Question

i current is flowing in a wire shown in figure. What will be the value of magnetic field at O of semi-circle :
(i) Due to each length \(l\) of straight portion, (ii) Due to radius R of semi-circle ?
\includegraphics[width=0.5\linewidth]{1.png}

Hint:

When calculating magnetic fields from complex wire shapes, always break the wire into simpler geometric segments (straight lines, arcs). Calculate the field from each segment and then add them vectorially. Remember that for any point lying on the extended line of a straight current-carrying wire, the magnetic field is zero.

Answer 1

The total magnetic field at point O is the vector sum of the magnetic fields produced by the three sections of the wire: the two straight portions and the semi-circular portion.
(i) Magnetic Field due to Straight Portions

Step 1: Understanding the Concept:
The magnetic field produced by a current-carrying wire can be calculated using the Biot-Savart Law. This law relates the magnetic field to the current element, its length, and the position vector to the point of interest.

Step 2: Key Formula or Approach:
The Biot-Savart Law is given by: \[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{i(d\vec{l} \times \hat{r})}{r^2} \] Here, \(d\vec{l}\) is the vector representing the current element and \(\hat{r}\) is the unit vector pointing from the current element to the point O where the field is being calculated.

Step 3: Detailed Explanation:
For both straight portions of the wire, the point O lies on the axis of the wire. This means for any current element \(d\vec{l}\) on the straight wires, the position vector \(\vec{r}\) pointing to O is either parallel (angle of 0\(^\circ\)) or anti-parallel (angle of 180\(^\circ\)) to \(d\vec{l}\).
The cross product \(d\vec{l} \times \hat{r}\) has a magnitude of \(|d\vec{l}| |\hat{r}| \sin\theta = dl \cdot 1 \cdot \sin\theta\).
Since \(\theta = 0^\circ\) or \(\theta = 180^\circ\), \(\sin\theta = 0\).
Therefore, the cross product is zero, and the magnetic field contribution from both straight portions at point O is zero.
\[ \vec{B}_{\text{straight}} = 0 \] (ii) Magnetic Field due to Semi-circular Portion

Step 1: Understanding the Concept:
A current flowing through a circular arc produces a magnetic field at the center of the arc. The magnitude of this field is proportional to the current and the angle subtended by the arc at the center.

Step 2: Key Formula or Approach:
The magnetic field at the center of a full circular loop of radius R carrying current i is: \[ B_{\text{full circle}} = \frac{\mu_0 i}{2R} \] A semi-circle is half of a full circle, so it subtends an angle of \(\pi\) radians at the center. The field it produces will be half of that of a full circle.

Step 3: Detailed Explanation:
The magnetic field due to the semi-circular portion of radius R is: \[ B_{\text{semi-circle}} = \frac{1}{2} \times B_{\text{full circle}} = \frac{1}{2} \left( \frac{\mu_0 i}{2R} \right) \] \[ B_{\text{semi-circle}} = \frac{\mu_0 i}{4R} \] Direction: To find the direction of the magnetic field, we use the Right-Hand Grip Rule. If we curl the fingers of our right hand in the direction of the current in the semi-circle (counter-clockwise in the diagram provided, although the arrows suggest a left-to-right flow), the thumb points in the direction of the magnetic field. For the current shown flowing from left to right, the thumb points into the plane of the page.

Step 4: Final Answer:
(i) The magnetic field at O due to each straight portion is zero.
(ii) The value of the magnetic field at O due to the semi-circle of radius R is \(\mathbf{\frac{\mu_0 i}{4R}}\), directed into the plane of the page.