TS EAMCET 18-July-2022 Shift 1 Solved Paper

The given situation is shown as

Magnitude of magnetic field due to straight current carrying wire AB is given as
B1=‌‌

µ0

4π

⋅‌

I

OM

(sin‌φ+sin‌φ)
=‌‌

µ0

4π

×‌

I

Rsin‌φ

(2sin‌φ)‌‌(∵OM=Rsin‌φ)
=‌‌

µ0

4π

×‌

2I

R

=10−7×‌

2×5

0.1

‌(∵R=100‌mm=0.1m‌ and ‌I=5A)
=‌10−5T
Magnitude of magnetic field due to current carrying circular arc,
B2‌=‌

µ0I

2R

×‌

(2π−2φ)

2π

‌=‌

4π×10−7×5

2×0.1

×‌

(2π− π 2 )

2π

‌‌(∵2φ=90∘=‌

π

2

)
‌=‌

4×314×10−7×5

0.2

×‌

3

4

‌B2=2.36×10−5⋅T
∴ Total magnetic field at 0 ,
B‌=B1+B2=10−5+2.36×10−5
‌=10−5(1+2.36)=3.36×10−5T
‌=33.6×10−6T
‌=33.6µT