Uniform magnetic fields of different strengths (mathrmB_1$(\mathrm{B}_1$ and mathrmB_2)$\mathrm{B}_2)$ , both normal to the plane of the paper exist as shown in the figure. A charged particle of mass m$m$ and charge q$q$ , at the interface at an instant, moves into the region 2 with velocity v$v$ and returns to the interface. It continues to move into region 1 and finally reaches the interface. What is the displacement of the particle during this movement along the interface?

(Consider the velocity of the particle to be normal to the magnetic field and mathrmB_2 > mathrmB_1$\mathrm{B}_2 > \mathrm{B}_1$ )

### 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

### Core Logic Analyzing motion from path phases mathrmA rightarrow mathrmB$\mathrm{A} \rightarrow \mathrm{B}$ and mathrmB rightarrow mathrmC$\mathrm{B} \rightarrow \mathrm{C}$ inside the coordinate field:

vecmathrmB = fracmu_0 mathrmI2pi mathrmr(-hatmathrmk)$$\vec{\mathrm{B}} = \frac{\mu_0 \mathrm{I}}{2\pi \mathrm{r}}(-\hat{\mathrm{k}})$$ The lorentz magnetic field acceleration rules dictate differential trajectory steps: int_v_0^0 fracv_x dv_xsqrtv_0^2 - v_x^2 = -fracmu_0 mathrmI q2pi mathrmm int_a^x_1 fracdrr$$\int_{v_0}^{0} \frac{v_x dv_x}{\sqrt{v_0^2 - v_x^2}} = -\frac{\mu_0 \mathrm{I q}}{2\pi \mathrm{m}} \int_{a}^{x_1} \frac{dr}{r}$$ Solving this integration step gives the position node parameter: x_1 = a e^-frac2pi m v_0mu_0 I q$$x_1 = a e^{-\frac{2\pi m v_0}{\mu_0 I q}}$$ Compounding this loop interaction for the turning path phase mathrmB rightarrow mathrmC$\mathrm{B} \rightarrow \mathrm{C}$ gives: ### Step 1: Final Solution Integration x = x_1 e^-frac2pi m v_0mu_0 I q = a e^-frac4pi m v_0mu_0 I q$$x = x_1 e^{-\frac{2\pi m v_0}{\mu_0 I q}} = a e^{-\frac{4\pi m v_0}{\mu_0 I q}}$$ Matches criteria for option (4). ### Pattern Recognition Variable magnetic field cross-products result in dual exponential scaling metrics. Remember the total velocity magnitude stays fixed under zero work magnetic operations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

### Related Formula Time Period of circular motion in a magnetic field: T = frac2pi mqB$$T = \frac{2\pi m}{qB}$$ ### Core Logic Convert given parameter values into SI baseline metrics: q = 1.6\ mutextC = 1.6 times 10^-6text C$$q = 1.6\ \mu\text{C} = 1.6 \times 10^{-6}\text{ C}$$ m = 16\ mutextg = 16 times 10^-9text kg$$m = 16\ \mu\text{g} = 16 \times 10^{-9}\text{ kg}$$ B = 6.28text T = 2pitext T$$B = 6.28\text{ T} = 2\pi\text{ T}$$ ### Step 1: Solve for Time Period T = frac2pi times 16 times 10^-91.6 times 10^-6 times 6.28 = frac6.28 times 16 times 10^-91.6 times 10^-6 times 6.28$$T = \frac{2\pi \times 16 \times 10^{-9}}{1.6 \times 10^{-6} \times 6.28} = \frac{6.28 \times 16 \times 10^{-9}}{1.6 \times 10^{-6} \times 6.28}$$ T = frac16 times 10^-91.6 times 10^-6 = 10 times 10^-3 = 0.01text seconds$$T = \frac{16 \times 10^{-9}}{1.6 \times 10^{-6}} = 10 \times 10^{-3} = 0.01\text{ seconds}$$ Rounding to the nearest integer as required for standard integer formatting gives **0**.

### Pattern Recognition Check your prefixes. Micrograms (mutextg$\mu\text{g}$) conversion introduces a factor of 10^-9text kg$10^{-9}\text{ kg}$, not 10^-6$10^{-6}$. If nearest integer is requested, 0.01text s$0.01\text{ s}$ rounds down cleanly to 0. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

### Related Formula vecF = q(vecv times vecB)$$\vec{F} = q(\vec{v} \times \vec{B})$$ ### Core Logic For a particle to move with a constant velocity in a straight line inside a magnetic field alone, the net magnetic force must be zero: vecF = 0 implies vecv parallel vecB$$\vec{F} = 0 \implies \vec{v} \parallel \vec{B}$$ This means the angle theta$\theta$ between the velocity vector and the magnetic field vector is either 0^circ$0^{\circ}$ or 180^circ$180^{\circ}$. Thus, if the magnetic field is along the direction of velocity, the force is zero, allowing unaccelerated straight-line motion. Both Assertion and Reason are true, and the Reason correctly explains the Assertion. ### Pattern Recognition Magnetic field cannot change the speed of a charged particle, but it changes direction unless vecv$\vec{v}$ is parallel or anti-parallel to vecB$\vec{B}$, in which case the magnetic force vanishes entirely. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

### Related Formula B_textin = fracmu_0 I r2pi a^2 implies B_textin propto r$$B_{\text{in}} = \frac{\mu_0 I r}{2\pi a^2} \implies B_{\text{in}} \propto r$$ B_textout = fracmu_0 I2pi r implies B_textout propto frac1r $$B_{\text{out}} = \frac{\mu_0 I}{2\pi r} \implies B_{\text{out}} \propto \frac{1}{r} $$ ### Core Logic Inside the wire (r < a$r < a$), the magnetic field grows linearly with distance r$r$ from the axis. At the surface (r = a$r = a$), it reaches its maximum value B_textmax = fracmu_0 I2pi a$B_{\text{\max}} = \frac{\mu_0 I}{2\pi a}$. Outside the wire (r > a$r > a$), it decays inversely with r$r$. This combination matches the curve shown in Graph (1).

### Pattern Recognition Solid cylinder current profile: linear inside (B propto r$B \propto r$), hyperbolic outside (B propto 1/r$B \propto 1/r$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism