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