TS EAMCET 20-July-2022 Shift 1 Solved Paper

Current carrying loop ABCD is shown below Current, I=‌

1.2

π

A
Magnetic field at centre O due to current carrying arc‌A‌D,
B1‌=‌

µ0I

21

⋅‌

θ

360∘

‌=‌

4π×10−7× 1.2 π

2×1×10−2

×‌

30∘

360∘

‌=24×10−5×‌

1

12

‌=2×10−6T‌ (upward direction) ‌
Similarly, magnetic field at centre ˙O due to current carrying arc BC,
B2‌=‌

µ0I

2r2

⋅‌

θ

360∘

‌=‌

4π×10−7× 1.2 π

2×2×10−2

×‌

30

360∘

=1×10−6T (downward direction)
∴ Net magnetic field at centre 0 ,
B‌net ‌‌=B1−B2=2×10−6−1×10−6
‌=1×10−6T=1µT‌ (upward) ‌