The current in a coil of self-inductance 2.0 H is increasing according to I = 2 sin(t2) A. The amount of energy spent during the period when the current changes from 0 to 2 A is _________ J.

Answer (Detailed Solution Below) 2

Free
JEE Main 04 April 2024 Shift 1
12.2 K Users
90 Questions 300 Marks 180 Mins

Detailed Solution

Download Solution PDF

CONCEPT:

  •  The self-inductance for e.m.f of the coil is written as;

            E = L \(\frac{dI}{dt}\)

            Here we have "I" as current and L is inductance.

  • The current is the ratio of change in charge per unit time we have;

           I = \(\frac{dq}{dt}\)

          Here q as the charge and "I" as the current.

CALCULATION:

Given: I =2sin(t2) A,  Self-inductance of the coil, L = 2.0 H

I = 2sin(t2)   ----(1)

Let us suppose the current is 2 A we have;

2 = 2 sin(t2)

⇒ 1 = sin (t2)

⇒ t2 = sin-1(1)

⇒ t2 = \(\frac{\pi}{2}\)

⇒ t = \(\sqrt\frac{\pi}{2}\)

As we have the self-induced for e.m.f we have;

E = L \(\frac{dI}{dt}\)

The change in the work done is written as;

⇒ W =  L \(\frac{dI}{dt}\) dq

The current is written as;

I = \(\frac{dq}{dt}\)

⇒ dq = Idt

Therefore, d L \(\frac{di}{dt}\) \(\times\)Idt

dW = LIdi

Now, Integrating above equation we have;

\(\int dW = \int_{0}^{t} LIdi\)

⇒ W = \(\int_{0}^{t}( 2 sin^2t) Ld(2sin^2t)\)

⇒ W = \(\int_{0}^{t}( 8 sin^2t)( Lcos^2t)dt\)

⇒ W = 8 \(\int_{0}^{t} Lsin^2t cos^2tdt\)

Now, by using sin 2t = 2sintcost, therefore;

⇒ W = 4\(\int_{0}^{t} Lsin2t ^2\)     ----(1)

Now, on putting 2t2 = x

⇒ 4t dt = dx

⇒ dt = \(\frac{dx}{4t}\)

and x = \(\sqrt \frac{\pi}{2}\) and x =0

Therefore equation (1) we have;

W = 4L\(\int_{0}^{\sqrt \frac{\pi}{2}} \frac{sinx}{4}dx\)

⇒ W = L \((-cos x )_0^\sqrt\frac{\pi}{2}\)

⇒ W = L \((-cos 2t^2 )_0^\sqrt\frac{\pi}{2}\)

⇒ W = L(-0+2)

⇒ W = 2 J

Hence, the amount of energy is 2 J

Latest JEE Main Updates

Last updated on May 23, 2025

-> JEE Main 2025 results for Paper-2 (B.Arch./ B.Planning) were made public on May 23, 2025.

-> Keep a printout of JEE Main Application Form 2025 handy for future use to check the result and document verification for admission.

-> JEE Main is a national-level engineering entrance examination conducted for 10+2 students seeking courses B.Tech, B.E, and B. Arch/B. Planning courses. 

-> JEE Mains marks are used to get into IITs, NITs, CFTIs, and other engineering institutions.

-> All the candidates can check the JEE Main Previous Year Question Papers, to score well in the JEE Main Exam 2025. 

More Inductors and Inductance Questions

More Electromagnetic Induction and Inductance Questions

Hot Links: teen patti real teen patti star teen patti casino download teen patti casino teen patti club