Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction B = B0e-t is established inside the coil. If the key (K) is closed, the electrical power developed right after closing the switch is equal to

P=ε2R;ε=−ddt(BA)=-AddtBoe−t=ABoe−t⇒P=1RABoe−t2=A2Bo2e−2tRAt the time of starting t=0 so P=A2Bo2R⇒P=πr22Bo2R=Bo2π2r4R

Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction B = B0e-t is established inside the coil. If the key (K) is closed, the electrical power developed right after closing the switch is equal to

P=ε2R;ε=−ddt(BA)=-AddtBoe−t=ABoe−t⇒P=1RABoe−t2=A2Bo2e−2tRAt the time of starting t=0 so P=A2Bo2R⇒P=πr22Bo2R=Bo2π2r4R

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction B = B₀e^-t is established inside the coil. If the key (K) is closed, the electrical power developed right after closing the switch is equal to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{B_{0}^{2} π r^{2}}{R}$

$\frac{B_{0} 10 r^{3}}{R}$

$\frac{B_{0}^{2} π^{2} r^{4}}{R}$

$\frac{B_{0}^{2} π^{2} r^{4} R}{5}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} P = \frac{ε^{2}}{R}; ε = - \frac{d}{d t} (B A) = - A \frac{d}{d t} (B_{o} e^{- t}) = A B_{o} e^{- t} \\ \Rightarrow P = \frac{1}{R} {(A B_{o} e^{- t})}^{2} = \frac{A^{2} B_{o}^{2} e^{- 2 t}}{R} \end{array}$

At the time of starting $t = 0 so P = \frac{A^{2} B_{o}^{2}}{R}$

$\Rightarrow P = \frac{{(π r^{2})}^{2} B_{o}^{2}}{R} = \frac{B_{o}^{2} π^{2} r^{4}}{R}$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

best study material, now at your finger tips!

live classes
progress tracking
24x7 mentored guidance
study plan analysis

download the app

Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction B = B0e-t is established inside the coil. If the key (K) is closed, the electrical power developed right after closing the switch is equal to

Best Courses for You

Detailed Solution

courses

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

download the app

Download the App

Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction B = B₀e^-t is established inside the coil. If the key (K) is closed, the electrical power developed right after closing the switch is equal to