Question

A wire has three different sections as shown in figure. The magnitude of the magnetic field produced at the centre '\(O\)' of the semicircle by three sections together is ( \(\mu_0 =\) permiability of free space)

• A. \(\frac{\mu_0 \text{I}}{4\text{R}}\)
• B. \(\frac{\mu_0 \text{I}}{2\text{R}}\)
• C. \(\frac{\mu_0 \text{I}}{4\pi\text{R}}\)
• D. \(\frac{\mu_0 I}{2\pi R}\)

Hint:

If a straight wire segment is in line with the point $O$, ignore it. For arcs, the field is $(\text{angle in radians}/2\pi) \times (\mu_0 I/2R)$. Semicircle $= \pi/2\pi = 1/2$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To find the total magnetic field at point \(O\) caused by a complex wire shape, we use the principle of superposition.
We will evaluate the magnetic field contribution from each distinct section of the wire independently and sum them up.
Step 2: Key Formula or Approach:
The Biot-Savart Law determines the magnetic field.
1. For a straight current-carrying wire, if the point of interest lies precisely on the line extending the wire, the angle between the current element \(d\vec{l}\) and the position vector \(\vec{r}\) is \(0^\circ\) or \(180^\circ\). Since \(\sin(0^\circ) = \sin(180^\circ) = 0\), the magnetic field contribution is zero.
2. For a circular arc of radius \(R\) subtending an angle \(\theta\) at the center, the magnetic field at the center is \(B = \frac{\mu_0 I}{4\pi R} \theta\).
Step 3: Detailed Explanation:
Let's systematically calculate the field from each of the three sections at center \(O\).
Section (i): This is a straight horizontal wire segment. As indicated by the dashed line in the figure, its extended axis passes directly through the point \(O\).
Consequently, its contribution to the magnetic field at \(O\) is zero: \(B_1 = 0\).
Section (iii): Similar to the first section, this is a straight wire segment whose extended axis also passes straight through \(O\).
Therefore, its contribution to the magnetic field at \(O\) is also zero: \(B_3 = 0\).
Section (ii): This section is a semicircular arc with radius \(R\).
A complete circle subtends an angle of \(2\pi\) radians, so a semicircle subtends an angle of \(\pi\) radians (\(\theta = \pi\)).
Applying the formula for a circular arc:
\[ B_2 = \frac{\mu_0 I}{4\pi R} \times \pi = \frac{\mu_0 I}{4R} \]
Alternatively, recalling that the field at the center of a full circular loop is \(\frac{\mu_0 I}{2R}\), a semicircle generates exactly half of this field:
\[ B_2 = \frac{1}{2} \left( \frac{\mu_0 I}{2R} \right) = \frac{\mu_0 I}{4R} \]
The total magnetic field is the vector sum of these contributions. Since all segments lie in the same plane, their fields (if non-zero) would be collinear.
\[ B_{\text{total}} = B_1 + B_2 + B_3 = 0 + \frac{\mu_0 I}{4R} + 0 = \frac{\mu_0 I}{4R} \]
Step 4: Final Answer:
The magnitude of the magnetic field at \(O\) is \(\frac{\mu_0 \text{I}}{4\text{R}}\).