As,
B=μ_onI/2a
a ->radius
B ∝1/a
B₁/B₂=a₂/a₁
B₁=2B₂
B₂=(1/2) x B₁,
Magnetic field is halved.

We know,
B = μoiπ/4πR
Here, 
Substituting the value of theta = π/4
So, Magnetic Field = μoIπ/π(4r x 4)r
 => B = μoI/16r

dB=μ_oidlsin90/4πr²=μ_oidl/4πr²
B_net= ∫dBsinθ
      = ∫μ_oidlR/4πr²r== (μ_oiR/4πr³) ∫dl
      = (μ_oiR/4πr³)2πR= μ_oiR²/2(R²+x²)^3/2
   B= μ_oiR²/2(R2+x²)^3/2
If x>>>R
B= μ_oiR²/2(x²)^3/2

[=B ∝ x^-3]

B₁=μ₀i/2r,
B2=μ₀i/2.2r=μ₀i/4r,
μ₀i/4r=1T
B0=B₁−B₂=μ₀i/4r=1T

From Biot-Savart law, magnetic field at a point p, B= (μ0​/4π)∫ [(Idl×r​)/ r3]
where r is the distance of point p from conductor and I is the current in the conductor.
Thus magnetic field due to current carrying conductor depends on the current flowing through conductor and distance from the conductor and length of the conductor.
 

B= 2μ₀​i⋅πr²/4π(r²+x²)^3/2
Magnetic field at centre= μ₀​i/2r
B= μ₀​i/2r
Put r= √3
B¹=2μ_0​i⋅πr²/4π(4r²)^3/2
=B/8
Hence, √3R distance from the centre magnetic field is equal to magnetic field at centre 
 

The magnetic field due to a circular loop is given by:
B= μ₀​​2πi​/4πr
=10⁻⁷×2π×2/0.0157 ​
=8×10⁻⁵ Wb/m²

The magnetic field at the center O due to the upper side of the semicircular current loop is equal and opposite to that due to the lower side of the loop.

 B=μ₀2πnIa²/4π (a²+x²)^3/2
=10⁻⁷×2×(22/7)×200×5×(0⋅1/2)²/ [(0⋅1/2)²+(0⋅15)²]^3/2
=39.74x10^-5
 

Let the radii be r₁​ and r₂​ respectively.
Since there are two turns of radius r₂​, r₁​=2r_2​
Magnetic field B at the centre of  the coil of radius r_1​ B_1​=​μo​i/2r₁​=​μ_o​i​/4r₂
Magnetic field B at the center of the coil of radius r_2​ B_2​=2×​μ_o​i​/2r₂
∴ B₂/B₁ =(2× μ_o​i/2r₂​)/(μ_o​i /4r₂​)​ ​​=4
Hence the answer is option C, four times its initial value.