Correct Answer is Option (D)

There is no current inside the pipe. Therefore

$\oint \bar{B}.\overline{d\ell }={\mu }_{0}I$

$I=0$

$\therefore$ $B=0$

Correct Answer is Option (D)

Here, current is uniformly distributed across the cross-section of the wire, therefore, current enclosed in the ampere-an path formed at a distance $r_1\left(=\frac{a}{2}\right)$

$=\left(\frac{\pi r_1^2}{\pi a^2}\right)\times I,$ where $I$ is total current

$\therefore$ Magnetic field at $P_1$ is

$B_1=\frac{{\mu }_0\times currentenclosed}{path}$

$\Rightarrow B_1=\frac{{\mu }_0\times \left(\frac{\pi r_1^2}{\pi a^2}\right)\times I}{2\pi r_1}$

$=\frac{{\mu }_0\times I{r}_1}{2\pi a^2}$

Now, magnetic field at point $P_2,$



$B_2=\frac{{\mu }_0}{2\pi }.\frac{I}{\left(2a\right)}=\frac{{\mu }_{0}I}{4\pi a}$

$\therefore$ Required ratio $=\frac{B_1}{B_2}=\frac{{\mu }_{0}I{r}_1}{2\pi a^2}\times \frac{4\pi a}{{\mu }_{0}I}$

$=\frac{2r_1}{a}=\frac{2\times \frac{a}{2}}{a}=1.$

Correct Answer is Option (A)

$\frac{B_2}{B_1}=\frac{{\mu }_0{n}_2{i}_2}{{\mu }_0{n}_1{i}_1}$

$\Rightarrow \frac{B_2}{6.28\times {10}^{-2}}=\frac{100\times \frac{i}{3}}{200\times i}$

$\Rightarrow B_2=\frac{6.28\times {10}^{-2}}{6}$

$=1.05\times {10}^{-2}Wb/m^2$

Correct Answer is Option (B)

Using Ampere's law at a distance $r$ from axis, $B$ is same from symmetry.

$\int B.dl={\mu }_{0}i$

i.e., $B\times 2\pi r={\mu }_{0}i$

Here $i$ is zero, for $r<R,$ whereas $R$ is the radius

$\therefore$ $B=0$

Correct Answer is Option (B)

Using Ampere's law at a distance $r$ from axis, $B$ is same from symmetry.

$\int B.dl={\mu }_{0}i$

i.e., $B\times 2\pi r={\mu }_{0}i$

Here $i$ is zero, for $r<R,$ whereas $R$ is the radius

$\therefore$ $B=0$