Question

Which one of the following options represents the magnetic field $\vec{B}$ at $O$ due to the current flowing in the given wire segments lying on the $x y$ plane?

 

• A. $\vec{B}=\frac{-\mu_0 I}{L}\left(\frac{3}{2}+\frac{1}{4 \sqrt{2} \pi}\right) \hat{k}$
• B. $\vec{B}=-\frac{\mu_0 I}{L}\left(\frac{3}{2}+\frac{1}{2 \sqrt{2} \pi}\right) \hat{k}$
• C. $\vec{B}=\frac{-\mu_0 I}{L}\left(1+\frac{1}{4 \sqrt{2} \pi}\right) \hat{k}$
• D. $\vec{B}=\frac{-\mu_0 I}{L}\left(1+\frac{1}{4 \pi}\right) \hat{k}$

Answer 1

The Correct Option is C

To solve this problem, we need to use Ampère’s law to find the magnetic field \( \vec{B} \) at the origin due to the given current-carrying wire segments.

1. Understanding the Setup:
We are given a wire that lies in the \( xy \)-plane with current flowing through it. The segments of the wire have different lengths and are arranged as shown in the diagram. We need to determine the magnetic field at the origin \( O \), where the wire segments are positioned symmetrically around the point.

2. Using Ampère's Law:
The magnetic field due to a current-carrying straight wire is given by the Biot-Savart law, which in this case simplifies to Ampère's law for straight segments of the wire. The magnetic field at a point due to a current \( I \) in a straight segment of wire is given by:

\[ B = \frac{\mu_0 I}{2\pi r} \]

where: - \( \mu_0 \) is the permeability of free space, - \( I \) is the current, - \( r \) is the perpendicular distance from the wire to the point of interest (here, the origin).

3. Magnetic Field Contribution from the First Segment:
The first segment of the wire has length \( \frac{L}{2} \), and the current is flowing along the \( x \)-axis. The magnetic field at the origin from this segment will be along the \( k \)-axis, as the current creates a circular magnetic field around the wire. Using the formula, the magnetic field at the origin due to this segment is:

\[ B_1 = \frac{\mu_0 I}{2\pi \frac{L}{2}} = \frac{\mu_0 I}{\pi L} \]

4. Magnetic Field Contribution from the Second Segment:
The second segment of the wire has length \( \frac{L}{4} \), and the current is flowing along the \( y \)-axis. Similarly, the magnetic field at the origin from this segment is:

\[ B_2 = \frac{\mu_0 I}{2\pi \frac{L}{4}} = \frac{\mu_0 I}{2\pi L} \]

5. Net Magnetic Field at the Origin:
The magnetic field from the two segments combine. Since both segments contribute to the magnetic field along the same axis, we add the magnitudes of their fields:

\[ B_{\text{total}} = B_1 + B_2 = \frac{\mu_0 I}{\pi L} + \frac{\mu_0 I}{2\pi L} = \frac{\mu_0 I}{L} \left( \frac{3}{2} \right) \]

6. Final Expression:
The total magnetic field at the origin is:

\[ \vec{B} = \frac{\mu_0 I}{L} \left( 1 + \frac{1}{4\pi} \right) \hat{k} \]

Final Answer:
The correct option is C.