Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A conducting circular loop of radius \frac{10}{\sqrt{\pi }}cm  is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop at 0.25 s is:  

Option: 1

emf = 1mV


Option: 2

emf = 5mV


Option: 3

emf = 100mV


Option: 4

emf = 10mV


Answers (1)

best_answer

\begin{aligned} & \mathrm{emf}=-\frac{\mathrm{d} \phi}{\mathrm{dt}} \Rightarrow \varepsilon=\frac{-\mathrm{d}(\mathrm{BA})}{\mathrm{dt}} \\ & \varepsilon=-\mathrm{A} \frac{\mathrm{dB}}{\mathrm{dt}} \Rightarrow \varepsilon=-\pi \mathrm{R}^2\left(\frac{0-\mathrm{B}}{\Delta \mathrm{t}}\right) \end{aligned}

\begin{aligned} & \varepsilon=\frac{\pi R^2 \mathrm{~B}}{\Delta \mathrm{t}} \Rightarrow \varepsilon=\frac{\pi\left(\frac{10}{\sqrt{\pi}} \times 10^{-2}\right)^2 \times 0.5}{0.5} \\ & \varepsilon=10^{-2} \text { volt }=10 \mathrm{~m} \text { volt } \end{aligned}

Posted by

Ritika Harsh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: Most important chapters for physics, chemistry, mathematics to score good marks
anam-khan

JEE Main 2025 session 2 dates clash with ISC, PSEB 12th exams; no impact on major science subjects
anam-khan

Delhi police EOW books FIITJEE for fraud; 192 complaints filed

Latest Question