The magnetic flux through each turn of a coil having 200 turns is given as (t^2-2 t) × 10^ -3 Wb , where t is in second. The emf induced in the coil at t=3 ~s is

Correct answer: C. 0.8 V Given magnetic flux φ=(t^2-2 t) × 10^ -3 On differentiating with recpect to time t (d φ)/(d t)=(2 t-2) × 10^ -3 At time t=3 s aligned |(d φ)/(d t)|_ t=3 s & =(2 × 3-2) × 10^ -3 & =4 × 10^ -3 Wb s^ -1 Now, |e| & =N|(d φ)/(d t)| & =200 × 4 × 10^ -3 & =800 × 10^ -3 =0.8 V aligned

JIPMER 2015

Exam: JIPMER
Year: 2015
Subject: Physics
Chapter: Electromagnetic Induction

JIPMER 2015 · Physics · Electromagnetic Induction

The magnetic flux through each turn of a coil having 200 turns is...

The magnetic flux through each turn of a coil having 200 turns is given as

\left(t^2-2 t\right) \times 10^{-3}

Wb , where

t

is in second. The emf induced in the coil at

t=3 \mathrm{~s}

Options

A0.7 V
B1.2 V
C0.8 V
D0.9 V

Correct answer

C. 0.8 V

Step-by-step solution

Given magnetic flux

\phi=\left(t^2-2 t\right) \times 10^{-3}

On differentiating with recpect to time

t \frac{d \phi}{d t}=(2 t-2) \times 10^{-3}

At time

\quad t=3 \mathrm{ s}

\begin{aligned}\left|\frac{d \phi}{d t}\right|_{t=3 \mathrm{ s}} & =(2 \times 3-2) \times 10^{-3} \\ & =4 \times 10^{-3} \mathrm{ Wb} \mathrm{ s}^{-1} \\ \text{ Now, }|e| & =N\left|\frac{d \phi}{d t}\right| \\ & =200 \times 4 \times 10^{-3} \\ & =800 \times 10^{-3}=0.8 \mathrm{ V}\end{aligned}

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