A thin wire carrying current i is bent to form a closed loop of one turn. The loop is placed in y-z plane with centre at origin. If R is the radius of the loop, then

Options

A. At a point (x, 0) on x-axis `B_(x) = (mu_(0)iR^(2))/(2(R^(2) + x^(2))^(3//2))`
B. `int_(0)^(oo) B_(x) dx = (mu_(0)i)/(2)`
C. `int_(0)^(-oo) B_(x) dx = mu_(0)i`
D. `int_(-oo)^(oo) B_(x) dx = 0`

Answer

To solve the problem of finding the magnetic field at a point on the x-axis due to a current-carrying loop in the y-z plane, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have a thin wire bent into a circular loop of radius \( R \) in the y-z plane, centered at the origin.
- We need to find the magnetic field at a point on the x-axis, specifically at the point \( (x, 0, 0) \).

2. **Magnetic Field Due to a Current Loop**:
- The magnetic field \( B \) at a distance \( x \) from the center of the loop can be derived using the Biot-Savart law.
- The magnetic field \( B_x \) at point \( P(x, 0, 0) \) due to the current \( I \) in the loop can be expressed as:
\[ B_x = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} \]
- Here, \( \mu_0 \) is the permeability of free space.

3. **Setting Up the Integral**:
- To find the total magnetic field, we need to integrate this expression over the entire loop. However, since the loop is symmetric, we can simplify our calculations.
- The magnetic field contribution from opposite sides of the loop will add up in the x-direction, while contributions in the y and z directions will cancel out.

4. **Integration**:
- The total magnetic field \( B_x \) can be obtained by integrating the expression for \( B_x \) from \( -\infty \) to \( +\infty \):
\[ B_x = \int_{-\infty}^{+\infty} \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} \, dx \]

5. **Evaluating the Integral**:
- To evaluate the integral, we can use the substitution \( x = R \tan(\theta) \), which transforms the integral limits and simplifies the expression.
- After performing the integration, we find:
\[ B_x = \frac{\mu_0 I}{2} \]

6. **Conclusion**:
- The magnetic field at the point \( (x, 0, 0) \) due to the current loop is:
\[ B_x = \frac{\mu_0 I}{2} \]