Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

An infinetly long wire carrying a current $I(t)= I_0 \cos (\omega t)$ is placed at a distance 'a' from a square loop of side 'a' as shown in the figure .

If the resistance of the loop is R , Then power dissipated in loop is
Option: 1
$\frac{\mu _0^{2}}{4 \pi ^2 R}\times a^2 I_{0}^{2}w^{2}\sin ^{2}(wt)(log 2)^{2}$

Option: 2
$\frac{\mu _0}{2 \pi R}\times a I_{0}w\sin wt(log 2)$

Option: 3
$\frac{\mu _0^{2}}{4 \pi ^2 }\times a^2 I_{0}^{2}w^{2}\sin ^{2}(wt)(log 2)^{2}$

Option: 4
$\frac{\mu _0}{2 \pi }\times a I_{0}w\sin (wt)(log 2)$

Answers (1)

best_answer

As we have learned

Induced Power -

$P= \frac{\varepsilon ^{2}}{R}= \frac{N^{2}}{R}\left ( \frac{d\phi }{dt} \right )^{2}$

- wherein

It depends on time and resistance

Consider a stirp of width dx at a distance x from the current carrying wire , the magnetic field at distance x is $B = \frac{\mu _0 I(t)}{2\pi x }$

Let $d\phi$ be the flux through the elemental area adx , then

$d\phi = Badx = \frac{\mu _0I(t)}{2\pi x}adx$

Total flux through the square loop is

$\phi = \frac{\mu _0I(t)}{2\pi }a\int_{a}^{2a}\frac{dx}{x}$

$\phi = \frac{\mu _0I(t)a}{2\pi } log 2$

$= \frac{\mu _0I_o \cos wta}{2\pi } log 2$

Now induced emf ,

$\varepsilon = \frac{-d\phi }{dt}=\frac{\mu 0}{2 \pi }a I_0w log 2 \sin wt$

Power dissipated in loop is ,

$p = \frac{\varepsilon ^2}{R}$

$p = \frac{\mu _{0}^{2}}{4\pi ^2}a^2I_{0}^{2}w^2\sin ^2 wt (log 2)^2\times \frac{1}{R}$

Posted by

vinayak

View full answer

NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

NEET UG Counselling 2024: MCC revises eligibility for stray vacancy round; adds seat

anam-khan

UP NEET UG Counselling 2024: Stray vacancy round schedule out; registration starts tomorrow

anam-khan

NEET UG 2024: SC grants two more weeks to expert panel to file report

Latest Question