Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Solve! - Electromagnetic Induction and Alternating currents - BITSAT

Figure shows a circular area of radius R where a uniform magnetic field \vec{B} is going into the plane of paper and increasing in magnitude at a constant rate. In that case, which of the following graphs, drawn schematically, correctly shows the variation of the induced electric field E(r) ?

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 

Answers (1)

best_answer

As we discussed in concept

Induced electric field -

\varepsilon = \oint \vec{E }\cdot \vec{dl}= \frac{-d\phi }{dt}

When

A \rightarrow constant

\varepsilon = \oint \vec{E }\cdot \vec{dl}=A\frac{dB}{dt}

B \rightarrow Varying

E= \frac{a^{2}}{2r}\, \frac{dB}{dt}

 

- wherein

E_{inclosed}\, \alpha \frac{1}{r}

a\rightarrow radius

 

 E = \frac{r}{2}\frac{dB}{dt}\: \: \: \: r < R \\

 

= \frac{R^2}{2r}\frac{dB}{dt}\: \: \: \: r < R \\

 

 

 

 

 

 

 

 

 


Option 1)

Option 2)

Option 3)

Option 4)

Posted by

Avinash

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 2026 session 2 application correction begins on jeemain.nta.nic.in; edit details by tomorrow
anam-khan

JEE Main 2026 Registration LIVE: Session 2 exam from April 2 to 9; shift timings, admit card, exam pattern
anam-khan

JEE Main 2026 session 2 registration ends today for BTech, BArch; apply at jeemain.nta.nic.in

Latest Question