A conducting \bar moves on two conducting rails as shown in the figure. A constant magnetic field B \exists into the page. The \bar starts to move from the vertex at time $t = 0$ $t = 0$ with a constant velocity. If the induced EMF is $E propto t^n$ $E \propto t^n$ , then value of $n$ $n$ is
A linear conductor \bar moving laterally across V-shaped intersecting conducting guide rails.

Numerical Answer Type:

Enter a numerical value Answer: 1 +4 marks

Solution & Explanation

### Related Formula The motional EMF induced across a moving conductor of instantaneous length

ell

inside a perpendicular uniform magnetic field is given by:

E = B cdot ell cdot v

### Core Logic Let the V-shaped guide rails form an \angle, so that the instantaneous length

ell

of the conducting \bar grows linearly with its horizontal position distance

x

from the vertex [cite: 782, 791]:

ell propto x

Since the \bar moves with a constant velocity

v

, its displacement position at any time

t

is :

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

Comparing this to the given relation

E propto t^n

gives the exponent[cite: 188, 791]:

n = 1

### Step 1: Geometric Analysis The expanding circuit loop configuration across time is shown below:

Motional EMF geometric analysis diagram for Q21 — A linear conductor \bar moving laterally across V-shaped intersecting conducting guide rails.

### Pattern Recognition For \parallel rails, the length

ell

remains constant, meaning induced EMF is independent of time (

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

). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction

Reference Study Guides

Electromagnetic Induction

More Electromagnetic Induction Previous-Year Questions

Q8 2025 Self Induction and Energy Stored

A solenoid having area

A

and length

l

is filled with a material having relative permeability 2. The magnetic energy stored in the solenoid is :

A. $fracB^2 A lmu_0$ $\frac{B^2 A l}{\mu_0}$
B. $fracB^2 A l2 mu_0$ $\frac{B^2 A l}{2 \mu_0}$
C. $B^2 A l$ $B^2 A l$
D. $fracB^2 A l4 mu_0$ $\frac{B^2 A l}{4 \mu_0}$

Solution

### Related Formula 1. Magnetic Energy Density (energy per unit volume):

u_m = fracB^22 mu = fracB^22 mu_r mu_0

2. Total Energy stored:

U = u_m times V = u_m times (A l)

### Core Logic We are given: - Relative permeability

mu_r = 2

- Permeability of the filled core medium

mu = mu_r mu_0 = 2 mu_0

Substitute

mu_r = 2

into the energy density expression:

u_m = fracB^22 (2 mu_0) = fracB^24 mu_0

Multiply by the total volume of the solenoid (

V = A l

U = u_m cdot V = fracB^24 mu_0 A l

### Pattern Recognition Sees: Magnetic energy stored in a solenoid with medium relative permeability. Trap: Placing the relative permeability

mu_r

in the numerator of the formula instead of the denominator. Shortcut: Magnetic energy density is always inversely proportional to permeability. With medium

mu = 2mu_0

, energy density is halved compared to free space, giving

fracB^24mu_0

. Multiply by volume

Al

to get

fracB^2 Al4mu_0

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction Class 12 Physics: Magnetism and Matter

Q25 2025 Motional Electromotive Force

Conductor wire ABCDE with each arm

10mathrm~cm

in length is placed in magnetic field of

frac1sqrt2mathrm~Tesla

, perpendicular to its plane. When conductor is pulled towards right with constant velocity of

10mathrm~cm/s

, induced emf between points A and E is ________ mV.

Conductor wire path geometry layout for Q25 - JEE Main 2025 Morning — The figure outlines a segmented conductive wire trail pulled laterally through an orthogonal inward magnetic field matrix region.

Numerical Answer. Answer: 10 to 10

Solution

### Related Formula Motional electromotive force formula:

epsilon = B cdot v cdot l_texteff

where

l_texteff

is the net straight-line displacement distance joining start point

A

to target endpoint

E

(

l_AE

). ### Core Logic Since the external magnetic induction field map remains completely uniform across space, any arbitrary zig-zag wire loop profile can be simplified into a straight line connector vector bridging its raw boundary tips.

