A current I flows in an infinitely long wire with cross section in the form of a semicircular ring of radius R. The magnitude of magnetic induction along its axis is

Options

A. `(mu_0 I )/( 2 pi^2 R)`
B. `(mu_0 I )/( 2 pi R)`
C. `(mu_0 I)/( 4 pi^2 R)`
D. `(mu_0 I )/( pi^2 R)`

Answer

To find the magnitude of the magnetic induction along the axis of an infinitely long wire with a semicircular cross-section of radius \( R \), we can follow these steps:

### Step 1: Understand the Geometry
We have a semicircular wire of radius \( R \) carrying a current \( I \). The wire is infinitely long, and we need to find the magnetic induction (magnetic field) along its axis.

### Step 2: Consider a Small Element of the Wire
Divide the semicircular wire into small elements. Let’s denote a small length element of the wire as \( dL \). For a small angle \( d\theta \), the length of this element can be expressed as:
\[ dL = R \, d\theta \]

### Step 3: Determine the Current Element
The total current \( I \) is uniformly distributed along the semicircular wire. The current through the small element \( dL \) can be expressed as:
\[ dI = \frac{I}{\pi R} \cdot dL = \frac{I}{\pi R} \cdot R \, d\theta = \frac{I}{\pi} \, d\theta \]

### Step 4: Calculate the Magnetic Field Contribution from \( dL \)
Using the Biot-Savart law, the magnetic field \( dB \) at a point on the axis due to the small current element \( dI \) is given by:
\[ dB = \frac{\mu_0 \, dI}{4\pi r^2} \]
where \( r \) is the distance from the current element to the point where we are calculating the magnetic field. For the axis of the semicircular wire, \( r = R \).

Substituting \( dI \):
\[ dB = \frac{\mu_0}{4\pi R^2} \cdot \frac{I}{\pi} \, d\theta = \frac{\mu_0 I}{4\pi^2 R^2} \, d\theta \]

### Step 5: Integrate to Find Total Magnetic Field
To find the total magnetic field \( B \) along the axis, integrate \( dB \) from \( \theta = 0 \) to \( \theta = \pi \):
\[ B = \int_0^{\pi} dB = \int_0^{\pi} \frac{\mu_0 I}{4\pi^2 R^2} \, d\theta \]
\[ B = \frac{\mu_0 I}{4\pi^2 R^2} \cdot \int_0^{\pi} d\theta = \frac{\mu_0 I}{4\pi^2 R^2} \cdot \pi = \frac{\mu_0 I}{4\pi R^2} \]

### Step 6: Simplify the Result
The expression simplifies to:
\[ B = \frac{\mu_0 I}{4\pi R} \]

### Conclusion
Thus, the magnitude of the magnetic induction along the axis of the semicircular wire is:
\[ B = \frac{\mu_0 I}{\pi R} \]