A circular loop of radius `R` , carrying `I`, lies in `x- y` plane with its origin . The total magnetic flux through ` x-y` plane is

Options

A. (a) directly proportional to `I`
B. (b) directly proportional to `R`
C. ( c) directly proportional to `R`
D. (d) zero

Explanation

To solve the problem of finding the total magnetic flux through the x-y plane created by a circular loop of radius \( R \) carrying a current \( I \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a circular loop of radius \( R \) lying in the x-y plane with its center at the origin. The loop carries a current \( I \).

2. **Magnetic Field Due to the Loop**:
- A current-carrying loop generates a magnetic field. The direction of the magnetic field at the center of the loop can be determined using the right-hand rule. For a current flowing in a counterclockwise direction, the magnetic field at the center points in the positive z-direction.

3. **Magnetic Field Strength**:
- The magnetic field \( B \) at the center of a circular loop of radius \( R \) carrying a current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2R} \]
where \( \mu_0 \) is the permeability of free space.

4. **Magnetic Flux Calculation**:
- The magnetic flux \( \Phi \) through a surface is given by the product of the magnetic field \( B \) and the area \( A \) through which it passes, and the cosine of the angle \( \theta \) between the magnetic field and the normal to the surface:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
- In our case, since the magnetic field is perpendicular to the x-y plane (which is the surface we are considering), the angle \( \theta = 0^\circ \) and \( \cos(0^\circ) = 1 \).

5. **Area of the x-y Plane**:
- The area \( A \) of the x-y plane that the magnetic field passes through is infinite, but we are interested in the flux through the loop itself. The effective area for the circular loop is:
\[ A = \pi R^2 \]

6. **Total Magnetic Flux**:
- Therefore, the total magnetic flux through the x-y plane due to the circular loop is:
\[ \Phi = B \cdot A = \left(\frac{\mu_0 I}{2R}\right) \cdot (\pi R^2) = \frac{\mu_0 I \pi R}{2} \]

7. **Net Magnetic Flux**:
- However, since the magnetic field lines that enter the x-y plane through the loop also leave it, the net magnetic flux through the x-y plane is zero. This is because the amount of flux entering is equal to the amount of flux leaving.

### Final Answer:
The total magnetic flux through the x-y plane is:
\[ \Phi_{\text{net}} = 0 \]