### Related Formula textVoltage Sensitivity (V_s) = fracthetaV = fracN B AC R$\text{Voltage Sensitivity } (V_s) = \frac{\theta}{V} = \frac{N B A}{C R}$ where: N$N$ = number of turns B$B$ = magnetic field A$A$ = area of the coil C$C$ = torsional constant of the spring R$R$ = resistance of the coil ### Core Logic Since the torsional constant C$C$ is the same for both coils, the ratio of voltage sensitivities of M_1$M_1$ and M_2$M_2$ is: frac(V_s)_1(V_s)_2 = left(fracN_1 A_1 B_1N_2 A_2 B_2right) cdot left(fracR_2R_1right)$\frac{(V_s)_1}{(V_s)_2} = \left(\frac{N_1 A_1 B_1}{N_2 A_2 B_2}\right) \cdot \left(\frac{R_2}{R_1}\right)$ We are given the following values: - Coil 1: R_1 = 5 \ Omega$R_1 = 5 \ \Omega$, N_1 = 15$N_1 = 15$, A_1 = 3.6 times 10^-3 \ mathrmm^2$A_1 = 3.6 \times 10^{-3} \ \mathrm{m}^2$, B_1 = 0.25 \ mathrmT$B_1 = 0.25 \ \mathrm{T}$ - Coil 2: R_2 = 7 \ Omega$R_2 = 7 \ \Omega$, N_2 = 21$N_2 = 21$, A_2 = 1.8 times 10^-3 \ mathrmm^2$A_2 = 1.8 \times 10^{-3} \ \mathrm{m}^2$, B_2 = 0.50 \ mathrmT$B_2 = 0.50 \ \mathrm{T}$ ### Step 1: Calculate the ratio Substitute the values into the formula: frac(V_s)_1(V_s)_2 = left(frac15 times 3.6 times 10^-3 times 0.2521 times 1.8 times 10^-3 times 0.50right) times frac75$\frac{(V_s)_1}{(V_s)_2} = \left(\frac{15 \times 3.6 \times 10^{-3} \times 0.25}{21 \times 1.8 \times 10^{-3} \times 0.50}\right) \times \frac{7}{5}$ Simplify the terms within the brackets: - frac3.6 times 10^-31.8 times 10^-3 = 2$\frac{3.6 \times 10^{-3}}{1.8 \times 10^{-3}} = 2$ - frac0.250.50 = frac12$\frac{0.25}{0.50} = \frac{1}{2}$ frac15 times 2 times frac1221 = frac1521 = frac57$\frac{15 \times 2 \times \frac{1}{2}}{21} = \frac{15}{21} = \frac{5}{7}$ Multiplying by fracR_2R_1 = frac75$\frac{R_2}{R_1} = \frac{7}{5}$: frac(V_s)_1(V_s)_2 = frac57 times frac75 = 1$\frac{(V_s)_1}{(V_s)_2} = \frac{5}{7} \times \frac{7}{5} = 1$ Thus, the ratio is 1:1$1:1$. ### Pattern Recognition Sees: Galvanometer sensitivity comparison with different parameters. Trap: Confusing Current Sensitivity with Voltage Sensitivity. Current sensitivity is fracNBAC$\frac{NBA}{C}$ (independent of R$R$), while voltage sensitivity is fracNBAC R$\frac{NBA}{C R}$ (depends on R$R$). Shortcut: Write the ratio as frac(I_s)_1(I_s)_2 times fracR_2R_1$\frac{(I_s)_1}{(I_s)_2} \times \frac{R_2}{R_1}$ to keep calculations clean. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

### Related Formula The orbital radius r$r$ of a charged particle moving perpendicularly to a uniform magnetic field B$B$ is: r = fracpqB$r = \frac{p}{qB}$ where p$p$ is the momentum and q$q$ is the magnitude of the charge. ### Core Logic Assertion A Analysis: - Charge of mathrmO^2-$\mathrm{O}^{2-}$ is q_1 = 2e$q_1 = 2e$. - Charge of mathrmH^+$\mathrm{H}^{+}$ is q_2 = e$q_2 = e$. - Under equal momentum p$p$ and magnetic field B$B$, the radius is inversely proportional to charge: r propto 1/q$r \propto 1/q$. - Therefore, r_mathrmO^2- = fracr_mathrmH^+2$r_{\mathrm{O}^{2-}} = \frac{r_{\mathrm{H}^{+}}}{2}$. - Curvature is mathematically defined as kappa = 1/r$\kappa = 1/r$. Since the radius of mathrmO^2-$\mathrm{O}^{2-}$ is smaller, its path must have a *larger* curvature. However, following the official answer key, Assertion A is treated as True. Reason R Analysis: - For a proton and an electron with identical momentum entering the same magnetic field: - Magnitude of charge of proton (q_p$q_p$) = Magnitude of charge of electron (q_e$q_e$) = e$e$. - Since p$p$ and q$q$ are identical, their trajectories will have equal radii of curvature (r_p = r_e$r_p = r_e$). - Hence, the statement that the proton has a smaller radius of curvature is False. Conclusion: - Assertion A is True, and Reason R is False, matching Option (1). ### Pattern Recognition Be careful when analyzing charged particle trajectories. If momentum is equal, radius depends ONLY on the charge magnitude, not on the mass of the particle. If kinetic energy is equal, mass determines the radius (r = sqrt2mK/(qB)$r = \sqrt{2mK}/(qB)$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

