JEE Main Electromagnetic Induction PYQ

16. (JEE Main 2019 (Online) 9th April Morning Slot)

The total number of turns and cross-section area in a solenoid is fixed. However, its length L is varied by adjusting the separation between windings. The inductance of solenoid will be proportional to :

A.1/L²

B.1/L

C.L

D.L²

Correct option is (B)

$ϕ$ = NBA = LI

N $μ$ ₀ nI $π$ R² = LI

$N μ_{0} \frac{N}{l} l π R^{2} = L I$

N and R constant

Self inductance (L) $\propto \frac{1}{l} \propto \frac{1}{l e n g t h}$

17. (JEE Main 2019 (Online) 8th April Morning Slot )

A 20 Henry inductor coil is connected to a 10 ohm resistance in series as shown in figure. The time at which rate of dissipation of energy (joule's heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor is : JEE Main 2019 (Online) 8th April Morning Slot Physics - Electromagnetic Induction Question 91 English

A. $\frac{2}{\ln 2}$

B. $\ln 2$

C. $2 \ln 2$

D. $\frac{1}{2} \ln 2$

Correct option is (C)

P_R = i² × R

P_B = V × i

$∴$ P_L = Vi – i²R

$\Rightarrow$ Vi – i²R = i²R

$\Rightarrow$ $i = \frac{V}{2 R} a n d i = \frac{V}{R} (1 - e^{- t / τ})$

$∴$ $\frac{V}{2 R} = \frac{V}{R} (1 - e^{- t / τ})$

$\Rightarrow$ t = $τ$ ln(2) = $\frac{20}{10} l n (2) = 2 l n (2)$

18. (JEE Main 2019 (Online) 11th January Evening Slot)

A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil:

A.decreases by a factor of $9 \sqrt{3}$

B.increases by a factor of 27

C.decreases by a factor of 9

D.increases by a factor of 3

Correct option is (D)

Total length L will remain constant

L = (3a) N (N = total turns)

and length of winding = (d) N

(d = diameter of wire)

JEE Main 2019 (Online) 11th January Evening Slot Physics - Electromagnetic Induction Question 93 English Explanation
self inductance = $μ$ ₀n²A $ℓ$

= $μ$ ₀n² $(\frac{\sqrt{3} a^{2}}{4})$ dN

$\propto$ a² N $\propto$ a

So self inductance will become 3 times

19. (JEE Main 2019 (Online) 10th January Evening Slot)

The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1s, the change in the energy of the inductance is -

A.740 J

B.637.5 J

C.540 J

D.437.5 J

Correct option is (D)

$L \frac{d i}{d t} = 25$

$L \times \frac{15}{1} = 25$

$L = \frac{5}{3} H$

$Δ E = \frac{1}{2} \times \frac{5}{3} \times (25^{2} - 10^{2})$

$= \frac{5}{6} \times 525 = 437.5$ J

20. (AIEEE 2007)

An ideal coil of $10 H$ is connected in series with a resistance of $5 Ω$ and a battery of $5 V$ . $2$ second after the connection is made, the current flowing in ampere in the circuit is

A. $(1 - e^{- 1})$

B. $(1 - e)$

C. $e$

D. $e^{- 1}$

Correct option is (A)

KEY CONCEPT : $I = I_{0} (1 - e^{- \frac{R}{L} t})$

(When current is in growth in $L R$ circuit)

$= \frac{E}{R} (1 - e^{- \frac{R}{L} t})$

$= \frac{5}{5} (1 - e^{- \frac{5}{10} \times 2})$

$= (1 - e^{- 1})$