A long cylindrical conductor with large cross section carries an electric current distributed uniformly over its cross-section. Magnetic field due to this current is: A. maximum at either ends of the conductor and minimum at the midpoint B. maximum at the axis of the conductor C. minimum at the surface of the conductor D. minimum at the axis of the conductor E. same at all points in the cross-section of the conductor Choose the correct answer from the options given below:

### 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 B = fracmu_0 I4pi d [sin theta_1 + sin theta_2]$$B = \frac{\mu_0 I}{4\pi d} [\sin \theta_1 + \sin \theta_2]$$ ### Core Logic

For an equilateral triangle, the perpendicular distance d$d$ from centroid to any side is given by d = fraca2sqrt3$d = \frac{a}{2\sqrt{3}}$, where a = 4sqrt3 \, mathrmcm$a = 4\sqrt{3} \, \mathrm{cm}$. d = frac4sqrt32sqrt3 = 2 \, mathrmcm = 2 times 10^-2 \, mathrmm$$d = \frac{4\sqrt{3}}{2\sqrt{3}} = 2 \, \mathrm{cm} = 2 \times 10^{-2} \, \mathrm{m}$$ The angles subtended by the side at the centroid are theta_1 = 60^circ$\theta_1 = 60^{\circ}$ and theta_2 = 60^circ$\theta_2 = 60^{\circ}$. ### Step 1: Field due to one side B_1 = fracmu_04pi cdot fracId (sin 60^circ + sin 60^circ)$$B_1 = \frac{\mu_0}{4\pi} \cdot \frac{I}{d} (\sin 60^{\circ} + \sin 60^{\circ})$$ B_1 = 10^-7 times frac22 times 10^-2 left( fracsqrt32 + fracsqrt32 right)$$B_1 = 10^{-7} \times \frac{2}{2 \times 10^{-2}} \left( \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \right)$$ B_1 = 10^-5 times sqrt3 \, mathrmT$$B_1 = 10^{-5} \times \sqrt{3} \, \mathrm{T}$$ ### Step 2: Total Magnetic Field Since there are 3 identical sides and their field vectors point in the same direction at the centroid: B_textnet = 3 times B_1 = 3 times sqrt3 times 10^-5 \, mathrmT$$B_{\text{net}} = 3 \times B_1 = 3 \times \sqrt{3} \times 10^{-5} \, \mathrm{T}$$ Comparing with alpha times 10^-5mathrmT$\alpha \times 10^{-5}\mathrm{T}$, we get alpha = 3sqrt3$\alpha = 3\sqrt{3}$. ### Pattern Recognition For regular polygons of n$n$ sides, B_textcentroid = n cdot B_textside$B_{\text{centroid}} = n \cdot B_{\text{side}}$. An equilateral triangle has n=3$n=3$, d = a/(2sqrt3)$d = a/(2\sqrt{3})$, and angles are always 60^circ$60^{\circ}$. ### 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 B = fracmu_0 i R^22(x^2 + R^2)^3/2$$B = \frac{\mu_0 i R^2}{2(x^2 + R^2)^{3/2}}$$ ### Core Logic The net magnetic field at O is the vector sum of fields from both loops. Since the currents are in the same relative orientation (clockwise from O), their magnetic fields at O will point in opposite directions. B_textnet = B_1 - B_2$$B_{\text{net}} = B_1 - B_2$$

### Step 1: Superposition of Fields Magnetic field due to loop Q (carrying 4I) towards Q, and due to loop P (carrying I) towards P. B_textnet = fracmu_0 (4mathrmi) R^22(R^2 + R^2)^3/2 - fracmu_0 (mathrmi) R^22(R^2 + R^2)^3/2$$B_{\text{net}} = \frac{\mu_0 (4\mathrm{i}) R^2}{2(R^2 + R^2)^{3/2}} - \frac{\mu_0 (\mathrm{i}) R^2}{2(R^2 + R^2)^{3/2}}$$ B_textnet = frac3mu_0 mathrmi R^22(2R^2)^3/2$$B_{\text{net}} = \frac{3\mu_0 \mathrm{i} R^2}{2(2R^2)^{3/2}}$$ ### Step 2: Final Calculation B_textnet = frac3mu_0 mathrmi R^22(2sqrt2R^3) = frac3mu_0 mathrmi4sqrt2R$$B_{\text{net}} = \frac{3\mu_0 \mathrm{i} R^2}{2(2\sqrt{2}R^3)} = \frac{3\mu_0 \mathrm{i}}{4\sqrt{2}R}$$ The direction is towards Q because the field from the loop carrying 4I$4I$ is dominant. ### Pattern Recognition When symmetrical coils carry opposing fields along their axis at equidistance, you simply subtract their current multipliers (4I - I = 3I) and apply the standard axial magnetic field formula once. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Moving Charges and Magnetism