A solenoid having area A$A$ and length l$l$ is filled with a material having relative permeability 2. The magnetic energy stored in the solenoid is :

### Related Formula textemf = -fracDeltaphiDelta t$$\text{emf} = -\frac{\Delta\phi}{\Delta t}$$ ### Core Logic Initial magnetic flux is given by phi_i = B A costheta$\phi_i = B A \cos\theta$, where A = 2.5 times 2 = 5text m^2$A = 2.5 \times 2 = 5\text{ m}^2$, B = 4text T$B = 4\text{ T}$, and theta = 60^circ$\theta = 60^{\circ}$ (angle aligned with the axis mapping context rules in the problem source text). phi_i = 4 times 5 times cos(60^circ) = 20 times 0.5 = 10text Wb$$\phi_i = 4 \times 5 \times \cos(60^{\circ}) = 20 \times 0.5 = 10\text{ Wb}$$ Final flux after removal phi_f = 0$\phi_f = 0$. ### Step 1: Compute Induced EMF textemf = -fracphi_f - phi_iDelta t = -frac0 - 1010 = +1text V$$\text{emf} = -\frac{\phi_f - \phi_i}{\Delta t} = -\frac{0 - 10}{10} = +1\text{ V}$$ ### Pattern Recognition Removal fields yield positive flux variants under canonical sign configurations due to the absolute reduction profile mapped by Lenz/Faraday relationships. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction

### Related Formula textemf = -M fracdidt$$\text{emf} = -M \frac{di}{dt}$$ ### Core Logic Differentiating the current expression with respect to time: fracdidt = fracddt(i_0 sinomega t) = i_0 omega cosomega t$$\frac{di}{dt} = \frac{d}{dt}(i_0 \sin\omega t) = i_0 \omega \cos\omega t$$ Hence, the expression for induced emf is: textemf = -M i_0 omega cosomega t$$\text{emf} = -M i_0 \omega \cos\omega t$$ ### Step 1: Isolate Maximum value The peak magnitude occurs when cosomega t = 1$\cos\omega t = 1$: textemf_textmax = M i_0 omega$$\text{emf}_{\text{max}} = M i_0 \omega$$ Substitute given numerical values: textemf_textmax = 0.002 times 5 times 50pi = 0.5pi = fracpi2text V$$\text{emf}_{\text{max}} = 0.002 \times 5 \times 50\pi = 0.5\pi = \frac{\pi}{2}\text{ V}$$ ### Step 2: Match with target variable Comparing fracpi2$\frac{\pi}{2}$ with fracpialpha$\frac{\pi}{\alpha}$ gives: alpha = 2$\alpha = 2$ ### Pattern Recognition Harmonic driving functions induce derivative cosine arrays whose peak amplitudes scale strictly as the product of primary parameters: M cdot i_0 cdot omega$M \cdot i_0 \cdot \omega$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction

### Related Formula According to Faraday's Law of Electromagnetic Induction, the magnitude of induced EMF (e$e$) is: e = fracDelta phiDelta t$$e = \frac{\Delta \phi}{\Delta t}$$ where magnetic flux phi$\phi$ is defined via dot product: phi = vecB cdot vecA = B A cos theta$$\phi = \vec{B} \cdot \vec{A} = B A \cos \theta$$ ### Core Logic Let the East-West plane lie vertical along the x-z plane. The normal vector to the loop points along the North direction (along hatj$\hat{j}$): vecA = (0.1 mathrm~m)^2 hatj = 0.01 hatj mathrm~m^2$$\vec{A} = (0.1 \mathrm{~m})^2 \hat{j} = 0.01 \hat{j} \mathrm{~m^2}$$ The uniform magnetic field points North-East, meaning it is at an angle of 45^circ$45^\circ$ to the North vector direction: vecB = B cos 45^circ hati + B sin 45^circ hatj = frac0.2sqrt2hati + frac0.2sqrt2hatj$$\vec{B} = B \cos 45^\circ \hat{i} + B \sin 45^\circ \hat{j} = \frac{0.2}{\sqrt{2}}\hat{i} + \frac{0.2}{\sqrt{2}}\hat{j}$$

### Step 1: Calculate Initial Flux $phi_i = vecB cdot vecA = left( frac0.2sqrt2 right) times 0.01 = frac2 times 10^-3sqrt2 = sqrt2 times 10^-3 mathrm~Wb$$$\phi_i = \vec{B} \cdot \vec{A} = \left( \frac{0.2}{\sqrt{2}} \right) \times 0.01 = \frac{2 \times 10^{-3}}{\sqrt{2}} = \sqrt{2} \times 10^{-3} \mathrm{~Wb}$$ ### Step 2: Find Induced EMF Since the field goes steadily to zero in $ ### Step 2: Find Induced EMF Since the field goes steadily to zero in $\Delta t = 1 \mathrm{~s}, the final flux value $, the final flux value $\phi_f = 0: $: $e = frac|0 - phi_i|1 = sqrt2 times 10^-3 mathrm~V$e = \frac{|0 - \phi_i|}{1} = \sqrt{2} \times 10^{-3} \mathrm{~V}$ ### Step 3: Extract x Comparing this with the target expression $ ### Step 3: Extract x Comparing this with the target expression $\sqrt{x} \times 10^{-3} \mathrm{~V}: $: $x = 2 ### Pattern Recognition Be careful when identifying angles between directional plane descriptors. An East-West plane has a normal axis directed North-South. A North-East field makes a clean $ ### Pattern Recognition Be careful when identifying angles between directional plane descriptors. An East-West plane has a normal axis directed North-South. A North-East field makes a clean $45^\circ$ angle relative to this structural normal line. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction

