JEE Main 30 June 2022 Shift 1 Solved Paper

In the width of r2−r1 total n turns presents.
∴ In 1 unit width ‌

n

r2−r1

turns presents.
∴ In the width of d r number of turns,
n′=‌

n

r2−r1

×dr
Magnetic field (dB) due to element of dr length is
dB=‌

µ0×I×n′

2r

=‌

µ0I

2r

×‌

n

(r2−r1)

∴ Total magnetic field due to entire coil is,
∫dB=

r2

∫

r1

‌

µ0I

2r

×‌

n

(r2−r1)

dr
⇒B=‌

µ0In

2(r2−r1)

‌

r2

∫

r1

‌

dr

r

=‌

µ0In

2(r2−r1)

×[loger]r2r1
=‌

µ0In

2(r2−r1)

×(loger2−loger1)
=‌

µ0In

2(r2−r1)

×loge‌

r2

r1