The minimum value of current \( I \) so that the loop starts tilting is:
\[ I = \frac{mg}{\pi r B_x} \]