Effective vector length translation resolution mapping for Q25 - JEE Main 2025 Morning — The figure outlines a segmented conductive wire trail pulled laterally through an orthogonal inward magnetic field matrix region.

### Step 1: Compute Effective Length From the geometric angle configurations:

l_texteff = 2 cdot (10 cdot sin 45^circ) = 2 cdot 10 cdot frac1sqrt2 = 10sqrt2mathrm~cm = 0.1sqrt2mathrm~m

(Alternatively tracking structural vector projections:

l_AB = 2 cdot 10sin(45^circ)mathrm~cm

). ### Step 2: Calculate Induced EMF Substitute the parameters into the induction formula:

epsilon = left( frac1sqrt2 ight) times left( 10 times 10^-2mathrm~m/s ight) times left( 10sqrt2 times 10^-2mathrm~m ight)

epsilon = frac1sqrt2 cdot (0.1) cdot (0.1sqrt2) = 0.01mathrm~V = 10mathrm~mV

### Pattern Recognition In uniform magnetic fields, intermediate paths do not affect motional EMF values. The output voltage depends strictly on the straight line distance joining the end tips perpendicular to the velocity direction. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction

Q1 2025 Motional EMF

A uniform magnetic field of

0.4 mathrmT

acts perpendicular to a circular copper disc

20 mathrm~cm

in radius. The disc is having a uniform angular velocity of

10 pi

rad

mathrms^-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

)

A. $text0.0628 V$ $\text{0.0628 V}$
B. $text0.5024 V$ $\text{0.5024 V}$
C. $text0.2512 V$ $\text{0.2512 V}$
D. $text0.1256 V$ $\text{0.1256 V}$

Solution

### Related Formula The induced electromotive force (EMF) developed between the center and the rim of a rotating disc in a perpendicular magnetic field is given by:

E = frac12 B omega R^2

### Core Logic Given parameters from the problem statement [cite: 655, 657, 658]: * Magnetic field,

B = 0.4 text T

* Radius of the disc,

R = 20 text cm = 0.2 text m

* Angular velocity, \omega = 10\pi \text{ rad s}^{-1} Substituting the values into the governing formula:

E = frac12 times 0.4 times (10 times 3.14) times (0.2)^2

E = 0.2 times 31.4 times 0.04

E = 0.2512 text V

### Step 1: Evaluation The potential difference developed between the axis of the disc and the rim is precisely

0.2512 text V

. ### Pattern Recognition For any rotating conductor of length

R

or a continuous disc rotating about its center in a perpendicular magnetic field, the induced EMF is mathematically equivalent to a single radial rod sweeping the area, leading directly to the formula

frac12Bomega R^2

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction

Q7 2025 Mutual Inductance

Consider

mathrmI_1

and

mathrmI_2

are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If

mathrmL_1 =

self inductance of coil 1,

mathrmM_12 =

mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be [cite: 1, 2]

A. $varepsilon_1 = -mathrmL_1fracmathrmdI_1mathrmdt +mathrmM_12fracmathrmdI_2mathrmdt$
B. $varepsilon_1 = -mathrmL_1fracmathrmdI_1mathrmdt -mathrmM_12fracmathrmdI_1mathrmdt$
C. $varepsilon_1 = -mathrmL_1fracmathrmdI_1mathrmdt -mathrmM_12fracmathrmdI_2mathrmdt$
D. $varepsilon_1 = -mathrmL_1fracmathrmdI_2mathrmdt -mathrmM_12fracmathrmdI_1mathrmdt$

Solution

### Related Formula

phi_1 = L_1 I_1 + M_12 I_2

varepsilon_1 = -fracmathrmdphi_1mathrmdt

### Core Logic The total flux linked with coil 1 is due to its own current

I_1

and the mutual influence of current

I_2

in the neighboring coil [cite: 2, 608]:

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

### 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

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