Question

If $B_1,B_2$ and $B_3$ are the magnetic fields at points $2R,R$ and $\frac{R}{2}$ from the axis respectively, then $B_1,B_2$ and $B_3$ are in the respective ratio of

• A. $4:2:1$
• B. $1:2:1$
• C. $1:2:4$
• D. $2:1:2$

Answer 1

The correct option is C $1:2:4$

For $r=2R$
Consider a ring of radius $r$ and width $dr$ as shown.
Current through the ring of width $dr:dI=\left(2\pi rdr\right)J=2\pi r\left(\frac{a}{r}{e}^{2\left(\frac{r}{R}-1\right)}\right)$
Net current $I={\int }_0^{R}2\pi r\left(\frac{a}{r}{e}^{2\left(\frac{r}{R}-1\right)}\right)dr=2\pi a{\int }_0^R{e}^{2\left(\frac{r}{R}-1\right)}dr=2\pi a\frac{R}{2e^2}(e^2-1)=\pi aR(1-\frac{1}{e^2})$
$\overrightarrow{B_1}.\overrightarrow{dl}={\mu }_0{i}_{enclosed}={\mu }_0\pi aR(1-\frac{1}{e^2})$
$\therefore B_1\times 2\pi r={\mu }_0\pi aR(1-\frac{1}{e^2})$
$\Rightarrow B_1=\frac{{\mu }_0\pi aR}{2\pi r}(1-\frac{1}{e^2})=\frac{{\mu }_0\pi aR}{2\pi \times 2R}(1-\frac{1}{e^2})=\frac{{\mu }_{0}a}{4}(1-\frac{1}{e^2})$
For $r=R$
$B_2\times 2\pi R={\mu }_0\pi aR(1-\frac{1}{e^2})$
$\Rightarrow B_2=\frac{{\mu }_{0}a}{2}(1-\frac{1}{e^2})$
For $r=\frac{R}{2}$
$I_{enclosed}={\int }_0^{\frac{R}{2}}2\pi r\left(\frac{a}{r}{e}^{2\left(\frac{r}{R}-1\right)}\right)dr=\pi aR(1-\frac{1}{e^2})$
$B_3\times 2\pi \frac{R}{2}={\mu }_0\pi aR(1-\frac{1}{e^2})$
$\Rightarrow B_3={\mu }_{0}a(1-\frac{1}{e^2})$
$\Rightarrow B_1:B_2:B_3=1:2:4$