Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

A circular ring of small cross-sectional area ‘a’ is placed horizontally in a region where a uniform vertical (upward) magnetic field B0 exists. Radius of the wire is R and current I0  in wire is in anticlockwise direction as shown in figure. The Young’s modulus of the material of the ring is Y. In steady state the elastic potential energy in the ring comes out to be U=I0pRqB0rπaY  then find the sum p + q + r. 

Question Image
 

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

answer is 7.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Question Image

 

 

 

 

 

 

 

 

 ΔI=2RθFi=2Tsinθ=2Tθ Fβout=i0ΔIB=i02Rθ.B0     2Tθ=2R i0θB0T=i0RB0 PE=12(stress)(stressY).a2πR =12(I0RB0a)2.1Ya.2πR   PE=I02R3B02πaY

Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring