The magnetic flux linked to a circular coil of radius R is: φ=2 t^3+4 t^2+2 t+5 ~Wb The magnitude of induced emf in the coil at t=5 ~s is:

Correct answer: D. 192 ~V Given: φ=2 t^3+4 t^2+2 t+5, & t=5 ~s By Faraday's law of EMI, we have, emf=-(d φ)/(d t) (where, the symbols have their usual meanings) aligned e=|(d φ)/(d t)| & =|(d)/(d t)(2 t^3+4 t^2+2 t+5)| & =6 t^2+8 t+2 aligned aligned ext At t & =5 s e & =6(5)^2+8(5)+2 & =192 ~V aligned

NEET 2022

Exam: NEET
Year: 2022
Subject: Physics
Chapter: Electromagnetic Induction

NEET 2022 · Physics · Electromagnetic Induction

The magnetic flux linked to a circular coil of radius R is: φ=2...

The magnetic flux linked to a circular coil of radius

R

is:

\phi=2 t^3+4 t^2+2 t+5 \mathrm{~Wb}

The magnitude of induced emf in the coil at

t=5 \mathrm{~s}

is:

Options

A $108 \mathrm{~V}$
B $197 \mathrm{~V}$
C $150 \mathrm{~V}$
D $192 \mathrm{~V}$

Correct answer

D. $192 \mathrm{~V}$

Step-by-step solution

Given:

\phi=2 t^3+4 t^2+2 t+5,

t=5 \mathrm{~s}

By Faraday's law of EMI, we have,

\mathrm{emf}=-\frac{d \phi}{d t}

(where, the symbols have their usual meanings)

\begin{aligned} e=\left|\frac{d \phi}{d t}\right| & =\left|\frac{d}{d t}\left(2 t^3+4 t^2+2 t+5\right)\right| \\ & =6 t^2+8 t+2 \end{aligned}

\begin{aligned} ext{At } t & =5 s \\ e & =6(5)^2+8(5)+2 \\ & =192 \mathrm{~V} \end{aligned}

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