Correct Answer is Option (C)

The magnetic field due to a current carrying wire can be calculated using Ampere's law. When the current is uniformly distributed across the cross-section of the wire, the situation will be different inside and outside the wire.

Inside the wire (r < a), the magnetic field is directly proportional to r (the distance from the center of the wire). This is because as you move away from the center of the wire, you enclose more current, so the magnetic field increases linearly with r.

Outside the wire (r > a), all the current in the wire is enclosed, so the magnetic field decreases with increasing r. This is a result of the magnetic field lines spreading out as they move away from the wire.

Correct Answer is Option (C)

Number of turns per meter $=7000$ turns per $\mathrm{m}$

$\begin{array}{rlr}& i=2\mathrm{A}& \\ \\ & B={\mu }_{0}ni=4\pi \times {10}^{-7}\times 7000\times 2\\ \\ & =56\pi \times {10}^{-4}\mathrm{T}\\ \\ & =56\times \frac{22}{7}\times {10}^{-4}\mathrm{T}\\ \\ & =176\times {10}^{-4}\mathrm{T}\end{array}$

Correct Answer is Option (D)

Inside a hollow cylindrical conductor with uniform current distribution net magnetic field is zero in hollow space.

But outside the cylindrical conductor $B\propto \frac{1}{r}$

$\Rightarrow$ Graph in option D would be a correct one

Correct Answer is Option (A)

$B={\mu }_{0}ni$

Now $i\to 2i$

And $n\to \frac{n}{2}$

$B^{\prime }={\mu }_0\frac{n}{2}\times 2i={\mu }_{0}ni=B$

Correct Answer is Option (B)

$\int \bar{B}.\overline{dl}={\mu }_0{I}_{in}$

$\Rightarrow B\times 2\pi r=\frac{{\mu }_{0}I}{\pi R^2}\times \pi r^2$

$\Rightarrow B\propto r$

Correct Answer is Option (C)

$B=\frac{{\mu }_{0}I}{2r}$

$B_a=\frac{{\mu }_{0}I{r}^2}{2\left(r^2+\frac{r^2}{4}\right)}$

$\Rightarrow \frac{B_a}{B}={\left(\frac{2}{\sqrt{5}}\right)}^3$

$\Rightarrow B_a={\left(\frac{2}{\sqrt{5}}\right)}^{3}B$

Correct Answer is Option (A)

when x < a

$B_1(2\pi x)={\mu }_0\left(\frac{i_0}{\pi a^2}\right)\pi x^2$

$B(2\pi x)=\frac{{\mu }_0{i}_0{x}^2}{a^2}$

$B_1=\frac{{\mu }_0{i}_{0}x}{2\pi a^2}$ .... (1)

when a < x < b

$B_2(2\pi x)={\mu }_0{i}_0$

$B_2=\frac{{\mu }_0{i}_0}{2\pi x}$ ..... (2)

$\frac{B_1}{B_2}=\frac{{\mu }_0{i}_0\frac{x}{2\pi a^2}}{\frac{{\mu }_0{i}_0}{2\pi x}}=\frac{x^2}{a^2}$

Correct Answer is Option (D)

$B_{axis}=\frac{{\mu }_{0}i{R}^2}{2{(R^2+x^2)}^{3/2}}$

$B_{centre}=\frac{{\mu }_{0}i}{2R}$

$\therefore$ $B_{centre}=\frac{{\mu }_{0}i}{2a}$

$\therefore$ $B_{axis}=\frac{{\mu }_{0}i{a}^2}{2{(a^2+r^2)}^{3/2}}$

$\therefore$ fractional change in magnetic field =

$\frac{\frac{{\mu }_{0}i}{2a}-\frac{{\mu }_{0}i{a}^2}{2{(a^2+r^2)}^{3/2}}}{\frac{{\mu }_{0}i}{2a}}=1-\frac{1}{{\left[1+\left(\frac{r^2}{a^2}\right)\right]}^{3/2}}$

$\approx 1-\left[1-\frac{3}{2}\frac{r^2}{a^2}\right]=\frac{3}{2}\frac{r^2}{a^2}$

Note : ${\left(1+\frac{r^2}{a^2}\right)}^{-3/2}\approx \left(1-\frac{3}{2}\frac{r^2}{a^2}\right)$

[True only if r << a]

Hence, option (d) is the most suitable option.

Correct Answer is Option (C)

Graph for wire of radius R :


As b > a

B_a > B_b

$B_a=\frac{{\mu }_{0}i}{2\pi a}$

$B_b=\frac{{\mu }_{0}i}{2\pi b}$

Correct Answer is Option (D)

$3I_1=i$

$\Rightarrow$ $I_1=\frac{i}{3}$

$B={\mu }_{0}ni=3T$

$B_C={\mu }_{0}n{I}_1$

$\Rightarrow$ $B_C={\mu }_{0}n\frac{i}{3}=\frac{3T}{3}=1T$

Correct Answer is Option (A)

B = $\mu$ n i

B = $\mu$_r $\mu$₀ n i

B = 500 $\times$ 4$\pi$ $\times$ 10^$-$7 $\times$ 10³ $\times$ 5

B = $\pi$ $\times$ 10^$-$3 $\times$ 10³

B = $\pi$ T

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

$\therefore$ $B=0$