Question 18

We calculate the magnetic field $\vec{B}$ at the origin $O$ by summing the contributions of each segment:

1. Vertical wire segment (Left): Located at distance $x = -L$, length $L$ from $y = -L$ to $y = 0$. Using the formula $B = \frac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2)$: $d = L$, $\theta_1 = 0^\circ$ (level with $O$), $\theta_2 = 45^\circ$ (since $\tan\theta = L/L = 1$). $\vec{B}_1 = \frac{\mu_0 I}{4\pi L}(\sin 45^\circ + \sin 0^\circ) (-\hat{k}) = \frac{\mu_0 I}{4\sqrt{2}\pi L} (-\hat{k})$.

2. Horizontal segments: Any straight wire segment lying on a line passing through the origin $O$ (like segments on the x-axis or the diagonal segment $y = -x$) contributes zero magnetic field at $O$ because $d\vec{l} \times \vec{r} = 0$.

3. Semi-circle arc: Radius $R_1 = L/2$. $\vec{B}_{arc1} = \frac{\mu_0 I}{4R_1} (-\hat{k}) = \frac{\mu_0 I}{4(L/2)} (-\hat{k}) = \frac{\mu_0 I}{2L} (-\hat{k})$.

4. Quarter-circle arc: Radius $R_2 = L/4$. $\vec{B}_{arc2} = \frac{\mu_0 I}{8R_2} (-\hat{k}) = \frac{\mu_0 I}{8(L/4)} (-\hat{k}) = \frac{\mu_0 I}{2L} (-\hat{k})$.

5. Total Magnetic Field: $\vec{B}_{net} = \vec{B}_1 + \vec{B}_{arc1} + \vec{B}_{arc2}$ $\vec{B}_{net} = \left( \frac{\mu_0 I}{4\sqrt{2}\pi L} + \frac{\mu_0 I}{2L} + \frac{\mu_0 I}{2L} \right) (-\hat{k})$ $\vec{B}_{net} = -\frac{\mu_0 I}{L} \left( 1 + \frac{1}{4\sqrt{2}\pi} \right) \hat{k}$.