TS EAMCET 14-Sep-2020 Shift 2 Solved Paper

For a solenoid,
I=20A,l=2m,B=20mT=20×10−3T
Diameter, d=3cm=3×10−2m
∴ Radius, r=‌

3

2

×10−2m
If n be the number of turns per unit length in the solenoid, then
B=µ0nI⇒n=‌

B

µ0I

=‌

20×10−3

4π×10−7×20

=‌

104

4π

∴ Total number of turns in the solenoid,
N=nl=‌

104

4π

×2⇒N=‌

104

2π

Length of wire =N× length of wire in one turn
‌‌=N×2πr=‌

104

2π

×2π×‌

3

2

×10−2
‌‌=1.5×102m=150m