JEE Main 2024 Question Paper With Solutions PDF

# Faraday's Law Of Induction - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Faraday's law of induction is considered one the most difficult concept.

• 48 Questions around this concept.

## Solve by difficulty

The flux linked with a coil at any instant t is given by $\dpi{100} \phi$ = 10t2 - 50t + 250. The induced emf (in Volts) at t = 3s is

Figure shows three regions of magnetic field, each of area A, and in each region magnitude of magnetic field decreases at a constant rate a. If  is induced electric field then value of line integral . along the given loop is equal to

A small circular loop of wire of radius a is located at the center of a much larger circular wire loop of radius b.  The two loops are in the same plane. The outer loop of radius b carries an alternating current I=Io cos (ωt).  The emf induced in the smaller inner loop is nearly :

In the figure shown a square loop PQRS of side 'a' and resistance 'r' is placed near an infinitely long wire carrying a constant current I. The sides PQ and RS are parallel to the wire. The wire and the loop are in the same plane. The loop is rotated by 180º about an axis parallel to the long wire and passing through the mid points of the side QR and PS. The total amount of charge which passes through any point of the loop during rotation is  :

## Concepts Covered - 1

Whenever the number of magnetic lines of force (Magnetic Flux) passing through a circuit changes an emf called induced emf is produced in the circuit. The induced emf persists only as long as there is a change of flux.

The induced emf is given by the rate of change of magnetic flux linked with the circuit.

i.e  Rate of change of magnetic Flux= $\varepsilon = \frac{-d\phi }{dt}$

where $d\phi\rightarrow \phi _{2}-\phi _{1}$= change in flux

And For N turns it is given as   $\varepsilon = \frac{-Nd\phi }{dt}$ where N= Number of turns in the Coil .

The negative sign indicates that induced emf (e) opposes the change of flux.
And this Flux may change with time in several ways

I.e As $\phi = BA\cos \Theta$  So  $\varepsilon = N\frac{-d}{dt}(BA\cos \Theta )$

1.If Area (A) change then $\varepsilon = -NB\cos \Theta \left ( \frac{dA}{dt} \right )$

2.If Magnetic field (B) change then $\varepsilon = -NA\cos \Theta \left ( \frac{dB}{dt} \right )$

3. If Angle (θ ) change then  $\varepsilon = -NAB\frac{d\left ( \cos \Theta \right )}{d\Theta }\times \frac{d\Theta }{dt}$   or  $\varepsilon =+NBA\; \omega \sin \Theta$

• Induced Current-

$I= \frac{\varepsilon }{R}=\frac{-N}{R}\frac{d\phi }{dt}$

where

$R\rightarrow$ Resistance

$\frac{d\phi }{dt}\rightarrow$ Rate of change of flux

• Induced Charge-

$dq= i.dt= \frac{-N}{R}\frac{d\phi }{dt}.dt$

$dq= \frac{-N}{R}d\phi$

I.e Induced Charge time-independent.

• Induced Power-

$P= \frac{\varepsilon ^{2}}{R}= \frac{N^{2}}{R}\left ( \frac{d\phi }{dt} \right )^{2}$

i.e - Induced Power depends on both time and resistance

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