Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A circular disc of radius 0.2 m is placed in a uniform magnetic field of induction <img alt="\mathrm {\mathrm{n}\left(\frac{1}{\pi}\right)Wbm^{-2}}" src="https://entrancecorner.oncodecogs.com/

A circular disc of radius 0.2 m is placed in a uniform magnetic field of induction \mathrm {\mathrm{n}\left(\frac{1}{\pi}\right)Wbm^{-2}}  in such a way that its axis makes an angle of 60° with B. The magnetic flux linked with the disc is

Option: 1

0.02 Wb


Option: 2

0.06 Wb


Option: 3

0.08 Wb


Option: 4

0.01 Wb


Answers (1)

best_answer

\mathrm{\text { Here, } \mathrm{r}=0.2 \mathrm{~m}, \mathrm{~B}=\frac{1}{\pi} \mathrm{Wbm}^{-2}, \theta=60^{\circ}}

\mathrm{\therefore \text { Magnetic flux, } \phi=\mathrm{BA} \cos \theta=\mathrm{B}\left(\pi \mathrm{r}^2\right) \cos \theta=\frac{1}{\pi} \pi(0.2)^2 \cos 60^{\circ}=0.02 \mathrm{~Wb}}

Posted by

Divya Prakash Singh

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 Maths 2025: Five tips to score high marks in ‘toughest’ subject; most important chapters
anam-khan

JEE Main 2025 exam city intimation slip for session 1 soon at jeemain.nta.nic.in
anam-khan

Focus on modern physics, formulas to solve tricky problems; 5 tips to tackle JEE Main physics 2025

Latest Question