Correct Answer is Option (A)

By given information

$32=4\times \frac{{\mu }_{0}i}{2r}$ ..... (i)

Also, $r^{\prime }=4r$ ...... (ii)

and $B^{\prime }=1\times \frac{{\mu }_{0}i}{2r^{\prime }}$ .... (iii)

$\Rightarrow B^{\prime }=\frac{{\mu }_{0}i}{2(4r)}=\frac{{\mu }_{0}i}{8r}=\frac{1}{8}\times 16=2T$

Correct Answer is Option (A)

Magnetic field due to wire $AB$ and wire $ED$ cancel each other. So, magnetic field due to BC and ET will be

$\begin{array}{rl}& B_1=\frac{{\mu }_{0}I}{4\pi (a/2)}\cdot (\sin \frac{\pi }{2}+\sin 0^{\circ })=\frac{{\mu }_{0}I}{2\pi a}\\ & B_2=\frac{{\mu }_{0}I}{4\pi (a/2)}\cdot (\sin \frac{\pi }{2}+\sin 0^{\circ })=\frac{{\mu }_{0}I}{2\pi a}\end{array}$

So, $B_{\text{net }}=B_1+B_2=\frac{{\mu }_{0}I}{\pi a}$

Correct Answer is Option (D)

$\because By$ at $P_1=0$ [option A and C are incorrect]

$By$ has the opposite direction for the +ve and $-vez$ axis

Correct Answer is Option (C)

$A\to B_0=\frac{-{\mu }_{0}I}{4\pi r}+\frac{{\mu }_{0}I}{2r}-\frac{{\mu }_{0}I}{4\pi r}$

$B_0=\frac{{\mu }_{0}I}{2\pi r}(\pi -1)A\to \text{ III }$

$B\to B_0=\frac{{\mu }_{0}I}{4\pi r}+\frac{{\mu }_{0}I}{4r}+\frac{{\mu }_{0}I}{4\pi r}$

$B_0=\frac{{\mu }_{0}I}{4\pi r}(\pi +2)B\to I$

$C\to B_0=\frac{{\mu }_{0}I}{4\pi r}+\frac{{\mu }_{0}I}{4r}+0$

$B_0=\frac{{\mu }_{0}I}{4\pi r}(\pi +1)\mathrm{C}\to \mathrm{IV}$

D. $\to B_0=\frac{{\mu }_{0}I}{4r}\mathrm{D}\to \mathrm{II}$

Correct Answer is 3

$B=\frac{{\mu }_{0}nI}{2R}$

$37.68\times {10}^{-4}=\frac{4\pi \times {10}^{-7}100I}{2\times 5\times {10}^{-2}}$

$I=\frac{300A}{100}$

$=3A$