Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A coil of resistance, 100 turns and radius 6 mm is connected to ammeter of resistance .

A coil of 40\Omega resistance, 100 turns and radius 6 mm is connected to ammeter of resistance 160\Omega . Coil is placed perpendicular to the magnetic field. When coil is taken out of the field, 32\mu C charge flows through it. The intensity of magnetic field will be:

Option: 1

6.55 T


Option: 2

5.66 T


Option: 3

2.55 T


Option: 4

0.566 T


Answers (1)

best_answer

\mathrm{We\: have, \Delta q=-\frac{N}{R} \Delta \phi\: and \: \Delta \phi=\Delta B \cdot A \cos \theta}

\mathrm{\therefore 32 \times 10^{-6}=-\frac{100}{(160+40)}(0-B) \times \pi \times\left(6 \times 10^{-3}\right)^2 \times \cos 0^{\circ} }

Intensity of magnetic field, B = 0.566 T

Posted by

Pankaj

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