A conducting circular loop is rotated about its diameter at a constant angular speed of 100 rad/s in a magnetic field of 0.5T perpendicular to the axis of rotation. When the loop is rotated by 30° from the horizontal position, the induced EMF is 15.4 mV. The radius of the loop is ____ mm. (Take pi = frac227$\pi = \frac{22}{7}$)

### Related Formula E = B l v$E = B l v$ i = fracER$$i = \frac{E}{R}$$ F_B = i l B = fracB^2 l^2 vR$$F_{B} = i l B = \frac{B^2 l^2 v}{R}$$ ### Core Logic To maintain a constant speed, the external force applied must balance the opposing magnetic force generated by the induced current. F_textext = F_B$$F_{\text{ext}} = F_{B}$$ ### Step 1: Calculate External Force Given values: B = 0.10text T$B = 0.10\text{ T}$ l = 1text m$l = 1\text{ m}$ v = 1.5text m/s$v = 1.5\text{ m/s}$ R = 2\, Omega$R = 2\, \Omega$ F_textext = fracB^2 l^2 vR$$F_{\text{ext}} = \frac{B^2 l^2 v}{R}$$ F_textext = frac(0.10)^2 times (1)^2 times 1.52$$F_{\text{ext}} = \frac{(0.10)^2 \times (1)^2 \times 1.5}{2}$$ F_textext = frac0.01 times 1.52 = frac0.0152 = 0.0075$$F_{\text{ext}} = \frac{0.01 \times 1.5}{2} = \frac{0.015}{2} = 0.0075$$ F_textext = 7.5 times 10^-3text N$$F_{\text{ext}} = 7.5 \times 10^{-3}\text{ N}$$

### Pattern Recognition Standard "sliding rod on rails" problem. The required mechanical force to maintain terminal velocity is always F = fracB^2 L^2 vR$F = \frac{B^2 L^2 v}{R}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction

### Related Formula phi = B cdot A$$\phi = B \cdot A$$ E = -fracdphidt$$E = -\frac{d\phi}{dt}$$ P = fracE^2R$$P = \frac{E^2}{R}$$ ### Core Logic Given area of the loop, A = 1text m^2$A = 1\text{ m}^2$ and magnetic field B = sin(100t)$B = \sin(100t)$. The magnetic flux passing through the loop is: phi = B cdot A = sin(100t) times 1 = sin(100t)$$\phi = B \cdot A = \sin(100t) \times 1 = \sin(100t)$$ Induced EMF E = left| fracdphidt right| = 100cos(100t)$E = \left| \frac{d\phi}{dt} \right| = 100\cos(100t)$ ### Step 1: Calculating Power and Energy Instantaneous power P = fracE^2R = frac100^2 cos^2(100t)100 = 100cos^2(100t)$P = \frac{E^2}{R} = \frac{100^2 \cos^2(100t)}{100} = 100\cos^2(100t)$. Thermal energy dissipated in one time period T$T$: Q = int_0^T P \, dt = int_0^T 100cos^2(100t) \, dt$$Q = \int_{0}^{T} P \, dt = \int_{0}^{T} 100\cos^2(100t) \, dt$$ The angular frequency omega = 100text rad/s$\omega = 100\text{ rad/s}$, so time period T = frac2piomega = frac2pi100 = fracpi50text sec$T = \frac{2\pi}{\omega} = \frac{2\pi}{100} = \frac{\pi}{50}\text{ sec}$. Q = 100 int_0^pi/50 cos^2(100t) \, dt = 100 int_0^pi/50 frac1 + cos(200t)2 \, dt$$Q = 100 \int_{0}^{\pi/50} \cos^2(100t) \, dt = 100 \int_{0}^{\pi/50} \frac{1 + \cos(200t)}{2} \, dt$$ Q = 50 left[ t + fracsin(200t)200 right]_0^pi/50$$Q = 50 \left[ t + \frac{\sin(200t)}{200} \right]_{0}^{\pi/50}$$ Q = 50 left( fracpi50 - 0 right) = pitext Joules$$Q = 50 \left( \frac{\pi}{50} - 0 \right) = \pi\text{ Joules}$$ ### Pattern Recognition For a sinusoidal signal, the integral of cos^2(omega t)$\cos^2(\omega t)$ over one full period T$T$ is always T/2$T/2$. Thus, int P \, dt = P_textmax times fracT2$\int P \, dt = P_{\text{max}} \times \frac{T}{2}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction Class 12 Physics: Alternating Current

