JEE Main 31-Aug-2021 Shift 1 Solved Paper

Given, number of turns = N
Inner radius = a
Outer radius = b
Current = I
Consider an element of thickness dr at a distance r from the centre.Now, number of turns in dr element, dN=

N

(b−a)

‌dr
Magnetic field due to this element at the centre,
dB=

μ0

2

dN.I

r

∴dB=

1

2

μ0I

r

N

(b−a)

‌dr
∴ Total magnetic field at the centre,
B=∫abdB=∫ab

μ0‌IN

2(b−a)

dr

r

=

N

2

μ0I

(b−a)

‌ln‌r|ab
=

N

2

μ0l

(b−a)

‌ln(

b

a

)