A uniform magnetic field of 0.4 mathrmT$0.4 \mathrm{T}$ acts perpendicular to a circular copper disc 20 mathrm~cm$20 \mathrm{~cm}$ in radius. The disc is having a uniform angular velocity of 10 pi$10 \pi$ rad mathrms^-1$\mathrm{s}^{-1}$ about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim? ( pi = 3.14$\pi = 3.14$ )

### Related Formula The motional EMF induced across a moving conductor of instantaneous length ell$\ell$ inside a perpendicular uniform magnetic field is given by: E = B cdot ell cdot v$$E = B \cdot \ell \cdot v$$ ### Core Logic Let the V-shaped guide rails form an \angle, so that the instantaneous length ell$\ell$ of the conducting \bar grows linearly with its horizontal position distance x$x$ from the vertex [cite: 782, 791]: ell propto x$\ell \propto x$ Since the \bar moves with a constant velocity v$v$, its displacement position at any time t$t$ is : x = v cdot t implies ell propto v cdot t$$x = v \cdot t \implies \ell \propto v \cdot t$$ Substituting this time-dependent length into the induced EMF expression : E = B cdot ell cdot v implies E propto B cdot (v cdot t) cdot v implies E propto t^1$$E = B \cdot \ell \cdot v \implies E \propto B \cdot (v \cdot t) \cdot v \implies E \propto t^1$$ Comparing this to the given relation E propto t^n$E \propto t^n$ gives the exponent[cite: 188, 791]: n = 1$n = 1$ ### Step 1: Geometric Analysis The expanding circuit loop configuration across time is shown below:

### Pattern Recognition For \parallel rails, the length ell$\ell$ remains constant, meaning induced EMF is independent of time (E propto t^0$E \propto t^0$). For V-shaped divergent rails, the effective length increases linearly with distance, making the induced EMF directly proportional to time (E propto t^1$E \propto t^1$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction

### Related Formula phi_1 = L_1 I_1 + M_12 I_2$$\phi_1 = L_1 I_1 + M_{12} I_2$$ varepsilon_1 = -fracmathrmdphi_1mathrmdt$$\varepsilon_1 = -\frac{\mathrm{d}\phi_1}{\mathrm{d}t}$$ ### Core Logic The total flux linked with coil 1 is due to its own current I_1$I_1$ and the mutual influence of current I_2$I_2$ in the neighboring coil : phi_1 = L_1 I_1 + M_12 I_2$$\phi_1 = L_1 I_1 + M_{12} I_2$$ Differentiating with respect to time according to Faraday\'s Law yields : varepsilon_1 = -L_1 fracmathrmdI_1mathrmdt - M_12 fracmathrmdI_2mathrmdt$$\varepsilon_1 = -L_1 \frac{\mathrm{d}I_1}{\mathrm{d}t} - M_{12} \frac{\mathrm{d}I_2}{\mathrm{d}t}$$ ### Pattern Recognition Total induced emf sums both self-induction and mutual induction effects additively with standard Lenz law negative signs. ### Chapter Mix Class 12 Physics: Electromagnetic Induction

### Related Formula phi = BAN cos(omega t)$$\phi = BAN \cos(\omega t)$$ varepsilon = BANomega sin(omega t)$$\varepsilon = BAN\omega \sin(\omega t)$$ ### Core Logic

When the magnetic field vector vecB$\vec{B}$ lines up parallel to the plane of the coil, the norm area vector stands perpendicular to vecB$\vec{B}$, yielding omega t = fracpi2$\omega t = \frac{\pi}{2}$. Thus : phi = BAN cosleft(fracpi2right) = 0$$\phi = BAN \cos\left(\frac{\pi}{2}\right) = 0$$ varepsilon = BANomega sinleft(fracpi2right) = NABomega$$\varepsilon = BAN\omega \sin\left(\frac{\pi}{2}\right) = NAB\omega$$ ### Pattern Recognition Flux is zero when the field lines are parallel to the coil surface, but the rate of change of flux (and thus emf) peaks to its absolute maximum. ### Chapter Mix Class 12 Physics: Electromagnetic Induction

