A rectangular loop of sides 12mathrm~cm$12\mathrm{~cm}$ and 5mathrm~cm$5\mathrm{~cm}$, with its sides parallel to the x-axis and y-axis respectively moves with a velocity of 5mathrm~cm/s$5\mathrm{~cm/s}$ in the positive x-axis direction, in a space containing a variable magnetic field in the positive z-direction. The field has a gradient of 10^-3mathrm~T/cm$10^{-3}\mathrm{~T/cm}$ along the negative x-direction and it is decreasing with time at the rate of 10^-3mathrm~T/s$10^{-3}\mathrm{~T/s}$. If the resistance of the loop is 6mathrm~mOmega$6\mathrm{~m\Omega}$, the power dissipated by the loop as heat is x times 10^-9mathrm~W$x \times 10^{-9}\mathrm{~W}$. The value of x$x$ is:
Numerical Answer Type:
Enter a numerical valueAnswer: 216 to 216+4 marks
Solution & Explanation
### 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
Coordinate mapping for spatial induction gradient solution layout Q52
Keywords:#Motional and time varying induced EMF power#JEE Main 2024 Morning Q52#Magnetic field gradient flux change loop#Electromagnetic induction energy dissipation
More Electromagnetic Induction Previous-Year Questions — Page 2
Qjee_main_2025_29_jan_morningAC Generator
A coil of area A and N turns is rotating with angular velocity omega$\omega$ in a uniform magnetic field vecB$\vec{B}$ about an axis perpendicular to vecB$\vec{B}$ . Magnetic flux varphi$\varphi$ and induced emf varepsilon$\varepsilon$ across it, at an instant when vecB$\vec{B}$ is parallel to the plane of coil, are:
### 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
AC Generator explanation diagram for Q12
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
Q52jee_main_2024_29_january_eveningMotional Electromotive Force
A horizontal straight wire 5text m$5\text{ m}$ long extending from east to west falling freely at right angle to horizontal component of earth's magnetic field 0.60 times 10^-4text Wb m^-2$0.60 \times 10^{-4}\text{ Wb m}^{-2}$. The instantaneous value of emf induced in the wire when its velocity is 10text ms^-1$10\text{ ms}^{-1}$ is x times 10^-3text V$x \times 10^{-3}\text{ V}$. The value of x$x$ is:
Numerical Answer.Answer: 3 to 3
Solution
### 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
Q42jee_main_2024_27_jan_morningFaraday's Law
A rectangular loop of length 2.5text m$2.5\text{ m}$ and width 2text m$2\text{ m}$ is placed at 60^circ$60^{\circ}$ to a magnetic field of 4text T$4\text{ T}$. The loop is removed from the field in 10text sec$10\text{ sec}$. The average emf induced in the loop during this time is:
A.-2text V$-2\text{ V}$
B.+2text V$+2\text{ V}$
C.+1text V$+1\text{ V}$
D.-1text V$-1\text{ V}$
Solution
### 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
Q53jee_main_2024_27_jan_morningMutual Induction
Two coils have mutual inductance 0.002text H$0.002\text{ H}$. The current changes in the first coil according to the relation i = i_0sinomega t$i = i_{0}\sin\omega t$, where i_0 = 5text A$i_{0} = 5\text{ A}$ and \omega = 50pitext rad/s$50\pi\text{ rad/s}$. The maximum value of emf in the second coil is fracpialphatext V$\frac{\pi}{\alpha}\text{ V}$. The value of alpha$\alpha$ is ______.
Numerical Answer.Answer: 2 to 2
Solution
### 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
More Electromagnetic Induction Questions — jee_main_2024_01_february_morning
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