### Related Formula eta = fracP_textoutP_textin$$\eta = \frac{P_{\text{out}}}{P_{\text{in}}}$$ P_textin = V_p I_p$$P_{\text{in}} = V_p I_p$$ P_textout = V_s I_s$$P_{\text{out}} = V_s I_s$$ ### Core Logic The efficiency eta$\eta$ of a transformer is the ratio of output power to input power. We can use this to find the secondary (output) current. ### Step 1: Calculate Input Power Given: V_p = 2.3 mathrm~kV = 2300 mathrm~V$V_p = 2.3 \mathrm{~kV} = 2300 \mathrm{~V}$ I_p = 5 mathrm~A$I_p = 5 \mathrm{~A}$ P_textin = 2300 times 5 mathrm~W$$P_{\text{in}} = 2300 \times 5 \mathrm{~W}$$ ### Step 2: Calculate Output Power and Current Efficiency eta = 90\% = 0.9$\eta = 90\% = 0.9$ P_textout = eta times P_textin = 0.9 times 2300 times 5$$P_{\text{out}} = \eta \times P_{\text{in}} = 0.9 \times 2300 \times 5$$ Since P_textout = V_s I_s$P_{\text{out}} = V_s I_s$ and V_s = 230 mathrm~V$V_s = 230 \mathrm{~V}$: V_s I_s = 230 times I_s = 0.9 times 2300 times 5$$V_s I_s = 230 \times I_s = 0.9 \times 2300 \times 5$$ I_s = frac0.9 times 2300 times 5230 = 0.9 times 10 times 5$$I_s = \frac{0.9 \times 2300 \times 5}{230} = 0.9 \times 10 \times 5$$ I_s = 9 times 5 = 45 mathrm~A$$I_s = 9 \times 5 = 45 \mathrm{~A}$$ ### Pattern Recognition Transformer equations are direct: eta V_p I_p = V_s I_s$\eta V_p I_p = V_s I_s$. The number of turns (3000$3000$) is distractor data not needed unless calculating the secondary turns. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Alternating Current

### Related Formula varepsilon = frac12 B_v omega ell^2$$\varepsilon = \frac{1}{2} B_v \omega \ell^2$$ B_v = B sin delta$$B_v = B \sin \delta$$ ### Core Logic For a horizontal fan rotating in the Earth's magnetic field, the blades cut only the vertical component of the magnetic field (B_v$B_v$). The number of blades is a distractor, as they are all in parallel, meaning the EMF developed across one blade is the same as the EMF across the whole fan setup. ### Step 1: Calculate Field and Omega Vertical component of magnetic field: B_v = B sin(30^circ) = (0.5 times 10^-4 mathrm~T) times frac12 = 0.25 times 10^-4 mathrm~T = frac14 times 10^-4 mathrm~T$$B_v = B \sin(30^\circ) = (0.5 \times 10^{-4} \mathrm{~T}) \times \frac{1}{2} = 0.25 \times 10^{-4} \mathrm{~T} = \frac{1}{4} \times 10^{-4} \mathrm{~T}$$ Angular velocity in rad/s: omega = 2pi f = 2pi left(frac120060right) = 40pi mathrm~rad/s$$\omega = 2\pi f = 2\pi \left(\frac{1200}{60}\right) = 40\pi \mathrm{~rad/s}$$ ### Step 2: Calculate Induced EMF Length of blade ell = 80 mathrm~cm = 0.8 mathrm~m$\ell = 80 \mathrm{~cm} = 0.8 \mathrm{~m}$. varepsilon = frac12 B_v omega ell^2$$\varepsilon = \frac{1}{2} B_v \omega \ell^2$$ varepsilon = frac12 left(frac14 times 10^-4right) (40pi) (0.8)^2$$\varepsilon = \frac{1}{2} \left(\frac{1}{4} \times 10^{-4}\right) (40\pi) (0.8)^2$$ varepsilon = frac18 times 10^-4 times 40pi times 0.64$$\varepsilon = \frac{1}{8} \times 10^{-4} \times 40\pi \times 0.64$$ varepsilon = 5pi times 10^-4 times 0.64$$\varepsilon = 5\pi \times 10^{-4} \times 0.64$$ varepsilon = 3.2pi times 10^-4 mathrm~V$$\varepsilon = 3.2\pi \times 10^{-4} \mathrm{~V}$$ varepsilon = 32pi times 10^-5 mathrm~V$$\varepsilon = 32\pi \times 10^{-5} \mathrm{~V}$$ ### Step 3: Extract N Given varepsilon = Npi times 10^-5 mathrm~V$\varepsilon = N\pi \times 10^{-5} \mathrm{~V}$, we can directly see that N = 32$N = 32$. ### Pattern Recognition Motional EMF for rotating rods acts like a battery. Multiple identical blades radiating from the center to the rim behave like multiple identical batteries in parallel; the total voltage does not stack. Isolate B_v$B_v$ for horizontal spinners. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction Class 12 Physics: Magnetism and Matter