### Related Formula Total induced EMF in a moving loop within a time-varying spatial field: varepsilon_textnet = varepsilon_textmotional + varepsilon_texttime$$\varepsilon_{\text{net}} = \varepsilon_{\text{motional}} + \varepsilon_{\text{time}}$$ varepsilon_textmotional = l cdot v cdot Delta B = l cdot v cdot left(fracdBdx cdot Delta xright)$$\varepsilon_{\text{motional}} = l \cdot v \cdot \Delta B = l \cdot v \cdot \left(\frac{dB}{dx} \cdot \Delta x\right)$$ varepsilon_texttime = A cdot fracdBdt$$\varepsilon_{\text{time}} = A \cdot \frac{dB}{dt}$$ ### Core Logic Loop dimensions: l = 5mathrm~cm = 0.05mathrm~m$l = 5\mathrm{~cm} = 0.05\mathrm{~m}$, Delta x = 12mathrm~cm = 0.12mathrm~m$\Delta x = 12\mathrm{~cm} = 0.12\mathrm{~m}$. Velocity v = 5mathrm~cm/s = 0.05mathrm~m/s$v = 5\mathrm{~cm/s} = 0.05\mathrm{~m/s}$. Spatial gradient fracdBdx = 10^-3mathrm~T/cm = 0.1mathrm~T/m$\frac{dB}{dx} = 10^{-3}\mathrm{~T/cm} = 0.1\mathrm{~T/m}$. Time decay rate fracdBdt = 10^-3mathrm~T/s$\frac{dB}{dt} = 10^{-3}\mathrm{~T/s}$. Calculate the motional component across the leading edges: varepsilon_textmotional = 300 times 10^-7mathrm~V$$\varepsilon_{\text{motional}} = 300 \times 10^{-7}\mathrm{~V}$$ Calculate the time-varying field induction across the loop area: A = 12 times 5 = 60mathrm~cm^2 = 60 times 10^-4mathrm~m^2$$A = 12 \times 5 = 60\mathrm{~cm}^2 = 60 \times 10^{-4}\mathrm{~m}^2$$ varepsilon_texttime = A cdot fracdBdt = 60 times 10^-4 times 10^-3 = 60 times 10^-7mathrm~V$$\varepsilon_{\text{time}} = A \cdot \frac{dB}{dt} = 60 \times 10^{-4} \times 10^{-3} = 60 \times 10^{-7}\mathrm{~V}$$ Both changes induce current in the same direction according to Lenz's law: varepsilon_textnet = 300 times 10^-7 + 60 times 10^-7 = 360 times 10^-7mathrm~V$$\varepsilon_{\text{net}} = 300 \times 10^{-7} + 60 \times 10^{-7} = 360 \times 10^{-7}\mathrm{~V}$$ ### Step 1: Calculate Dissipated Power Given loop resistance R = 6mathrm~mOmega = 6 times 10^-3mathrm~Omega$R = 6\mathrm{~m\Omega} = 6 \times 10^{-3}\mathrm{~\Omega}$: P = fracvarepsilon_textnet^2R = frac(360 times 10^-7)^26 times 10^-3 = frac129600 times 10^-146 times 10^-3$$P = \frac{\varepsilon_{\text{net}}^2}{R} = \frac{(360 \times 10^{-7})^2}{6 \times 10^{-3}} = \frac{129600 \times 10^{-14}}{6 \times 10^{-3}}$$ P = 21600 times 10^-11 = 216 times 10^-9mathrm~W$$P = 21600 \times 10^{-11} = 216 \times 10^{-9}\mathrm{~W}$$ Therefore, x = 216$x = 216$. ### Pattern Recognition When a loop moves through a field that changes in both space and time, the total induced EMF is the sum of the motional EMF (vfracpartial Bpartial x$v\frac{\partial B}{\partial x}$) and the transformer EMF (Afracpartial Bpartial t$A\frac{\partial B}{\partial t}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction

### Related Formula The motional electromotive force (emf) induced in a conductor of length L$L$ moving with velocity v$v$ perpendicular to a magnetic field B$B$ is: e = B v L$e = B v L$ ### Core Logic Given parameters: * Length of wire, L = 5text m$L = 5\text{ m}$ * Horizontal magnetic field component, B_H = 0.60 times 10^-4text Wb m^-2$B_H = 0.60 \times 10^{-4}\text{ Wb m}^{-2}$ * Velocity of fall, v = 10text ms^-1$v = 10\text{ ms}^{-1}$ ### Step 1: Calculate the Induced EMF Substitute the parameters directly into the motional emf formula: e = B_H v L$e = B_H v L$ e = (0.60 times 10^-4text Wb m^-2) times (10text ms^-1) times (5text m)$$e = (0.60 \times 10^{-4}\text{ Wb m}^{-2}) \times (10\text{ ms}^{-1}) \times (5\text{ m})$$ e = 3 times 10^-3text V$$e = 3 \times 10^{-3}\text{ V}$$ Comparing this to x times 10^-3text V$x \times 10^{-3}\text{ V}$, we find: x = 3$x = 3$ ### Pattern Recognition Motional EMF is directly the product of field, velocity, and length (e = B v L$e = B v L$) when they are mutually perpendicular. A simple multiplication is all that is required here. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction