### Related Formula B_1 = fracmu_0 I2R$B_1 = \frac{\mu_0 I}{2R}$ B_2 = fracmu_0 I R^22(R^2 + x^2)^3/2$B_2 = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$ B_2 = B_1 sin^3theta quad textwhere sintheta = fracRsqrtR^2 + x^2$B_2 = B_1 \sin^3\theta \quad \text{where } \sin\theta = \frac{R}{\sqrt{R^2 + x^2}}$ ### Core Logic Given the ratio x : R = 3 : 4$x : R = 3 : 4$. Let x = 3k$x = 3k$ and R = 4k$R = 4k$. The distance to the element is: sqrtR^2 + x^2 = sqrt(4k)^2 + (3k)^2 = 5k$\sqrt{R^2 + x^2} = \sqrt{(4k)^2 + (3k)^2} = 5k$ The sine of the semi-vertical angle subtended at the axial point is: sintheta = fracRsqrtR^2 + x^2 = frac4k5k = frac45$\sin\theta = \frac{R}{\sqrt{R^2 + x^2}} = \frac{4k}{5k} = \frac{4}{5}$ We know that the axial magnetic field relates to the central magnetic field as follows: fracB_2B_1 = sin^3theta = left(frac45 ight)^3 = frac64125$\frac{B_2}{B_1} = \sin^3\theta = \left(\frac{4}{5} ight)^3 = \frac{64}{125}$ ### Step 1: Final Conclusion The ratio $ ### Step 1: Final Conclusion The ratio $\frac{B_2}{B_1} is $ is $64:125. ### Pattern Recognition For axial magnetic field problems, bypass manual calculation of $. ### Pattern Recognition For axial magnetic field problems, bypass manual calculation of $(R^2+x^2)^{3/2} by writing the relation directly in terms of the subtended angle: $ by writing the relation directly in terms of the subtended angle: $B_{\text{axial}} = B_{\text{center}} \sin^3\theta$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Magnetic Effects of Current

### Related Formula B_textwire = fracmu_0 I4pi d left(sintheta_1 + sintheta_2right)$B_{\text{wire}} = \frac{\mu_0 I}{4\pi d} \left(\sin\theta_1 + \sin\theta_2\right)$ where, B_textwire$B_{\text{wire}}$ = magnetic field due to a straight wire segment d$d$ = perpendicular distance from center to wire segment (d = a/2$d = a/2$) theta_1, theta_2$\theta_1, \theta_2$ = subtended angles at the center ### Core Logic The loop parts ABC and ADC are in parallel. Let total current entering corner A and leaving corner C be I$I$. - Resistance of path ABC = r$r$ - Resistance of path ADC = 2r$2r$ Using current divider rule: - Current in path ABC, I_1 = frac2rr + 2r I = frac23 I$I_1 = \frac{2r}{r + 2r} I = \frac{2}{3} I$ - Current in path ADC, I_2 = fracrr + 2r I = frac13 I$I_2 = \frac{r}{r + 2r} I = \frac{1}{3} I$ Now, calculate the magnetic field contribution at center O$O$: - Distance of center from each of the 4 wire segments is d = fraca2$d = \frac{a}{2}$. - For each segment, the angles are theta_1 = theta_2 = 45^circ$\theta_1 = \theta_2 = 45^\circ$: B_textsegment = fracmu_0 i4pi (a/2) left(sin 45^circ + sin 45^circright) = fracmu_0 i2pi a sqrt2$B_{\text{segment}} = \frac{\mu_0 i}{4\pi (a/2)} \left(\sin 45^\circ + \sin 45^\circ\right) = \frac{\mu_0 i}{2\pi a} \sqrt{2}$ ### Step 1: Summing the Fields at the Center Using right-hand rule to find the direction of magnetic fields: - Paths AB and BC (carrying I_1$I_1$ clockwise) create fields pointing into the page (-hatk$-\hat{k}$). - Paths AD and DC (carrying I_2$I_2$ counter-clockwise) create fields pointing out of the page (+hatk$+\hat{k}$). B_textnet = 2 B_textsegment(I_2) - 2 B_textsegment(I_1)$B_{\text{net}} = 2 B_{\text{segment}}(I_2) - 2 B_{\text{segment}}(I_1)$ B_textnet = 2 left[ fracmu_0 (I/3)2pi a sqrt2 right] - 2 left[ fracmu_0 (2I/3)2pi a sqrt2 right]$B_{\text{net}} = 2 \left[ \frac{\mu_0 (I/3)}{2\pi a} \sqrt{2} \right] - 2 \left[ \frac{\mu_0 (2I/3)}{2\pi a} \sqrt{2} \right]$ B_textnet = fracsqrt2mu_0 Ipi a left( frac13 - frac23 right) = -fracsqrt2mu_0 I3pi ahatk$B_{\text{net}} = \frac{\sqrt{2}\mu_0 I}{\pi a} \left( \frac{1}{3} - \frac{2}{3} \right) = -\frac{\sqrt{2}\mu_0 I}{3\pi a}\hat{k}$ Taking the magnitude of the field: |B_textnet| = fracsqrt2mu_mathrmo I3pi a$|B_{\text{net}}| = \frac{\sqrt{2}\mu_{\mathrm{o}} I}{3\pi a}$ ### Pattern Recognition Sees: Current divides into parallel paths of a symmetric loop. Trap: In a completely symmetric square loop (equal resistances), the net magnetic field at the center is exactly 0$0$. However, here the paths have unequal resistances (r$r$ and 2r$2r$), which prevents total cancellation. Shortcut: Find the net effective current imbalance Delta I = I_1 - I_2 = frac23I - frac13I = frac13I$\Delta I = I_1 - I_2 = \frac{2}{3}I - \frac{1}{3}I = \frac{1}{3}I$. The magnitude is simply the magnetic field contribution of two sides carrying this net imbalance. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Magnetic Effects of Current

### Related Formula Magnetic field at the center of a circular segment of angle theta$\theta$: B = fracmu_0 I4pi R theta$B = \frac{\mu_0 I}{4\pi R} \theta$ For a semicircle (theta = pi$\theta = \pi$): B_textsemi = fracmu_0 I4 R$B_{\text{semi}} = \frac{\mu_0 I}{4 R}$ ### Core Logic Let's analyze the contributions from each part of the loop to the magnetic field at the center O$O$: - The straight wire segments AB$AB$ and CD$CD$ lie along radial lines passing directly through O$O$. Since dvecl parallel vecr$d\vec{l} \parallel \vec{r}$, their magnetic field contribution is zero: B_textAB = B_textCD = 0$B_{\text{AB}} = B_{\text{CD}} = 0$ - Semicircular segment of radius R_2 = 4pimathrm~m$R_2 = 4\pi\mathrm{~m}$ carries current creating a field pointing out of the page (by right-hand rule): B_R2 = fracmu_0 I4 R_2 quad (textOut of page)$B_{R2} = \frac{\mu_0 I}{4 R_2} \quad (\text{Out of page})$ - Semicircular segment of radius R_1 = 6pimathrm~m$R_1 = 6\pi\mathrm{~m}$ carries current creating a field pointing into the page: B_R1 = fracmu_0 I4 R_1 quad (textInto page)$B_{R1} = \frac{\mu_0 I}{4 R_1} \quad (\text{Into page})$ ### Step 1: Calculating Resultant Field Since R_2 < R_1$R_2 < R_1$, the field B_R2$B_{R2}$ is stronger. The net magnetic field is: B_textnet = B_R2 - B_R1 = fracmu_0 I4left(frac1R_2 - frac1R_1right)$B_{\text{net}} = B_{R2} - B_{R1} = \frac{\mu_0 I}{4}\left(\frac{1}{R_2} - \frac{1}{R_1}\right)$ Substitute the given values (I = 12mathrm~A$I = 12\mathrm{~A}$, R_2 = 4pi$R_2 = 4\pi$, R_1 = 6pi$R_1 = 6\pi$, mu_0 = 4pi times 10^-7$\mu_0 = 4\pi \times 10^{-7}$): B_textnet = frac(4pi times 10^-7) times 124 left(frac14pi - frac16piright)$B_{\text{net}} = \frac{(4\pi \times 10^{-7}) \times 12}{4} \left(\frac{1}{4\pi} - \frac{1}{6\pi}\right)$ B_textnet = 12pi times 10^-7 times left(frac6pi - 4pi24pi^2right)$B_{\text{net}} = 12\pi \times 10^{-7} \times \left(\frac{6\pi - 4\pi}{24\pi^2}\right)$ B_textnet = 12pi times 10^-7 times frac2pi24pi^2$B_{\text{net}} = 12\pi \times 10^{-7} \times \frac{2\pi}{24\pi^2}$ B_textnet = 12pi times 10^-7 times frac112pi = 1 times 10^-7mathrm~T$B_{\text{net}} = 12\pi \times 10^{-7} \times \frac{1}{12\pi} = 1 \times 10^{-7}\mathrm{~T}$ Comparing this to k times 10^-7mathrm~T$k \times 10^{-7}\mathrm{~T}$: k = 1$k = 1$ ### Pattern Recognition Notice how the radial straight wires never contribute to the magnetic field at the center. Semicircles with opposite currents simply subtract. Symmetrizing the math early by factoring out fracmu_0 I4pi$\frac{\mu_0 I}{4\pi}$ makes the calculations incredibly neat and quick. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Magnetic Effects of Current: Biot-Savart Law