Question

An element $\Delta l=\Delta x\hat{i}$ is placed at the origin and carries a large current $I=10\text{A}$, The magnetic field on the $y-\text{axis}$ at a distance of $0.5\text{m}$ is
$[\text{Take,}\Delta x=1\text{cm}]$

• A. $\overrightarrow{dB}=4\times {10}^{-8}\text{T}\hat{k}$
• B. $\overrightarrow{dB}=2\times {10}^{-8}\text{T}\hat{k}$
• C. $\overrightarrow{dB}=1\times {10}^{-8}\text{T}\hat{k}$
• D. $\overrightarrow{dB}=16\times {10}^{-8}\text{T}\hat{k}$

Answer 1

The correct option is A $\overrightarrow{dB}=4\times {10}^{-8}\text{T}\hat{k}$

Applying Biot- Savart's law,

$dB=\frac{{\mu }_{0}I}{4\pi }\frac{dl\sin \theta }{r^2}........\left(1\right)$

Here, $dl=\Delta x={10}^{-2}\text{m};I=10\text{A}$
$r=0.5\text{m};\theta ={90}^{\circ }$

Putting the values in eq. $\left(1\right)$

$dB=\frac{{10}^{-7}\times 10\times {10}^{-2}\sin {90}^{\circ }}{25\times {10}^{-2}}$

$dB=4\times {10}^{-8}\text{T}$

For direction of $dB$,

$\hat{n}=\frac{\overrightarrow{dl}\times \overrightarrow{r}}{|\overrightarrow{dl}||\overrightarrow{r}|}=\frac{\Delta x\hat{i}\times r\hat{j}}{\Delta xr}=\hat{k}$

$\therefore \overrightarrow{dB}=4\times {10}^{-8}\text{T}\hat{k}$

Hence, option $\left(a\right)$ is correct.