Let B_1$B_1$ be the magnitude of magnetic field at center of a circular coil of radius R$R$ carrying current I$I$. Let B_2$B_2$ be the magnitude of magnetic field at an axial distance 'x$x$' from the center. For x:R = 3:4$x:R = 3:4$, fracB_2B_1$\frac{B_2}{B_1}$ is:

### 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 force on a current-carrying straight wire: F = I (vecL times vecB) = I L B sin theta$$F = I (\vec{L} \times \vec{B}) = I L B \sin \theta$$ where, I$I$ = current, L$L$ = length of the wire, B$B$ = magnetic field strength, theta$\theta$ = angle between current direction and magnetic field. ### Core Logic Given values: - Length, L = 4.0mathrm~cm = 0.04mathrm~m$L = 4.0\mathrm{~cm} = 0.04\mathrm{~m}$ - Current, I = 8mathrm~A$I = 8\mathrm{~A}$ - Magnetic field strength, B = 0.15mathrm~T$B = 0.15\mathrm{~T}$ - Since the wire is placed perpendicular to the field: theta = 90^circ implies sin 90^circ = 1$\theta = 90^{\circ} \implies \sin 90^{\circ} = 1$ ### Step 1: Substitution and Calculation Substitute values into the force formula: F = 8 times 0.04 times 0.15 times 1$$F = 8 \times 0.04 \times 0.15 \times 1$$ F = 0.32 times 0.15 = 0.048mathrm~N$$F = 0.32 \times 0.15 = 0.048\mathrm{~N}$$ Convert this force into millinewtons (mathrmmN$\mathrm{mN}$): F = 0.048 times 1000 = 48mathrm~mN$$F = 0.048 \times 1000 = 48\mathrm{~mN}$$ ### Pattern Recognition Simple, direct application of Bil-sin-theta! Ensure units are converted to standard SI (meters, amperes, tesla) first, and then converted back into millinewtons at the very end to prevent decimal errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

### 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

### Related Formula The magnetic field contributions from unique structural line elements are given by: * **Semi-infinite straight wire segment** at a distance perpendicular to its \end tip: B_textstraight = fracmu_0 I4pi a$$B_{\text{straight}} = \frac{\mu_0 I}{4\pi a}$$ * **Circular arc path segment** subtending \angle theta$\theta$ at the center: B_textarc = fracmu_0 I4pi a theta$$B_{\text{arc}} = \frac{\mu_0 I}{4\pi a} \theta$$ ### Core Logic Let us decompose the structure into three functional parts as mapped out below[cite: 763, 764]:

1. **Segment 1 (Incoming semi-infinite line)**: The straight line extends to infinity, with its terminating tip at a perpendicular distance a$a$ from origin O$O$. Using the \right-hand grip rule, the direction points into the page: B_1 = fracmu_0 I4pi a quad (otimes)$$B_1 = \frac{\mu_0 I}{4\pi a} \quad (\otimes)$$ 2. **Segment 2 (Three-quarter circular loop)**: The loop forms an \angle of theta = frac3pi2$\theta = \frac{3\pi}{2}$ radians around O$O$. The field points into the page: B_2 = fracmu_0 I4pi a left(frac3pi2right) quad (otimes)$$B_2 = \frac{\mu_0 I}{4\pi a} \left(\frac{3\pi}{2}\right) \quad (\otimes)$$ 3. **Segment 3 (Outgoing semi-infinite line)**: This line aligns perfectly with the origin O$O$ along its vector axis, making sintheta = 0$\sin\theta = 0$ : B_3 = 0$B_3 = 0$ Summing the total fields via superposition: B = B_1 + B_2 + B_3 = fracmu_0 I4pi a + fracmu_0 I4pi aleft(frac3pi2 ight)$$B = B_1 + B_2 + B_3 = \frac{\mu_0 I}{4\pi a} + \frac{\mu_0 I}{4\pi a}\left(\frac{3\pi}{2} ight)$$ B = fracmu_0 I4pi a left[frac3pi2 + 1right]$$B = \frac{\mu_0 I}{4\pi a} \left[\frac{3\pi}{2} + 1\right]$$ ### Pattern Recognition Always check the axis alignment first. Any straight wire segment whose extended line passes directly through the field point contributes exactly zero to the total magnetic field value. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Magnetic Effects of Current and Magnetism