An infinitely long straight wire carrying current I is bent in a planer shape as shown in the diagram. The radius of the circular part is r. The magnetic field at the centre O of the circular loop is :

### Related Formula For a semi-infinite wire segment at distance r$r$: B = fracmu_0 I4pi r$$B = \frac{\mu_0 I}{4\pi r}$$ For a full circular loop at its center: B = fracmu_0 I2r$$B = \frac{\mu_0 I}{2r}$$ ### Core Logic

The total magnetic field at O is the vector sum of fields from three segments: 1. The incoming semi-infinite wire (AB) 2. The outgoing semi-infinite wire (DE) 3. The nearly full circular loop (BCD) Note: Based on the diagram, the loop is not fully closed, but geometrically it acts as a full circle subtracted by the gap. Typically this standard shape treats the circular part as a full circle and the straight wires as two semi-infinite wires. vecB_O = vecB_AB + vecB_DE + vecB_BCD$$\vec{B}_O = \vec{B}_{AB} + \vec{B}_{DE} + \vec{B}_{BCD}$$ ### Step 1: Adding the Vector Components Applying the Right Hand Rule: - Segment AB: current flows along +x$+x$, position vector to O is +y$+y$. dvecl times vecr = hati times hatj = hatk$d\vec{l} \times \vec{r} = \hat{i} \times \hat{j} = \hat{k}$. Wait, the diagram shows the loop in the x-y$x-y$ plane. Let's re-examine axes. Based on standard convention, if current is in xy$xy$ plane, field is in z$z$ (hatk$\hat{k}$) direction. The solution shows vectors in hati$\hat{i}$. This means the axes are drawn such that the loop is in the y-z$y-z$ plane. Yes, the provided axes show x$x$ pointing out, y$y$ to the right, z$z$ upwards. - Segment AB (current along y$y$ axis): B$B$ at origin is along +x$+x$ (hati$\hat{i}$). - Segment DE (current along y$y$ axis): B$B$ at origin is along +x$+x$ (hati$\hat{i}$). - Circular Loop (current clockwise in y-z$y-z$ plane): B$B$ at origin points inwards, i.e., -x$-x$ (-hati$-\hat{i}$). vecB_AB = fracmu_0 I4pi r hati $$ \vec{B}_{AB} = \frac{\mu_0 I}{4\pi r} \hat{i} $$ vecB_DE = fracmu_0 I4pi r hati $$ \vec{B}_{DE} = \frac{\mu_0 I}{4\pi r} \hat{i} $$ vecB_BCD = - fracmu_0 I2r hati $$ \vec{B}_{BCD} = - \frac{\mu_0 I}{2r} \hat{i} $$ ### Step 2: Final Conclusion vecB_O = fracmu_0 I4pi r hati + fracmu_0 I4pi r hati - fracmu_0 I2r hati$$\vec{B}_O = \frac{\mu_0 I}{4\pi r} \hat{i} + \frac{\mu_0 I}{4\pi r} \hat{i} - \frac{\mu_0 I}{2r} \hat{i}$$ vecB_O = fracmu_0 I2pi r hati - fracmu_0 I2r hati$$\vec{B}_O = \frac{\mu_0 I}{2\pi r} \hat{i} - \frac{\mu_0 I}{2r} \hat{i}$$ vecB_O = fracmu_0 I2pi r (1 - pi) hati$$\vec{B}_O = \frac{\mu_0 I}{2\pi r} (1 - \pi) \hat{i}$$ vecB_O = -fracmu_0 I2pi r (pi - 1) hati$$\vec{B}_O = -\frac{\mu_0 I}{2\pi r} (\pi - 1) \hat{i}$$ ### Pattern Recognition Always separate complex wire geometries into standard segments: infinite wires, semi-infinite wires, and arcs. Use the Right-Hand Rule carefully with the given explicit coordinate frame to avoid sign errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism