A ceiling fan having 3 blades of length 80 mathrm~cm$80 \mathrm{~cm}$ each is rotating with an angular velocity of 1200 mathrm~rpm$1200 \mathrm{~rpm}$. The magnetic field of earth in that region is 0.5 mathrm~G$0.5 \mathrm{~G}$ and angle of dip is 30^circ$30^{\circ}$. The emf induced across the blades is N pi times 10^-5 mathrm~V$N \pi \times 10^{-5} \mathrm{~V}$. The value of N$N$ is _______.
Numerical Answer Type:
Enter a numerical valueAnswer: 32 to 32+4 marks
Solution & Explanation
### 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
Keywords:#Motional EMF in Rotating Blades#JEE Main 2024 Morning Q60#Electromagnetic Induction JEE Main 2024#Induced emf fan
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_01_february_morningInduced EMF
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.Answer: 216 to 216
Solution
### 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
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_30_jan_morning
We Map Every Repeating Question in Competitive Exams.
Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.
Select Your Target Exam
Choose an exam track below to find formulas per chapter and patterns.
Syncing Exam Intelligence
Mapping formulas and patterns across all tracks…
PATH A — FULL LENGTH PRACTICE
Full Mock Test Hub
Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.
Under Development
PATH B — TARGETED PRACTICE
Topic-wise Practice Hub
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