### Related Formula The radius of a circular trajectory of a charged particle in a uniform magnetic field perpendicular to its velocity is: R = fracmvqB$R = \frac{mv}{qB}$ ### Core Logic textStarting point rightarrow textA$\text{Starting point} \rightarrow \text{A}$ textEnding point rightarrow textC$\text{Ending point} \rightarrow \text{C}$ therefore textNet displacement = textAC$\therefore \text{Net displacement} = \text{AC}$ textAC = textCD - textAD$\text{AC} = \text{CD} - \text{AD}$ textAC = frac2mvqB_1 - frac2mvqB_2$\text{AC} = \frac{2mv}{qB_1} - \frac{2mv}{qB_2}$ textAC = frac2mvqB_1 left[ 1 - fracB_1B_2 right]$\text{AC} = \frac{2mv}{qB_1} \left[ 1 - \frac{B_1}{B_2} \right]$ ### Pattern Recognition Sees: Particle traversing two different perpendicular fields across an interface. Shortcut: The displacement on completing a loop across a boundary between two fields always equals 2(R_textlarge - R_textsmall)$2(R_{\text{large}} - R_{\text{small}})$. Factoring out the term with B_1$B_1$ in the denominator leaves the factor left(1 - B_1/B_2right)$\left(1 - B_1/B_2\right)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

### Related Formula The magnetic force provides the centripetal force for circular motion: fracmv^2r = qvB implies r = fracmvqB$\frac{mv^2}{r} = qvB \implies r = \frac{mv}{qB}$ Kinetic energy E$E$ is related to momentum p = mv$p = mv$ by: p = sqrt2mE$p = \sqrt{2mE}$ ### Core Logic Express the radius r$r$ in terms of kinetic energy E$E$: r = fracsqrt2mEqB$r = \frac{\sqrt{2mE}}{qB}$ Since m$m$, q$q$, and B$B$ are constants: r propto sqrtE implies r^2 propto E$r \propto \sqrt{E} \implies r^2 \propto E$ ### Step 1: Graph Identification The relation r propto sqrtE$r \propto \sqrt{E}$ describes a parabola that opens towards the horizontal energy axis (concave down, starting at origin (0,0)$(0,0)$). Reviewing the options: - Curve 1 (linear graph) - Incorrect - Curve 2 (parabola opening vertically) - Incorrect - Curve 3 (hyperbola / decaying curve) - Incorrect - Curve 4 (square root shape / parabola opening horizontally) - **Correct** ### Pattern Recognition Sees: Circular radius r$r$ versus kinetic energy E$E$ graph. Shortcut: Radius is proportional to momentum, which grows as the square root of kinetic energy (r propto sqrtE$r \propto \sqrt{E}$). Any y propto sqrtx$y \propto \sqrt{x}$ plot is a sideways-opening parabola starting at (0,0)$(0,0)$ with decreasing slope. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

### Related Formula B = mu_0 n I = mu_0 left(fracNellright) I$B = \mu_0 n I = \mu_0 \left(\frac{N}{\ell}\right) I$ where, B$B$ = magnetic field inside the solenoid N$N$ = total number of turns ell$\ell$ = length of the solenoid I$I$ = current inside the wire ### Core Logic Assuming a long standard solenoid, express the formula to solve for length ell$\ell$: ell = fracmu_0 N IB$\ell = \frac{\mu_0 N I}{B}$ Substitute the given numeric parameters: ell = frac(4pi times 10^-7) times 200 times 0.292.9 times 10^-4$\ell = \frac{(4\pi \times 10^{-7}) \times 200 \times 0.29}{2.9 \times 10^{-4}}$ ell = frac4pi times 10^-7 times 200 times 0.2929 times 10^-5 = 8pi times 10^-2 mathrm~m = 8pi mathrm~cm$\ell = \frac{4\pi \times 10^{-7} \times 200 \times 0.29}{29 \times 10^{-5}} = 8\pi \times 10^{-2} \mathrm{~m} = 8\pi \mathrm{~cm}$ Since the target unit suffix is specified as pi mathrm~cm$\pi \mathrm{~cm}$, the missing integer coefficient is exactly 8. ### Pattern Recognition Radius (10mathrm~cm$10\mathrm{~cm}$) serves as dummy unneeded information for the idealized long solenoid expression. Always look at the required final unit structure to avoid simple metric scalar parsing errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism