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. a.`(mg)/(pirsqrt(B_(x)^(2)+B_(z)^(2)))`
B. b. `(mg)/(pirB_(x))`
C. c. `(mg)/(pirB_(z))`
D. d. `(mg)/(pirsqrt(B_(x)B_(z)))`

Answer

To solve the problem, we need to determine the value of the current \( I \) that will cause the rigid circular loop to start tilting in the presence of the Earth's magnetic field. Here’s the step-by-step solution:

### Step 1: Understand the forces acting on the loop
The loop is in the XY plane and has a current \( I \) flowing through it. The Earth's magnetic field is given by \( \vec{B} = B_x \hat{i} + B_z \hat{k} \).

### Step 2: Calculate the torque due to the magnetic field
The torque \( \tau \) acting on the loop due to the magnetic field can be expressed as:
\[ \tau = m \cdot B \]
Where \( m \) is the magnetic moment of the loop. The magnetic moment \( m \) for a current loop is given by:
\[ m = I \cdot A \]
Here, \( A \) is the area of the loop. For a circular loop of radius \( r \):
\[ A = \pi r^2 \]
Thus, the magnetic moment becomes:
\[ m = I \cdot \pi r^2 \]

### Step 3: Calculate the torque due to the magnetic moment in the magnetic field
The torque acting on the loop can be calculated as:
\[ \tau = m \cdot B_x \]
Substituting the expression for \( m \):
\[ \tau = (I \cdot \pi r^2) \cdot B_x \]

### Step 4: Calculate the gravitational torque
The gravitational torque acting on the loop when it starts tilting is given by:
\[ \tau_{gravity} = mg \cdot r \]
Where \( m \) is the mass of the loop and \( g \) is the acceleration due to gravity.

### Step 5: Set the torques equal to find the condition for tilting
For the loop to start tilting, the torque due to the magnetic field must equal the gravitational torque:
\[ I \cdot \pi r^2 \cdot B_x = mg \cdot r \]

### Step 6: Solve for the current \( I \)
Rearranging the equation to solve for \( I \):
\[ I = \frac{mg \cdot r}{\pi r^2 B_x} \]
This simplifies to:
\[ I = \frac{mg}{\pi r B_x} \]

### Final Answer
Thus, the value of the current \( I \) that will cause the loop to start tilting is:
\[ I = \frac{mg}{\pi r B_x} \]

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