In a moving coil galvanometer, two moving coils M_1$M_1$ and M_2$M_2$ have the following particulars: beginaligned R_1 &= 5 \ Omega, quad N_1 = 15, quad A_1 = 3.6 times 10^-3 \ mathrmm^2, quad B_1 = 0.25 \ mathrmT \\ R_2 &= 7 \ Omega, quad N_2 = 21, quad A_2 = 1.8 times 10^-3 \ mathrmm^2, quad B_2 = 0.50 \ mathrmT endaligned$$\begin{aligned} R_1 &= 5 \ \Omega, \quad N_1 = 15, \quad A_1 = 3.6 \times 10^{-3} \ \mathrm{m}^2, \quad B_1 = 0.25 \ \mathrm{T} \\ R_2 &= 7 \ \Omega, \quad N_2 = 21, \quad A_2 = 1.8 \times 10^{-3} \ \mathrm{m}^2, \quad B_2 = 0.50 \ \mathrm{T} \end{aligned}$$ Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of M_1$M_1$ and M_2$M_2$ ?

### Related Formula I = fracqT = fracqomega2pi$$I = \frac{q}{T} = \frac{q\omega}{2\pi}$$ ### Core Logic The loop A encloses one of the moving point charges as it moves past, giving a current contribution localized to that cross-sectional segment intersection: I_A = fracNqleft(frac2piomega ight) = fracNqomega2pi$$I_A = \frac{Nq}{\left(\frac{2\pi}{\omega} ight)} = \frac{Nq\omega}{2\pi}$$ Loop B encloses the entire loop surface coplanar or enclosing the ring structure fully without clipping individual passing current tracks perpendicularly in the same directional fashion, resulting in zero net cross-surface passing enclosed current: I_B = 0$I_B = 0$ Therefore, the difference is: I_A - I_B = fracNqomega2pi$$I_A - I_B = \frac{Nq\omega}{2\pi}$$

### Pattern Recognition A current loop has net passing current across a large overarching bounding box equal to zero if it doesn't cross the boundary surfaces symmetrically. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

### Related Formula Time period of a revolving charge in a magnetic field: T = frac2pi mqB$$T = \frac{2\pi m}{qB}$$ Magnetic field inside a long solenoid: B = mu_0 n I$B = \mu_0 n I$ ### Core Logic Combining the expressions to isolate n$n$ (turns per meter): T = frac2pi mq(mu_0 n I)$$T = \frac{2\pi m}{q(\mu_0 n I)}$$ Substituting the given constants: 75 times 10^-9 = frac2pi times 9 times 10^-311.6 times 10^-19 times 4pi times 10^-7 times n times 1.5$$75 \times 10^{-9} = \frac{2\pi \times 9 \times 10^{-31}}{1.6 \times 10^{-19} \times 4\pi \times 10^{-7} \times n \times 1.5}$$ Simplifying terms: 75 times 10^-9 = frac18pi times 10^-319.6pi times 10^-26 times n = frac1.875 times 10^-5n$$75 \times 10^{-9} = \frac{18\pi \times 10^{-31}}{9.6\pi \times 10^{-26} \times n} = \frac{1.875 \times 10^{-5}}{n}$$ n = frac1.875 times 10^-575 times 10^-9 = 250$$n = \frac{1.875 \times 10^{-5}}{75 \times 10^{-9}} = 250$$ ### Pattern Recognition The circular motion time period depends exclusively on the field magnitude B$B$, completely independent of the orbit's velocity or radius. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

### Related Formula The magnetic field B_1$B_{1}$ produced by a straight wire segment carrying current I$I$ at a perpendicular distance d$d$ is given by the Biot-Savart relation: B_1 = fracmu_0I4pi d(sintheta_1 + sintheta_2)$$B_{1} = \frac{\mu_{0}I}{4\pi d}(\sin\theta_{1} + \sin\theta_{2})$$ ### Core Logic As shown in the square geometric layout

, the perpendicular distance from any side to the central origin point is exactly half the total side length : d = fraca2 = frac12sqrt2text m$$d = \frac{a}{2} = \frac{1}{2\sqrt{2}}\text{ m}$$ Connecting the ends of a side to the center forms internal angles of theta_1 = theta_2 = 45^circ$\theta_{1} = \theta_{2} = 45^{\circ}$. ### Step 1: Summing the Contributions Calculate the magnetic field contribution from a single side : B_1 = frac10^-7 times 5frac12sqrt2 left(sin 45^circ + sin 45^circ ight) = 10^-7 times 10sqrt2 times left(frac2sqrt2 ight) = 2 times 10^-6text T$$B_{1} = \frac{10^{-7} \times 5}{\frac{1}{2\sqrt{2}}} \left(\sin 45^{\circ} + \sin 45^{\circ} ight) = 10^{-7} \times 10\sqrt{2} \times \left(\frac{2}{\sqrt{2}} ight) = 2 \times 10^{-6}\text{ T}$$ Since the current flows in the same rotational direction along all four sides, their individual magnetic fields add constructively at the center : B_textnet = 4 times B_1 = 4 times (2 times 10^-6text T) = 8 times 10^-6text T$$B_{\text{net}} = 4 \times B_{1} = 4 \times (2 \times 10^{-6}\text{ T}) = 8 \times 10^{-6}\text{ T}$$ Comparing this with p times 10^-6text T$p \times 10^{-6}\text{ T}$ , we get: p = 8$p = 8$ ### Pattern Recognition The magnetic field at the center of any square loop simplifies to the standard formula: B = frac2sqrt2mu_0Ipi a$B = \frac{2\sqrt{2}\mu_{0}I}{\pi a}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

### Related Formula B_max = fracmu_0 I2pi a$$B_{\max} = \frac{\mu_0 I}{2\pi a}$$ B_textin = fracmu_0 I r2pi a^2, quad B_textout = fracmu_0 I2pi r$$B_{\text{in}} = \frac{\mu_0 I r}{2\pi a^2}, \quad B_{\text{out}} = \frac{\mu_0 I}{2\pi r}$$ ### Core Logic The maximum magnetic field occurs right at the wire\'s outer boundary surface (r=a$r=a$) : B_max = fracmu_0 I2pi a$$B_{\max} = \frac{\mu_0 I}{2\pi a}$$ We need positions where B = fracB_max2 = fracmu_0 I4pi a$B = \frac{B_{\max}}{2} = \frac{\mu_0 I}{4\pi a}$. ### Step 1: Calculate Inside Distance fracmu_0 I r2pi a^2 = fracmu_0 I4pi a implies r = fraca2$$\frac{\mu_0 I r}{2\pi a^2} = \frac{\mu_0 I}{4\pi a} \implies r = \frac{a}{2}$$ ### Step 2: Calculate Outside Distance fracmu_0 I2pi r = fracmu_0 I4pi a implies r = 2a$$\frac{\mu_0 I}{2\pi r} = \frac{\mu_0 I}{4\pi a} \implies r = 2a$$ ### Pattern Recognition Inside the wire, field scales linearly with radius; outside, it falls inversely with radius. ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

### Related Formula Voltmeter series conversion formula: V = I_g(R_g + R)$$V = I_g(R_g + R)$$ R = fracVI_g - R_g$$R = \frac{V}{I_g} - R_g$$ ### Core Logic Given data: R_g = 50mathrm~Omega$R_g = 50\mathrm{~\Omega}$, I_g = 5mathrm~mA = 5 times 10^-3mathrm~A$I_g = 5\mathrm{~mA} = 5 \times 10^{-3}\mathrm{~A}$, target voltage range V = 100mathrm~V$V = 100\mathrm{~V}$. Substitute values: R = frac1005 times 10^-3 - 50$$R = \frac{100}{5 \times 10^{-3}} - 50$$ ### Step 1: Complete Arithmetic Evaluation R = 20000 - 50 = 19950mathrm~Omega$R = 20000 - 50 = 19950\mathrm{~\Omega}$ ### Pattern Recognition Voltmeter resistance is always high because it is connected in parallel to circuits to prevent current drawing leaks. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism Class 12 Physics: Current Electricity