JEE Main 2025 Jan 29 Shift 1, Physics Q12: Faraday’s Law of Induction

JEE Main2025Jan 29, Shift 1Physics

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
$φ = A B, ε = 0$
B
$φ = 0, ε = N A B ω$
C
$φ = 0, ε = 0$
D
$φ = A B, ε = N A B ω$

Solution

Take the angle between $B$ and the area-vector $\overset{n}{^}$ to be $ω t .$ Then the flux per turn is $Φ = B A cos (ω t)$ and the total linked flux is $φ = N B A cos (ω t) .$ The induced emf is

$ε = - \frac{d φ}{d t} = N B A ω sin (ω t) .$

When $B$ is parallel to the plane of the coil, $B ⊥ \overset{n}{^}$ , so $ω t = π /2.$

Therefore $φ = 0$ and $ε = N B A ω .$

Concept behind this question

Faraday’s Law of InductionNotes, formulas and examples →

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 29 Shift 1, Physics Q12: Faraday’s Law of Induction

Concept behind this question

More previous year questions on Faraday’s Law of Induction

Practice more Physics