### Related Formula The radius R$R$ of the path of a charged particle moving perpendicular to a magnetic field B$B$ is: R = fracmvqB = fracpqB$R = \frac{mv}{qB} = \frac{p}{qB}$ In terms of kinetic energy K$K$: R = fracsqrt2mKqB$R = \frac{\sqrt{2mK}}{qB}$ Since the particle is accelerated through potential V$V$, kinetic energy K = qV$K = qV$: R = fracsqrt2mqVqB$R = \frac{\sqrt{2mqV}}{qB}$ R = frac1B sqrtfrac2mVq$R = \frac{1}{B} \sqrt{\frac{2mV}{q}}$ ### Core Logic For both particles X and Y, the following parameters are the same: * Potential Difference, V$V$ * Magnetic Field, B$B$ * Charge, q$q$ Therefore, we have the proportionality: R propto sqrtm implies R^2 propto m$R \propto \sqrt{m} \implies R^2 \propto m$ ### Step 1: Calculate Mass Ratio Using the proportionality relationship: fracm_1m_2 = left( fracR_1R_2 right)^2$\frac{m_1}{m_2} = \left( \frac{R_1}{R_2} \right)^2$ Thus, the mass ratio of X and Y is left(fracR_1R_2right)^2$\left(\frac{R_1}{R_2}\right)^2$. ### Pattern Recognition Shortcut: Whenever charges and potential differences are equal, the radius of orbit in a magnetic field scales as R propto sqrtm$R \propto \sqrt{m}$. Squaring both sides yields m propto R^2$m \propto R^2$ instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

### Related Formula The magnetic force on a moving charge is given by the Lorentz force equation: vecF = q (vecv times vecB)$\vec{F} = q (\vec{v} \times \vec{B})$ ### Core Logic Given data: * Charge, q = 4.0\ mutextC = 4.0 times 10^-6text C$q = 4.0\ \mu\text{C} = 4.0 \times 10^{-6}\text{ C}$ * Velocity vector, vecv = (4.0 times 10^6hatj)text ms^-1$\vec{v} = (4.0 \times 10^{6}\hat{j})\text{ ms}^{-1}$ * Magnetic field vector, vecB = 2hatktext T$\vec{B} = 2\hat{k}\text{ T}$ ### Step 1: Calculate the Force Vector Substitute the values into the force equation: vecF = (4.0 times 10^-6) left[ (4.0 times 10^6hatj) times (2hatk) right]$\vec{F} = (4.0 \times 10^{-6}) \left[ (4.0 \times 10^{6}\hat{j}) \times (2\hat{k}) \right]$ vecF = 4.0 times 10^-6 times 8.0 times 10^6 (hatj times hatk)$\vec{F} = 4.0 \times 10^{-6} \times 8.0 \times 10^{6} (\hat{j} \times \hat{k})$ Since hatj times hatk = hati$\hat{j} \times \hat{k} = \hat{i}$: vecF = 32hatitext N$\vec{F} = 32\hat{i}\text{ N}$ Comparing this to xhatitext N$x\hat{i}\text{ N}$, we find: x = 32$x = 32$ ### Pattern Recognition Cross-product check: hatj times hatk = hati$\hat{j} \times \hat{k} = \hat{i}$. The product of the scalar terms (4.0 times 10^-6) times (4.0 times 10^6) times 2$(4.0 \times 10^{-6}) \times (4.0 \times 10^6) \times 2$ immediately simplifies to 4 times 4 times 2 = 32$4 \times 4 \times 2 = 32$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

