A square loop of side 10 ~cm$10 \mathrm{~cm}$ and resistance 0.7 Ω$0.7 \, \Omega$ is placed vertically in east-west plane. A uniform magnetic field of 0.20 ~T$0.20 \mathrm{~T}$ is set up across the plane in north east direction. The magnetic field is decreased to zero in 1 ~s$1 \mathrm{~s}$ at a steady rate. Then, magnitude of induced emf is x × 10⁻³ ~V$\sqrt{\mathrm{x}} \times 10^{-3} \mathrm{~V}$. The value of x is
Numerical Answer Type:
Enter a numerical valueAnswer: 2 to 2+4 marks
Solution & Explanation
Related Formula
According to Faraday's Law of Electromagnetic Induction, the magnitude of induced EMF (e$e$) is:
e = (Δ φ)/(Δ t)$$e = \frac{\Delta \phi}{\Delta t}$$
where magnetic flux φ$\phi$ is defined via dot product:
φ = B · A = B A θ$$\phi = \vec{B} \cdot \vec{A} = B A \cos \theta$$
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 j$\hat{j}$):
The uniform magnetic field points North-East, meaning it is at an angle of 45^°$45^\circ$ to the North vector direction:
B = B 45^° i + B 45^° j = 0.2√(2) i + 0.2√(2) j$$\vec{B} = B \cos 45^\circ \hat{i} + B \sin 45^\circ \hat{j} = \frac{0.2}{\sqrt{2}}\hat{i} + \frac{0.2}{\sqrt{2}}\hat{j}$$
Vector reference showing orientation of loop area and North-East magnetic field vectors for Q53
\Delta t = 1 \mathrm{~s}, the final flux value$, the final flux value $\phi_f = 0:$:
$e = (|0 - φᵢ|)/(1) = √(2) × 10⁻³ ~V$e = \frac{|0 - \phi_i|}{1} = \sqrt{2} \times 10^{-3} \mathrm{~V}$
Step 3: Extract x
Comparing this with the target expression
$
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
$
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.
Chapter Mix
Class 12 Physics: Electromagnetic Induction
Vector reference showing orientation of loop area and North-East magnetic field vectors for Q53
More Electromagnetic Induction Previous-Year Questions — Page 2
Q50jee_main_2026_22_january_morningEnergy Density in Inductor
Inductance of a coil with 10⁴$10^{4}$ turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of 10~Ω$10~\Omega$. The energy density in the inductor when the current reaches ( 1e)$\left(\frac{1}{\mathrm{e}}\right)$ of its maximum value is α π × 1e² ~J / m³$\alpha \pi \times \frac{1}{\mathrm{e}^2} \mathrm{~J} / \mathrm{m}^3$. The value of α$\alpha$ is \_\_\_\_. (μ₀ = 4π × 10⁻⁷ Tm / A$\mu_0 = 4\pi \times 10^{-7} \mathrm{Tm} / \mathrm{A}$)
Numerical Answer.Answer: 20 to 20
Solution
Related Formula
Ed = (B²)/(2μ₀), B = μ₀ n I$$E_d = \frac{B^2}{2\mu_0}, \quad B = \mu_0 n I$$
Core Logic
Solution inductor energy density diagram for Q50 - JEE Main 2026 Morning
Sees: Energy density in inductor at exponential current growth stage.
Shortcut: Express magnetic field B$B$ in terms of turns density n$n$ and current I$I$, substitute into energy density formula Ed = (B²)/(2μ₀)$E_d = \frac{B^2}{2\mu_0}$.
Check: Numerical answer is 20. ✓
Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q38jee_main_2026_22_january_eveningTransient Behavior in LR Circuits
Figure shows the circuit that contains three resistances (9 Ω$\Omega$ each) and two inductors (4 mH each). The reading of ammeter at the moment switch K is turned ON, is ____ A.
The figure shows a circuit powered by a 9V battery with three 9-ohm resistors and two 4mH inductors connected in parallel branches.
A.1$1$
B. zero
C.3$3$
D.2$2$
Solution
Related Formula
I(t=0^+) for an ideal inductor = 0 (Open Circuit)$$I(t=0^+) \text{ for an ideal inductor} = 0 \quad (\text{Open Circuit})$$
Core Logic
At t = 0$t = 0$ (immediately after closing switch K), inductors oppose any instant change in current and behave as open circuits (IL = 0$I_L = 0$).
Removing the branches containing the 4 ~mH$4 \mathrm{~mH}$ inductors leaves only the middle branch containing a single 9 Ω$9 \Omega$ resistor connected to the 9 ~V$9 \mathrm{~V}$ battery.
Calculating total initial current I$I$ measured by the ammeter:
The figure shows a circuit powered by a 9V battery with three 9-ohm resistors and two 4mH inductors connected in parallel branches.
Step 1: Final Conclusion
The reading of the ammeter at the moment switch K is turned ON is 1 ~A$1 \mathrm{~A}$.
Pattern Recognition
LR Circuit Transient Rule: At t=0$t=0$, Replace Inductor arrow$\rightarrow$ Open Circuit. At t=∞$t=\infty$, Replace Inductor arrow$\rightarrow$ Short Circuit wire.
Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q46jee_main_2026_22_january_eveningInduced EMF in Rotating Loop
A conducting circular loop is rotated about its diameter at a constant angular speed of 100 rad/s in a magnetic field of 0.5T perpendicular to the axis of rotation. When the loop is rotated by 30° from the horizontal position, the induced EMF is 15.4 mV. The radius of the loop is ____ mm.
(Take π = (22)/(7)$\pi = \frac{22}{7}$)
Numerical Answer.Answer: 14 to 14
Solution
Related Formula
E = B A ω (ω t)$$\mathcal{E} = B A \omega \sin(\omega t)$$
Core Logic
Given B = 0.5 ~T, ω = 100 ~rad/s, θ = ω t = 30^°$B = 0.5 \mathrm{~T}, \omega = 100 \mathrm{~rad/s}, \theta = \omega t = 30^\circ$ and E = 15.4 × 10⁻³ ~V$\mathcal{E} = 15.4 \times 10^{-3} \mathrm{~V}$:
The radius of the circular loop is 14 ~mm$14 \mathrm{~mm}$.
Pattern Recognition
AC Generator induced EMF formula: E = B A ω θ$\mathcal{E} = B A \omega \sin\theta$.
Substitute 30^° = 1/2$\sin 30^\circ = 1/2$ and solve directly for radius r$r$.
Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q41jee_main_2026_23_january_morningMotional EMF
A 20 m long uniform copper wire held horizontally is allowed to fall under the gravity (g = 10 m/s²$^{2}$) through a uniform horizontal magnetic field of 0.5 Gauss perpendicular to the length of the wire. The induced EMF across the wire it travels a vertical distance of 200 m is ____ mV.
A.0.2√(10)$0.2\sqrt{10}$
B.20√(10)$20\sqrt{10}$
C.2 √(10)$2 \sqrt{10}$
D.200√(10)$200\sqrt{10}$
Solution
Related Formula
v = √(2gh)$v = \sqrt{2gh}$
ε = Bvl$$\varepsilon = Bvl$$
Step 1: Calculate Velocity
Falling freely under gravity for 200 m$200\text{ m}$:
v = √(2gh)$v = \sqrt{2gh}$
Convert magnetic field to Tesla: 0.5 Gauss = 0.5 × 10⁻⁴ T$0.5 \text{ Gauss} = 0.5 \times 10^{-4} \text{ T}$.
The velocity, magnetic field, and length are mutually perpendicular.
Sees: "wire falling in horizontal magnetic field" → find velocity using kinematics v = √(2gh)$v = \sqrt{2gh}$, then just plug into standard motional EMF ε = Bvl$\varepsilon = Bvl$. Remember to convert Gauss to Tesla.
Chapter Mix
Class 12 Physics: Electromagnetic Induction
Class 11 Physics: Motion in a Straight Line
Q46jee_main_2026_23_january_morningMotional EMF
A simple pendulum made of mass 10 g and a metallic wire of length 10 cm is suspended vertically in a uniform magnetic field of 2 T. The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of 60°$60^{\circ}$ with vertical, then maximum induced EMF between the point of suspension and point of oscillation is ____ mV. (Take g = 10 m/s²$g = 10 \text{ m/s}^{2}$)
The metallic string of the pendulum cuts the perpendicular magnetic field lines as it swings. The induced EMF along the length of a rotating rod is maximum when its angular velocity ω$\omega$ is maximum. This maximum ω$\omega$ occurs at the lowest point of the swing.
Step 1: Find Maximum Angular Velocity
Using conservation of mechanical energy from extreme to mean position:
Sees: "pendulum wire in magnetic field" → It's just a rotating rod! EMF is (1)/(2) B ω l²$\frac{1}{2} B \omega l^2$. Find max ω$\omega$ via energy conservation at the lowest point.
Chapter Mix
Class 12 Physics: Electromagnetic Induction
Class 11 Physics: Work, Energy and Power
More Electromagnetic Induction Questions — jee_main_2024_29_jan_morning
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.
