The current passing through a conducting loop in the form of equilateral triangle of side 4sqrt3 \, mathrmcm$4\sqrt{3} \, \mathrm{cm}$ is 2A. The magnetic field at its centroid is alpha times 10^-5mathrmT$\alpha \times 10^{-5}\mathrm{T}$ . The value of alpha$\alpha$ is ____. (Given: mu_mathrmo = 4pi times 10^-7$\mu_{\mathrm{o}} = 4\pi \times 10^{-7}$ SI units)

### Related Formula vecF_B = q(vecv times vecB)$$\vec{F}_{B} = q(\vec{v} \times \vec{B})$$ vecF_net = mveca$$\vec{F}_{net} = m\vec{a}$$ ### Core Logic Since the solenoid is placed vertically, the magnetic field vecB$\vec{B}$ inside the solenoid will be parallel or anti-parallel to the vertical axis (either +y$+y$ or -y$-y$ axis). When the charged particle is released from rest, gravity pulls it vertically downward, meaning it gains velocity vecv$\vec{v}$ strictly along the y$y$-axis (parallel or anti-parallel to vecB$\vec{B}$). Because velocity and magnetic field are collinear (vecv parallel vecB$\vec{v} \parallel \vec{B}$ or vecv parallel -vecB$\vec{v} \parallel -\vec{B}$), the cross product vecv times vecB = 0$\vec{v} \times \vec{B} = 0$. Therefore, the magnetic force vecF_B = 0$\vec{F}_B = 0$. ### Step 1: Calculating Net Acceleration The only force acting on the particle is gravity. vecF_net = mvecg$$\vec{F}_{net} = m\vec{g}$$ a_net = g$a_{net} = g$

### Pattern Recognition Whenever a charged particle moves parallel to a magnetic field lines (like moving along the axis of a solenoid), the magnetic force is absolutely zero. It behaves like free fall. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

### Related Formula For a semi-infinite wire segment at distance r$r$: B = fracmu_0 I4pi r$$B = \frac{\mu_0 I}{4\pi r}$$ For a full circular loop at its center: B = fracmu_0 I2r$$B = \frac{\mu_0 I}{2r}$$ ### Core Logic

The total magnetic field at O is the vector sum of fields from three segments: 1. The incoming semi-infinite wire (AB) 2. The outgoing semi-infinite wire (DE) 3. The nearly full circular loop (BCD) Note: Based on the diagram, the loop is not fully closed, but geometrically it acts as a full circle subtracted by the gap. Typically this standard shape treats the circular part as a full circle and the straight wires as two semi-infinite wires. vecB_O = vecB_AB + vecB_DE + vecB_BCD$$\vec{B}_O = \vec{B}_{AB} + \vec{B}_{DE} + \vec{B}_{BCD}$$ ### Step 1: Adding the Vector Components Applying the Right Hand Rule: - Segment AB: current flows along +x$+x$, position vector to O is +y$+y$. dvecl times vecr = hati times hatj = hatk$d\vec{l} \times \vec{r} = \hat{i} \times \hat{j} = \hat{k}$. Wait, the diagram shows the loop in the x-y$x-y$ plane. Let's re-examine axes. Based on standard convention, if current is in xy$xy$ plane, field is in z$z$ (hatk$\hat{k}$) direction. The solution shows vectors in hati$\hat{i}$. This means the axes are drawn such that the loop is in the y-z$y-z$ plane. Yes, the provided axes show x$x$ pointing out, y$y$ to the right, z$z$ upwards. - Segment AB (current along y$y$ axis): B$B$ at origin is along +x$+x$ (hati$\hat{i}$). - Segment DE (current along y$y$ axis): B$B$ at origin is along +x$+x$ (hati$\hat{i}$). - Circular Loop (current clockwise in y-z$y-z$ plane): B$B$ at origin points inwards, i.e., -x$-x$ (-hati$-\hat{i}$). vecB_AB = fracmu_0 I4pi r hati $$ \vec{B}_{AB} = \frac{\mu_0 I}{4\pi r} \hat{i} $$ vecB_DE = fracmu_0 I4pi r hati $$ \vec{B}_{DE} = \frac{\mu_0 I}{4\pi r} \hat{i} $$ vecB_BCD = - fracmu_0 I2r hati $$ \vec{B}_{BCD} = - \frac{\mu_0 I}{2r} \hat{i} $$ ### Step 2: Final Conclusion vecB_O = fracmu_0 I4pi r hati + fracmu_0 I4pi r hati - fracmu_0 I2r hati$$\vec{B}_O = \frac{\mu_0 I}{4\pi r} \hat{i} + \frac{\mu_0 I}{4\pi r} \hat{i} - \frac{\mu_0 I}{2r} \hat{i}$$ vecB_O = fracmu_0 I2pi r hati - fracmu_0 I2r hati$$\vec{B}_O = \frac{\mu_0 I}{2\pi r} \hat{i} - \frac{\mu_0 I}{2r} \hat{i}$$ vecB_O = fracmu_0 I2pi r (1 - pi) hati$$\vec{B}_O = \frac{\mu_0 I}{2\pi r} (1 - \pi) \hat{i}$$ vecB_O = -fracmu_0 I2pi r (pi - 1) hati$$\vec{B}_O = -\frac{\mu_0 I}{2\pi r} (\pi - 1) \hat{i}$$ ### Pattern Recognition Always separate complex wire geometries into standard segments: infinite wires, semi-infinite wires, and arcs. Use the Right-Hand Rule carefully with the given explicit coordinate frame to avoid sign errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism

### Related Formula F = qvB$F = qvB$ phi = B cdot textArea$$\phi = B \cdot \text{Area}$$ U = frac12 L I^2$$U = \frac{1}{2} L I^{2}$$ ### Core Logic For Magnetic induction (B): [B] = left[ fracFqv right] = [MT^-2A^-1]$$[B] = \left[ \frac{F}{qv} \right] = [MT^{-2}A^{-1}]$$ So, A matches III. For Magnetic Flux (phi$\phi$): [phi] = [B] cdot [textArea] = [ML^2T^-2A^-1]$$[\phi] = [B] \cdot [\text{Area}] = [ML^{2}T^{-2}A^{-1}]$$ So, B matches IV. For Magnetic Permeability (mu$\mu$): [mu] = [MLT^-2A^-2]$$[\mu] = [MLT^{-2}A^{-2}]$$ So, C matches I. For Self inductance (L): Using U = frac12 LI^2$U = \frac{1}{2} LI^{2}$, [L] = [ML^2T^-2A^-1]$$[L] = [ML^{2}T^{-2}A^{-1}]$$ Wait, the given option II is [ML^2 T^-2 A^-2]$[ML^2 T^{-2} A^{-2}]$. Let's re-verify: Energy U = [ML^2 T^-2]$U = [ML^2 T^{-2}]$. I^2 = [A^2]$I^2 = [A^2]$. So L = [ML^2 T^-2 A^-2]$L = [ML^2 T^{-2} A^{-2}]$. So, D matches II.

### Step 1: Final Conclusion A-III, B-IV, C-I, D-II. Option (4) is correct. ### Pattern Recognition Dimensional analysis of electromagnetic quantities frequently hinges on knowing formulas for force, flux, and energy. Deriving from F=qvB$F=qvB$ and U=frac12LI^2$U=\frac{1}{2}LI^2$ is the fastest approach. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism Class 12 Physics: Electromagnetic Induction

### 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