JEE Main Electromagnetic Induction PYQ

46. (JEE Main 2016 (Online) 10th April Morning Slot )

A conducting metal circular-wire-loop of radius r is placed perpendicular to a magnetic field which varies with time as
B = B₀e $^{\frac{- t}{r}}$ , where B₀ and $τ$ are constants, at time t = 0. If the resistance of the loop is R then the heat generated in the loop after a long time (t $\to$ $\infty$ ) is :

A. $\frac{π^{2} r^{4} B_{0}^{4}}{2 τ R}$

B. $\frac{π^{2} r^{4} B_{0}^{2}}{2 τ R}$

C. $\frac{π^{2} r^{4} B_{0}^{2} R}{τ}$

D. $\frac{π^{2} r^{4} B_{0}^{2}}{τ R}$

Correct option is (B)

Given,

B = B₀e $^{- \frac{t}{τ}}$

Area of the circular loop, A = $π$ r²

$∴$ Flux $ϕ$ = BA = $π$ r² B₀ e $^{- \frac{t}{τ}}$

Induced emf in the loop,

$ε$ = $-$ $\frac{d ϕ}{d t}$ = $π$ r²B₀ $\frac{1}{τ}$ e $^{- \frac{t}{τ}}$

Heat generated

= $\int_{0}^{\propto} i^{2} R d t$

= $\int_{0}^{\propto} \frac{ε^{2}}{R} d t$

= $\frac{1}{R} \frac{π^{2} r^{4} B_{0}^{2}}{τ^{2}} \int_{0}^{\propto} e^{- \frac{2 t}{τ}} d t$

= $\frac{π^{2} r^{4} B_{0}^{2}}{τ^{2} R} \times \frac{1}{(- \frac{2}{τ})} {[e^{- \frac{2 t}{τ}}]}_{0}^{\propto}$

= $\frac{- π^{2} r^{4} B_{0}^{2}}{2 τ^{2} R} \times τ (0 - 1)$

= $\frac{π^{2} r^{4} B_{0}^{2}}{2 τ R}$

47. (JEE Main 2013 (Offline) )

A circular loop of radius $0.3$ $c m$ lies center of the small loop is on the axis of the bigger loop. The distance between their centers is $15$ $c m .$ If a current of $2.0$ $A$ flows through the smaller loop, than the flux linked with bigger loop is

A. $9.1 \times 10^{- 11}$ weber

B. $6 \times 10^{- 11}$ weber

C. $3.3 \times 10^{- 11}$ weber

D. $6.6 \times 10^{- 9}$ weber

Correct option is (A)

As we know, Magnetic flux, $ϕ = B . A$

$\frac{μ_{0} (2) {(20 \times 10^{- 2})}^{2}}{2 [{(0.2)}^{2} + {(0.15)}^{2}]} \times π {(0.3 \times 10^{- 2})}^{2}$

On solving

$= 9.216 \times 10^{- 11} = 9.2 \times 10^{- 11}$ weber

48. (AIEEE 2012 )

A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating; It is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to :

A.development of air current when the plate is placed

B.induction of electrical charge on the plate

C.shielding of magnetic lines of force as aluminium is a para-magnetic material.

D.electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.

Correct option is (D)

Because of the Lenz's law of conservation of energy.

49. (AIEEE 2006 )

The flux linked with a coil at any instant $^{'} t^{'}$ is given by
$ϕ = 10 t^{2} - 50 t + 250$
The induced $e m f$ at $t = 3 s$ is

A. $- 190$ $V$

B. $- 10$ $V$

C. $- 10$ $V$

D. $190$ $V$

Correct option is (B)

$ϕ = 10 t^{2} - 50 t + 250$

$e = - \frac{d ϕ}{d t} = - (20 t - 50)$

$e_{t = 3} = - 10 V$

50. ( AIEEE 2004)

A coil having $n$ turns and resistance $R Ω$ is connected with a galvanometer of resistance $4 R Ω .$ This combination is moved in time $t$ seconds from a magnetic field $W_{1}$ weber to $W_{2}$ weber. The induced current in the circuit is

A. $\frac{(W_{2} - W_{1})}{R n t}$

B. $- \frac{n (W_{2} - W_{1})}{5 R t}$

C. $- \frac{n (W_{2} - W_{1})}{5 R t}$

D. $- \frac{n (W_{2} - W_{1})}{R t}$

Correct option is (B)

$\frac{Δ ϕ}{Δ t} = \frac{(W_{2} - W_{1})}{t}$

$R_{t o t} = (R + 4 R) Ω = 5 R Ω$

$i = \frac{n d ϕ}{R_{t o t} d t} = \frac{- n (W_{2} - W_{1})}{5 R t}$

( as $W_{2} & W_{1}$ are magnetic flux )