Q.A current carrying loop consists of 3 identical quarter circles of radius $R$, lying in the positive quadrants of the $x$-$y$, $y$-$z$ and $z$-$x$ planes with their centres at the origin, joined together. Find the direction and magnitude of $\vec{B}$ at the origin.

Concept: Magnetic Force Balance — The net field at the centre is the vector sum of the fields from each quarter‑circle arc.

Reasoning:

1. For a full circular loop of radius $R$ carrying current $I$, the field at the centre is $B = \frac{\mu_0 I}{2R}$, directed perpendicular to the plane of the loop (right‑hand rule).
2. A quarter‑circle arc subtends $90^\circ$, so it contributes one‑quarter of that field: $B_{\text{arc}} = \frac{1}{4} \cdot \frac{\mu_0 I}{2R} = \frac{\mu_0 I}{8R}$.
3. Each arc lies in a different coordinate plane, so its field at the origin points along the axis perpendicular to that plane:
   • Arc in $x$-$y$ plane → field along $+z$ axis.
   • Arc in $y$-$z$ plane → field along $+x$ axis.
   • Arc in $z$-$x$ plane → field along $+y$ axis. …

Why magnetic force balance works here

The Biot–Savart law tells us that a current element $I\,d\vec{l}$ produces a field $d\vec{B} \propto d\vec{l} \times \hat{r}/r^2$. For a circular arc centred at the origin, every element $d\vec{l}$ is perpendicular to the radius vector $\vec{r}$ (which points from the origin to the element), and the cross product always points along the axis of the arc. Because the geometry is symmetric, the field at the centre of a full circle is $\mu_0 I/(2R)$; a quarter‑circle gives exactly one‑quarter of that — but only if the arc lies in a single plane.

The trick: each quarter‑circle lies in a different coordinate plane, so the three fields are not in the same direction. They are perpendicular to each other. The net field is therefore the vector sum of three orthogonal vectors of equal magnitude.

Step‑by‑step calculation

1. Field from a single quarter‑circle arc

For a full circular loop of radius $R$ carrying current $I$, the field at the centre is

directed perpendicular to the plane of the loop (right‑hand rule).

A quarter‑circle is one‑fourth of the loop, so the field magnitude is one‑fourth:

2. Direction of each quarter‑circle’s field

• Arc in the $x$‑$y$ plane (positive quadrant: $x\ge0$, $y\ge0$, $z=0$):

  The axis is the $z$‑axis. By the right‑hand rule, if the current flows from the positive $x$‑axis toward the positive $y$‑axis (counter‑clockwise when viewed from above), the field at the origin points along $+\hat{k}$.

• Arc in the $y$‑$z$ plane (positive quadrant: $y\ge0$, $z\ge0$, $x=0$):

  The axis is the $x$‑axis. Current flowing from $+y$ to $+z$ gives field along $+\hat{i}$.

• Arc in the $z$‑$x$ plane (positive quadrant: $z\ge0$, $x\ge0$, $y=0$):

  The axis is the $y$‑axis. Current flowing from $+z$ to $+x$ gives field along $+\hat{j}$.

Thus the three field vectors are: …

Method: Field at the Center From an Arc, Combined by Vector Addition Across Planes

General technique for any "loop made of arcs lying in different planes" field-at-a-point problem.

Steps

Step 1: Find the field from a full circular loop, then scale by the fraction of the circle the arc spans

This shortcut works only when the arc is centred exactly at the point where you want the field — every element then contributes in the same direction.

Step 2: Assign the DIRECTION of each arc's contribution one plane at a time, using the right-hand rule

Each planar arc's field points along the axis perpendicular to its own plane; the sign follows from curling your right-hand fingers along the arc's given current direction.

Step 3: Add the contributions as VECTORS, not magnitudes, when the arcs lie in different planes …