Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A circular coil of one turn of radius 5.0 cm is rotated about a diameter with a constant angular speed of 80 revolutions per minute. A uniform magnetic field B = 0.01 T exists in a direction perpen

A circular coil of one turn of radius 5.0 cm is rotated about a diameter with a constant angular speed of 80 revolutions per minute. A uniform magnetic field B = 0.01 T exists in a direction perpendicular to the axis of rotation, the maximum emf induced, the average emf induced in the coil over a long period and the average of the squares of emf induced over a long period is:

Option: 1

\mathrm{6.4 \times 10^{-4} \mathrm{~V}, zero, 2.2 \times 10^{-7} \mathrm{~V}^2}


Option: 2

\mathrm{6.6 \times 10^{-4} \mathrm{~V}, zero, 2.0 \times 10^{-7} \mathrm{~V}^2}


Option: 3

\mathrm{6.8 \times 10^{-4} \mathrm{~V}, zero, 2.5 \times 10^{-7} \mathrm{~V}^2}


Option: 4

\mathrm{6.4 \times 10^{-4} \mathrm{~V}, zero, 2.0 \times 10^{-6} \mathrm{~V}^2}


Answers (1)

best_answer

Let at any time, normal to loop makes an angle \theta with magnetic field

\mathrm{\begin{aligned} & \phi_{\mathrm{B}}=\mathrm{NBS} \cos \theta-\mathrm{NBS} \cos \omega \mathrm{t} \\ & \mathrm{e}=-\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}=\mathrm{NBS} \omega \sin \omega \mathrm{t}=\mathrm{e}_0 \sin \omega \mathrm{t} \\ & \Rightarrow \mathrm{e}_{\max }=\mathrm{e}_0=\mathrm{NSB} \omega=1 \times 0.01 \times \pi(5)^2 \times 10^{-4} \times \frac{80 \times 2 \pi}{60}=6.6 \times 10^{-4} \mathrm{~V} \end{aligned}}

Over a long period, i.e., one time period, average of induced emf is

\mathrm{\begin{aligned} & \overline{\mathrm{e}}=\frac{\int_0^{\mathrm{T}} \mathrm{edt}}{\mathrm{T}}=0 \\ & \overline{\mathrm{e}^2}=\frac{\int_0^{\mathrm{T}} \mathrm{e}^2 \mathrm{dt}}{\mathrm{T}}=\frac{\mathrm{e}_0^2}{2}=2.0 \times 10^{-7} \mathrm{~V}^2 \end{aligned}}

 

Posted by

sudhir kumar

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

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus
anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question