### Related Formula vecF_textnet = qvecE + q(vecv times vecB)$\vec{F}_{\text{net}} = q\vec{E} + q(\vec{v} \times \vec{B})$ ### Core Logic For velocity to remain constant, the net force must be zero: qvecE + q(vecv times vecB) = 0$q\vec{E} + q(\vec{v} \times \vec{B}) = 0$ Let's evaluate the cases: - Case (A): If E=0$E=0$ and B=0$B=0$, vecF = 0$\vec{F} = 0$. (Possible) - Case (B): If E=0$E=0$ and B neq 0$B \neq 0$, the magnetic force is zero if vecv$\vec{v}$ is parallel or antiparallel to vecB$\vec{B}$ (i.e., vecv times vecB = 0$\vec{v} \times \vec{B} = 0$). (Possible) - Case (C): If E neq 0$E \neq 0$ and B=0$B=0$, vecF = qvecE neq 0$\vec{F} = q\vec{E} \neq 0$, velocity must change. (Not possible) - Case (D): If E neq 0$E \neq 0$ and B neq 0$B \neq 0$, the electric and magnetic forces can balance each other perfectly if qvecE = -q(vecv times vecB)$q\vec{E} = -q(\vec{v} \times \vec{B})$. (Possible) ### Step 1: Conclusion Hence, statements (A), (B), and (D) represent valid situations where velocity can remain constant. ### Pattern Recognition Velocity filter / velocity selector setups utilize crossed fields where electric fields perfectly balance magnetic components, an archetype of Case (D). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

### Related Formula B = fracmu_0 i2pi r$B = \frac{\mu_0 i}{2\pi r}$ ### Core Logic Using the right-hand grip rule, both anti-parallel wire systems generate field arrays pointing in the exact same direction at the central midway coordinate. Hence, their field contributions add up directly: B_textnet = 2 B_1 = 2 left(fracmu_0 i2pi rright) = fracmu_0 ipi r$B_{\text{net}} = 2 B_1 = 2 \left(\frac{\mu_0 i}{2\pi r}\right) = \frac{\mu_0 i}{\pi r}$ Where i = 10text A$i = 10\text{ A}$, total distance = 5text cm implies r = 2.5text cm = 2.5 times 10^-2text m$= 5\text{ cm} \implies r = 2.5\text{ cm} = 2.5 \times 10^{-2}\text{ m}$. ### Step 1: Compute numeric value B_textnet = frac4pi times 10^-7 times 10pi times (2.5 times 10^-2) = frac4 times 10^-62.5 times 10^-2$B_{\text{net}} = \frac{4\pi \times 10^{-7} \times 10}{\pi \times (2.5 \times 10^{-2})} = \frac{4 \times 10^{-6}}{2.5 \times 10^{-2}}$ B_textnet = 1.6 times 10^-4text T = 160 times 10^-6text T = 160\ mutextT$B_{\text{net}} = 1.6 \times 10^{-4}\text{ T} = 160 \times 10^{-6}\text{ T} = 160\ \mu\text{T}$ ### Pattern Recognition Anti-parallel current pairs generate collaborative field additions inside their spatial boundary zone, rather than destructive structural cancellations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

### Related Formula To convert a galvanometer into an ammeter, a small resistance called a shunt (S$S$) is connected in parallel with the galvanometer (G$G$): I_g G = (I - I_g) S implies S = fracI_g GI - I_g$I_g G = (I - I_g) S \implies S = \frac{I_g G}{I - I_g}$ where, G$G$ = resistance of the galvanometer coil I_g$I_g$ = full-scale deflection current of the galvanometer I$I$ = total current to be measured S$S$ = shunt resistance ### Core Logic Given values: G = 10 \, Omega$G = 10 \, \Omega$ I_g = 3 mathrm~mA = 3 times 10^-3 mathrm~A$I_g = 3 \mathrm{~mA} = 3 \times 10^{-3} \mathrm{~A}$ I = 8 mathrm~A$I = 8 \mathrm{~A}$ ### Step 1: Calculate Shunt Resistance Substituting these values into the shunt formula: S = frac(3 times 10^-3) times 108 - 0.003$S = \frac{(3 \times 10^{-3}) \times 10}{8 - 0.003}$ S = frac0.037.997$S = \frac{0.03}{7.997}$ S approx 3.75 times 10^-3 \, Omega$S \approx 3.75 \times 10^{-3} \, \Omega$ Therefore, the required value of the shunt resistance is 3.75 times 10^-3 \, Omega$3.75 \times 10^{-3} \, \Omega$.

### Pattern Recognition Since I gg I_g$I \gg I_g$, the value I - I_g$I - I_g$ in the denominator can be approximated directly as I$I$ for a very accurate quick shortcut in choice selection: S approx fracI_g GI = frac0.038 = 3.75 times 10^-3 \, Omega$S \approx \frac{I_g G}{I} = \frac{0.03}{8} = 3.75 \times 10^{-3} \, \Omega$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism