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

Q.

Conductor of length l has shape of a semi-cylinder of radius R(<<l). Cross section of the conductor is shown in Fig. Thickness of the conductor is t(<<R) and conductivity of its material varies with angle θ only, according to the law  σ=σ0 cosθ. If a battery of emf V and of negligible internal resistance is connected across its end faces, calculate magnetic induction at mid point O of the axis of the semi-cylinder is μ0σ0bVctdal find  a2+b2+c2+d2

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 19.

(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

Element is long straight conductor of cross section area = (Rdo.)t
Its resistance   =pla
   =lσa
Current through this element 
di=Vresistance           di=Vσal
So magnetic field at O
dB=μ2πRdi

dBnet=2dBcosθ            Bnet=π2  2dBcosθ=μ0σ0Vt4l

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