### Related Formula For LC oscillations, charge varies as: Q(t) = Q_0 cos(omega t)$$Q(t) = Q_0 \cos(\omega t)$$ Where omega = frac1sqrtLC$\omega = \frac{1}{\sqrt{LC}}$ Energy in capacitor: U_C = fracQ^22C$$U_C = \frac{Q^2}{2C}$$ ### Core Logic Since 25% of the initial energy is transferred to the inductor, the remaining energy in the capacitor is 75% of its initial value. U_C_f = 75\% text of U_C_i$$U_{C_f} = 75\% \text{ of } U_{C_i}$$ fracQ_f^22C = frac34 times fracQ_0^22C$$\frac{Q_f^2}{2C} = \frac{3}{4} \times \frac{Q_0^2}{2C}$$ Q_f^2 = frac34 Q_0^2$$Q_f^2 = \frac{3}{4} Q_0^2$$ Q_f = fracsqrt32 Q_0$$Q_f = \frac{\sqrt{3}}{2} Q_0$$ ### Step 1: Finding Time Using the equation for charge variation: Q_0 cos(omega t) = fracsqrt32 Q_0$$Q_0 \cos(\omega t) = \frac{\sqrt{3}}{2} Q_0$$ cos(omega t) = fracsqrt32$$\cos(\omega t) = \frac{\sqrt{3}}{2}$$ omega t = fracpi6$$\omega t = \frac{\pi}{6}$$ ### Step 2: Final Conclusion Substitute omega = frac1sqrtLC$\omega = \frac{1}{\sqrt{LC}}$: frac1sqrtLC t = fracpi6$$\frac{1}{\sqrt{LC}} t = \frac{\pi}{6}$$ t = fracpi sqrtLC6$$t = \frac{\pi \sqrt{LC}}{6}$$ ### Pattern Recognition Energy is proportional to charge squared. 75\%$75\%$ energy remaining means charge is sqrt0.75 = fracsqrt32$\sqrt{0.75} = \frac{\sqrt{3}}{2}$ of the original. Cosine of pi/6$\pi/6$ yields this exact ratio. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction Class 12 Physics: Alternating Current

### Related Formula e = Bvl, quad i = fraceR$$e = Bvl, \quad i = \frac{e}{R}$$ ### Core Logic

At equilibrium (for terminal velocity): mg = iBl implies mg = left(fracBvlRright)Bl$$mg = iBl \implies mg = \left(\frac{Bvl}{R}\right)Bl$$ v = fracmgRB^2l^2$$v = \frac{mgR}{B^2l^2}$$ ### Pattern Recognition Sees: Sliding rod in magnetic field reaching terminal speed. Shortcut: Equate gravitational force with magnetic force iBl$iBl$ at terminal velocity. Check: Matches option (4). ✓ ### Chapter Mix Class 12 Physics: Electromagnetic Induction

### Related Formula vecB_textnet = vecB_C_2 - vecB_C_1, quad varepsilon = -fracdphidt$$\vec{B}_{\text{net}} = \vec{B}_{C_2} - \vec{B}_{C_1}, \quad \varepsilon = -\frac{d\phi}{dt}$$ ### Core Logic

Applying Lenz's law and magnetic field superposition: for induced current in C_2$C_2$ to be clockwise, the net magnetic flux through C_2$C_2$ must change accordingly. When C_1$C_1$ moves towards C_2$C_2$ and C_3$C_3$ moves away from C_2$C_2$, the net field change induces the specified clockwise current. ### Pattern Recognition Sees: Coaxial current-carrying coils in relative motion. Shortcut: Analyze net magnetic field variation at middle coil C_2$C_2$ using Lenz's law. Check: Matches option (2). ✓ ### Chapter Mix Class 12 Physics: Electromagnetic Induction