TS EAMCET 11-Sep-2020 Shift 1 Solved Paper

Consider a small circular element of thickness dr at a distance r from the centre of spiral as shown in the figure,Total number of turns in this element =‌

N

b−a

dr
Current passing through this element,
i=‌

N

b−a

drI
∴ Magnetic field at the centre of spiral due to this element is given as
dB=‌

µ0i

2r

=‌

µ0

2r

⋅‌

NI

b−a

dr
∴ Total magnetic field,
B=∫dB=

b

∫

a

‌

µ0NI

2(b−a)

⋅‌

dr

r

⇒‌‌B=‌

µ0NI

2(b−a)

‌ln(‌

b

a

)
Given, N=100,I=1A,a=1cm‌‌=10−2m
b‌‌=2cm=2×10−2m
=2π×10−3‌ln(2)
=2π‌ln(2)×10−3T
=2π‌ln(2)‌mT