### 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 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

### Related Formula varepsilon = -fracdphidt$\varepsilon = -\frac{d\phi}{dt}$ I = frac|varepsilon|R$I = \frac{|\varepsilon|}{R}$ ### Core Logic Calculate the time derivative of the magnetic flux to find the induced EMF. Evaluate it at the requested time, and then apply Ohm's law to find the current magnitude. ### Step 1: Calculate Induced EMF varepsilon = -fracddt (5t^2 - 36t + 1)$\varepsilon = -\frac{d}{dt} (5t^2 - 36t + 1)$ varepsilon = -(10t - 36)$\varepsilon = -(10t - 36)$ At t = 2 text s$t = 2 \text{ s}$: varepsilon = - (10 times 2 - 36)$\varepsilon = - (10 \times 2 - 36)$ varepsilon = -(20 - 36) = 16 text V$\varepsilon = -(20 - 36) = 16 \text{ V}$ ### Step 2: Calculate Induced Current I = fracvarepsilonR = frac168 = 2 text A$I = \frac{\varepsilon}{R} = \frac{16}{8} = 2 \text{ A}$ ### Pattern Recognition Flux polynomials (At^2 - Bt + C$At^2 - Bt + C$) instantly trigger a simple derivative test. Remember to drop the negative sign for final current magnitude unless direction is specifically asked. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction

### Related Formula varepsilon = N left| fracDeltaphiDelta t right|$\varepsilon = N \left| \frac{\Delta\phi}{\Delta t} \right|$ Deltaphi = (Delta B) A costheta$\Delta\phi = (\Delta B) A \cos\theta$ ### Core Logic Given data: Initial Magnetic Field, B_i = 5000mathrm\,T$B_i = 5000\mathrm{\,T}$ Final Magnetic Field, B_f = 3000mathrm\,T$B_f = 3000\mathrm{\,T}$ Time interval, Delta t = 2mathrm\,s$\Delta t = 2\mathrm{\,s}$ Diameter, d = 0.02mathrm\,m Rightarrow r = 0.01mathrm\,m$d = 0.02\mathrm{\,m} \Rightarrow r = 0.01\mathrm{\,m}$ Induced emf, varepsilon = 22mathrm\,V$\varepsilon = 22\mathrm{\,V}$ Change in magnetic field magnitude |Delta B| = 5000 - 3000 = 2000mathrm\,T$|\Delta B| = 5000 - 3000 = 2000\mathrm{\,T}$. Area of the coil A = pi r^2 = pi (0.01)^2 = 10^-4pi mathrm\,m^2$A = \pi r^2 = \pi (0.01)^2 = 10^{-4}\pi \mathrm{\,m^2}$. ### Step 2: Equation Evaluation Deltaphi = |Delta B| A = (2000) pi (0.01)^2 = 0.2pi$\Delta\phi = |\Delta B| A = (2000) \pi (0.01)^2 = 0.2\pi$ Using Faraday's Law: 22 = N left( frac0.2pi2 right)$22 = N \left( \frac{0.2\pi}{2} \right)$ 22 = N (0.1pi)$22 = N (0.1\pi)$ Taking pi approx 22/7$\pi \approx 22/7$: 22 = N left( 0.1 times frac227 right)$22 = N \left( 0.1 \times \frac{22}{7} \right)$ 1 = fracN70 Rightarrow N = 70$1 = \frac{N}{70} \Rightarrow N = 70$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction