Electromagnetic Induction Previous Year Questions — JEE Main Physics
28 past-year Electromagnetic Induction questions from JEE Main (Physics) — page 3.
Q1643 (2024)
A square loop of side $10 \mathrm{~cm}$ and resistance $0.7 \, \Omega$ is placed vertically in east-west plane. A uniform magnetic field of $0.20 \mathrm{~T}$ is set up across the plane in north east direction. The magnetic field is decreased to zero in $1 \mathrm{~s}$ at a steady rate. Then, magnitude of induced emf is $\sqrt{\mathrm{x}} \times 10^{-3} \mathrm{~V}$. The value of x is
### Related Formula
According to Faraday's Law of Electromagnetic Induction, the magnitude of induced EMF ($e$) is:
$$e = \frac{\Delta \phi}{\Delta t}$$
where magnetic flux $\phi$ is defined via dot product:
$$\phi = \vec{B} \cdot \vec{A} = B A \cos \theta$$
### Core Logic
Let the East-West plane lie vertical along the x-z plane. The normal vector to the loop points along the North direction (along $\hat{j}$):
$$\vec{A} = (0.1 \mathrm{~m})^2 \hat{j} = 0.01 \hat{j} \mathrm{~m^2}$$
The uniform magnetic field points North-East, meaning it is at an angle of $45^\circ$ to the North vector direction:
$$\vec{B} = B \cos 45^\circ \hat{i} + B \sin 45^\circ \hat{j} = \frac{0.2}{\sqrt{2}}\hat{i} + \frac{0.2}{\sqrt{2}}\hat{j}$$
{{SOL_IMG1}}
### Step 1: Calculate Initial Flux
$$$\phi_i = \vec{B} \cdot \vec{A} = \left( \frac{0.2}{\sqrt{2}} \right) \times 0.01 = \frac{2 \times 10^{-3}}{\sqrt{2}} = \sqrt{2} \times 10^{-3} \mathrm{~Wb}$$$
### Step 2: Find Induced EMF
Since the field goes steadily to zero in $\Delta t = 1 \mathrm{~s}$, the final flux value $\phi_f = 0$:
$$e = \frac{|0 - \phi_i|}{1} = \sqrt{2} \times 10^{-3} \mathrm{~V}$$
### Step 3: Extract x
Comparing this with the target expression $\sqrt{x} \times 10^{-3} \mathrm{~V}$:
$x = 2$
### Pattern Recognition
Be careful when identifying angles between directional plane descriptors. An East-West plane has a normal axis directed North-South. A North-East field makes a clean $45^\circ$ angle relative to this structural normal line.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q51 (2024)
A power transmission line feeds input power at $2.3 \mathrm{~kV}$ to a step down transformer with its primary winding having $3000$ turns. The output power is delivered at $230 \mathrm{~V}$ by the transformer. The current in the primary of the transformer is $5 \mathrm{A}$ and its efficiency is $90\%$. The winding of transformer is made of copper. The output current of transformer is ________ $\mathrm{A}$.
### Related Formula
$$\eta = \frac{P_{\text{out}}}{P_{\text{in}}}$$
$$P_{\text{in}} = V_p I_p$$
$$P_{\text{out}} = V_s I_s$$
### Core Logic
The efficiency $\eta$ of a transformer is the ratio of output power to input power. We can use this to find the secondary (output) current.
### Step 1: Calculate Input Power
Given:
$V_p = 2.3 \mathrm{~kV} = 2300 \mathrm{~V}$
$I_p = 5 \mathrm{~A}$
$$P_{\text{in}} = 2300 \times 5 \mathrm{~W}$$
### Step 2: Calculate Output Power and Current
Efficiency $\eta = 90\% = 0.9$
$$P_{\text{out}} = \eta \times P_{\text{in}} = 0.9 \times 2300 \times 5$$
Since $P_{\text{out}} = V_s I_s$ and $V_s = 230 \mathrm{~V}$:
$$V_s I_s = 230 \times I_s = 0.9 \times 2300 \times 5$$
$$I_s = \frac{0.9 \times 2300 \times 5}{230} = 0.9 \times 10 \times 5$$
$$I_s = 9 \times 5 = 45 \mathrm{~A}$$
### Pattern Recognition
Transformer equations are direct: $\eta V_p I_p = V_s I_s$. The number of turns ($3000$) is distractor data not needed unless calculating the secondary turns.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Alternating Current
Q60 (2024)
A ceiling fan having 3 blades of length $80 \mathrm{~cm}$ each is rotating with an angular velocity of $1200 \mathrm{~rpm}$. The magnetic field of earth in that region is $0.5 \mathrm{~G}$ and angle of dip is $30^{\circ}$. The emf induced across the blades is $N \pi \times 10^{-5} \mathrm{~V}$. The value of $N$ is _______.
### Related Formula
$$\varepsilon = \frac{1}{2} B_v \omega \ell^2$$
$$B_v = B \sin \delta$$
### Core Logic
For a horizontal fan rotating in the Earth's magnetic field, the blades cut only the vertical component of the magnetic field ($B_v$). 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:
$$B_v = B \sin(30^\circ) = (0.5 \times 10^{-4} \mathrm{~T}) \times \frac{1}{2} = 0.25 \times 10^{-4} \mathrm{~T} = \frac{1}{4} \times 10^{-4} \mathrm{~T}$$
Angular velocity in 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 $\ell = 80 \mathrm{~cm} = 0.8 \mathrm{~m}$.
$$\varepsilon = \frac{1}{2} B_v \omega \ell^2$$
$$\varepsilon = \frac{1}{2} \left(\frac{1}{4} \times 10^{-4}\right) (40\pi) (0.8)^2$$
$$\varepsilon = \frac{1}{8} \times 10^{-4} \times 40\pi \times 0.64$$
$$\varepsilon = 5\pi \times 10^{-4} \times 0.64$$
$$\varepsilon = 3.2\pi \times 10^{-4} \mathrm{~V}$$
$$\varepsilon = 32\pi \times 10^{-5} \mathrm{~V}$$
### Step 3: Extract N
Given $\varepsilon = N\pi \times 10^{-5} \mathrm{~V}$, we can directly see that $N = 32$.
### 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 $B_v$ for horizontal spinners.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Class 12 Physics: Magnetism and Matter
Q53 (2024)
The magnetic flux $\phi$ (in weber) linked with a closed circuit of resistance $8 \, \Omega$ varies with time (in seconds) as $\phi = 5t^2 - 36t + 1$. The induced current in the circuit at $t = 2 \text{ s}$ is ________ A.
### Related Formula
$$\varepsilon = -\frac{d\phi}{dt}$$
$$I = \frac{|\varepsilon|}{R}$$
### Core Logic
Calculate the time derivative of the magnetic flux to find the induced EMF. Evaluate it at the requested time, and then apply Ohm's law to find the current magnitude.
### Step 1: Calculate Induced EMF
$$\varepsilon = -\frac{d}{dt} (5t^2 - 36t + 1)$$
$$\varepsilon = -(10t - 36)$$
At $t = 2 \text{ s}$:
$$\varepsilon = - (10 \times 2 - 36)$$
$$\varepsilon = -(20 - 36) = 16 \text{ V}$$
### Step 2: Calculate Induced Current
$$I = \frac{\varepsilon}{R} = \frac{16}{8} = 2 \text{ A}$$
### Pattern Recognition
Flux polynomials ($At^2 - Bt + C$) instantly trigger a simple derivative test. Remember to drop the negative sign for final current magnitude unless direction is specifically asked.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q2007 (2024)
A small square loop of wire of side $\ell$ is placed inside a large square loop of wire of side $L$ ($L = \ell^2$). The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is $\sqrt{x} \times 10^{-7}\mathrm{\ H}$, where $x =$
### Related Formula
$$M = \frac{\phi_2}{i_1}$$
$$B_{\text{straight wire segment}} = \frac{\mu_0 i}{4\pi d} (\sin\theta_1 + \sin\theta_2)$$
### Core Logic
{{SOL_IMG1}}
Assume a current $i$ flows through the larger square loop of side $L$. The magnetic field generated by it at its center acts as a uniform field across the very small inner loop of side $\ell$.
The magnetic field at the center of the large square loop (distance $d = L/2$ from each side, angles $45^\circ$):
$$B = 4 \times \left[ \frac{\mu_0 i}{4\pi (L/2)} (\sin 45^\circ + \sin 45^\circ) \right]$$
$$B = \frac{\mu_0 i}{\pi (L/2)} \left( \frac{2}{\sqrt{2}} \right)$$
$$B = \frac{2\sqrt{2} \mu_0 i}{\pi L}$$
### Step 2: Mutual Inductance Calculation
Flux linkage for the inner loop:
$$\phi = B \cdot \ell^2$$
$$\phi = \frac{2\sqrt{2} \mu_0 i}{\pi L} \ell^2$$
Given $L = \ell^2$:
$$\phi = \frac{2\sqrt{2} \mu_0 i}{\pi (\ell^2)} \ell^2 = \frac{2\sqrt{2} \mu_0 i}{\pi}$$
Mutual inductance $M$:
$$M = \frac{\phi}{i} = \frac{2\sqrt{2} \mu_0}{\pi}$$
Using $\mu_0 = 4\pi \times 10^{-7}$:
$$M = \frac{2\sqrt{2} \times 4\pi \times 10^{-7}}{\pi}$$
$$M = 8\sqrt{2} \times 10^{-7}\mathrm{\,H}$$
$$M = \sqrt{128} \times 10^{-7}\mathrm{\,H}$$
Comparing with $\sqrt{x} \times 10^{-7}$, we get $x = 128$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
