Consider I1 and I2 are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L1 = self inductance of coil 1, M12 = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be

### Related Formula ### Core Logic The total flux linked with coil 1 is due to its own current I1 and the mutual influence of current I2 in the neighboring coil : Differentiating with respect to time according to Faraday's Law yields : ### 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

Consider I1 and I2 are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L1 = self inductance of coil 1, M12 = mutual inductance of coil 1 with respect to coil 2, then th

### Related Formula ### Core Logic The total flux linked with coil 1 is due to its own current I1 and the mutual influence of current I2 in the neighboring coil : Differentiating with respect to time according to Faraday's Law yields : ### 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

Consider I1 and I2 are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L1 = self inductance of coil 1, M12 = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be

### Related Formula ### Core Logic The total flux linked with coil 1 is due to its own current I1 and the mutual influence of current I2 in the neighboring coil : Differentiating with respect to time according to Faraday's Law yields : ### 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

Consider I1 and I2 are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L1 = self inductance of coil 1, M12 = mutual inductance of coil 1 with respect to coil 2, then th

### Related Formula ### Core Logic The total flux linked with coil 1 is due to its own current I1 and the mutual influence of current I2 in the neighboring coil : Differentiating with respect to time according to Faraday's Law yields : ### 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

Consider I1 and I2 are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L1 = self…

Year	2026	2025	2024	Total
Questions	12	6	10	28

More Electromagnetic Induction Previous-Year Questions — Page 4

Q21 jee_main_2025_28_jan_evening Motional EMF

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

with a constant velocity. If the induced EMF is

E \propto tⁿ

, then value of

n

Motional EMF diagram for Q21 - JEE Main 2025 Evening — A linear conductor \bar moving laterally across V-shaped intersecting conducting guide rails.

Numerical Answer. Answer: 1

Solution

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 \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 $x$ $x$ from the vertex [cite: 782, 791]:

$\propto x$ $\ell \propto x$

Since the \bar moves with a constant velocity $v$ $v$, its displacement position at any time $t$ $t$ is :

x = v \cdot t \propto v \cdot t

Substituting this time-dependent length into the induced EMF expression :

E = B \cdot \cdot v E \propto B \cdot (v \cdot t) \cdot v E \propto t¹

Comparing this to the given relation $E \propto tⁿ$ $E \propto t^n$ gives the exponent[cite: 188, 791]:

$n = 1$ $n = 1$

Step 1: Geometric Analysis

The expanding circuit loop configuration across time is shown below:

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 ( $E \propto t⁰$ $E \propto t^0$). For V-shaped divergent rails, the effective length increases linearly with distance, making the induced EMF directly proportional to time ( $E \propto t¹$ $E \propto t^1$).

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Q jee_main_2025_29_jan_morning AC Generator

A coil of area A and N turns is rotating with angular velocity

ω

in a uniform magnetic field

B

about an axis perpendicular to

B

. Magnetic flux

φ

and induced emf

ε

across it, at an instant when

B

is parallel to the plane of coil, are:

A. $φ = AB,ε = 0$ $\varphi = \mathrm{AB},\varepsilon = 0$
B. $φ = 0, ε = NABω$
C. $φ = 0, ε = 0$ $\varphi = 0, \varepsilon = 0$
D. $φ = AB,ε = NABω$

Solution

Related Formula

φ = BAN (ω t)

ε = BANω (ω t)

Core Logic

AC Generator explanation diagram for Q12

When the magnetic field vector $B$ $\vec{B}$ lines up parallel to the plane of the coil, the norm area vector stands perpendicular to $B$ $\vec{B}$, yielding $ω t = (π)/(2)$ $\omega t = \frac{\pi}{2}$. Thus :

φ = BAN ((π)/(2)) = 0

ε = BANω ((π)/(2)) = NABω

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

Q jee_main_2024_01_february_morning Induced EMF

A rectangular loop of sides

12~cm

and

5~cm

, with its sides parallel to the x-axis and y-axis respectively moves with a velocity of

5~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

10⁻³~T/cm

along the negative x-direction and it is decreasing with time at the rate of

10⁻³~T/s

. If the resistance of the loop is

6~mΩ

, the power dissipated by the loop as heat is

x \times 10⁻⁹~W

. The value of

x

is:

Numerical Answer. Answer: 216 to 216

Solution

Related Formula

Total induced EMF in a moving loop within a time-varying spatial field:

εₙₑₜ = εmotional + εtime

εmotional = l \cdot v \cdot Δ B = l \cdot v \cdot ((dB)/(dx) \cdot Δ x)

εtime = A \cdot (dB)/(dt)

Core Logic

Loop dimensions: $l = 5~cm = 0.05~m$ $l = 5\mathrm{~cm} = 0.05\mathrm{~m}$, $Δ x = 12~cm = 0.12~m$ . Velocity $v = 5~cm/s = 0.05~m/s$ . Spatial gradient $(dB)/(dx) = 10⁻³~T/cm = 0.1~T/m$ . Time decay rate $(dB)/(dt) = 10⁻³~T/s$ $\frac{dB}{dt} = 10^{-3}\mathrm{~T/s}$.

Calculate the motional component across the leading edges:

εmotional = 300 \times 10⁻⁷~V

Calculate the time-varying field induction across the loop area:

A = 12 \times 5 = 60~cm² = 60 \times 10⁻⁴~m²

εtime = A \cdot (dB)/(dt) = 60 \times 10⁻⁴ \times 10⁻³ = 60 \times 10⁻⁷~V

Both changes induce current in the same direction according to Lenz's law:

εₙₑₜ = 300 \times 10⁻⁷ + 60 \times 10⁻⁷ = 360 \times 10⁻⁷~V

Step 1: Calculate Dissipated Power

Given loop resistance $R = 6~mΩ = 6 \times 10⁻³~Ω$ :

P = εₙₑₜ²R = (360 \times 10⁻⁷)²6 \times 10⁻³ = 129600 \times 10⁻¹⁴6 \times 10⁻³

P = 21600 \times 10⁻¹¹ = 216 \times 10⁻⁹~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 ( $v(\partial B)/(\partial x)$ $v\frac{\partial B}{\partial x}$) and the transformer EMF ( $A(\partial B)/(\partial t)$ $A\frac{\partial B}{\partial t}$).

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Q52 jee_main_2024_29_january_evening Motional Electromotive Force

A horizontal straight wire

5 m

long extending from east to west falling freely at right angle to horizontal component of earth's magnetic field

0.60 \times 10⁻⁴ Wb m⁻²

. The instantaneous value of emf induced in the wire when its velocity is

10 ms⁻¹

x \times 10⁻³ V

. The value of

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, $L = 5 m$ $L = 5\text{ m}$
Horizontal magnetic field component, $BH = 0.60 \times 10⁻⁴ Wb m⁻²$
Velocity of fall, $v = 10 ms⁻¹$ $v = 10\text{ ms}^{-1}$

Step 1: Calculate the Induced EMF

Substitute the parameters directly into the motional emf formula:

$e = BH v L$ $e = B_H v L$

e = (0.60 \times 10⁻⁴ Wb m⁻²) \times (10 ms⁻¹) \times (5 m)

e = 3 \times 10⁻³ V

Comparing this to $x \times 10⁻³ V$ $x \times 10^{-3}\text{ V}$, 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.

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

More Electromagnetic Induction Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Geometric Analysis

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Dissipated Power

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate the Induced EMF

Pattern Recognition

Chapter Mix

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