A power transmission line feeds input power at 2.3 mathrm~kV$2.3 \mathrm{~kV}$ to a step down transformer with its primary winding having 3000$3000$ turns. The output power is delivered at 230 mathrm~V$230 \mathrm{~V}$ by the transformer. The current in the primary of the transformer is 5 mathrmA$5 \mathrm{A}$ and its efficiency is 90\%$90\%$. The winding of transformer is made of copper. The output current of transformer is ________ mathrmA$\mathrm{A}$.

### 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

### Related Formula M = fracphi_2i_1$$M = \frac{\phi_2}{i_1}$$ B_textstraight wire segment = fracmu_0 i4pi d (sintheta_1 + sintheta_2)$$B_{\text{straight wire segment}} = \frac{\mu_0 i}{4\pi d} (\sin\theta_1 + \sin\theta_2)$$ ### Core Logic

Assume a current i$i$ flows through the larger square loop of side L$L$. The magnetic field generated by it at its center acts as a uniform field across the very small inner loop of side ell$\ell$. The magnetic field at the center of the large square loop (distance d = L/2$d = L/2$ from each side, angles 45^circ$45^\circ$): B = 4 times left[ fracmu_0 i4pi (L/2) (sin 45^circ + sin 45^circ) right]$$B = 4 \times \left[ \frac{\mu_0 i}{4\pi (L/2)} (\sin 45^\circ + \sin 45^\circ) \right]$$ B = fracmu_0 ipi (L/2) left( frac2sqrt2 right)$$B = \frac{\mu_0 i}{\pi (L/2)} \left( \frac{2}{\sqrt{2}} \right)$$ B = frac2sqrt2 mu_0 ipi L$$B = \frac{2\sqrt{2} \mu_0 i}{\pi L}$$ ### Step 2: Mutual Inductance Calculation Flux linkage for the inner loop: phi = B cdot ell^2$$\phi = B \cdot \ell^2$$ phi = frac2sqrt2 mu_0 ipi L ell^2$$\phi = \frac{2\sqrt{2} \mu_0 i}{\pi L} \ell^2$$ Given L = ell^2$L = \ell^2$: phi = frac2sqrt2 mu_0 ipi (ell^2) ell^2 = frac2sqrt2 mu_0 ipi$$\phi = \frac{2\sqrt{2} \mu_0 i}{\pi (\ell^2)} \ell^2 = \frac{2\sqrt{2} \mu_0 i}{\pi}$$ Mutual inductance M$M$: M = fracphii = frac2sqrt2 mu_0pi$$M = \frac{\phi}{i} = \frac{2\sqrt{2} \mu_0}{\pi}$$ Using mu_0 = 4pi times 10^-7$\mu_0 = 4\pi \times 10^{-7}$: M = frac2sqrt2 times 4pi times 10^-7pi$$M = \frac{2\sqrt{2} \times 4\pi \times 10^{-7}}{\pi}$$ M = 8sqrt2 times 10^-7mathrm\,H$$M = 8\sqrt{2} \times 10^{-7}\mathrm{\,H}$$ M = sqrt128 times 10^-7mathrm\,H$$M = \sqrt{128} \times 10^{-7}\mathrm{\,H}$$ Comparing with sqrtx times 10^-7$\sqrt{x} \times 10^{-7}$, we get x = 128$x = 128$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction