Verified Solution • 2025–26 Session moderate

QJEEM22Ph65eb: JEE Main 2022 Physics - Class 12 JEEM22Ph65eb Solution: The magnetic flux through a coil perpendicular to its plane is…

The magnetic flux through a coil perpendicular to its plane is varying according to the relation ϕ=(5t3+4t2+2t−5) Weber. If the resistance of the coil is 5 ohm, then the induced current through the coil at t = 2 s will be,

Multiple Choice Options

  • A 15.6 A
  • B 16.6 A
  • C 17.6 A
  • D 18.6 A

Step-by-Step Verified Solution

ϕ=5t3+4t2+2t−5\phi=5 \mathrm{t}^3+4 \mathrm{t}^2+2 \mathrm{t}-5

∣e∣=dϕdt=15t2+8t+2|\mathrm{e}|=\frac{\mathrm{d} \phi}{\mathrm{dt}}=15 \mathrm{t}^2+8 \mathrm{t}+2

At t=2,∣e∣=15×22+8×2+2\mathrm{t}=2,|\mathrm{e}|=15 \times 2^2+8 \times 2+2

⇒e=78 V\Rightarrow \mathrm{e}=78 \mathrm{~V}

⇒I=eR=785=15.60\Rightarrow \mathrm{I}=\frac{\mathrm{e}}{\mathrm{R}}=\frac{78}{5}=15.60

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring