Question

A current carrying conductor having a length of $1\text{mm}$ produces a magnetic field of $5\times {10}^{-12}\text{T}$ at the point $\text{P}$. The value of current flowing in the conductor is $10\text{A}$ and the distance between the conductor and the point $\text{P}$ is $1\text{m}$. Find the angle between the length vector of conductor and the position vector of the point $\text{P}$.

• A. ${\sin }^{-1}\left(1/3\right)$
• B. ${\sin }^{-1}\left(1/4\right)$
• C. ${\sin }^{-1}\left(2/3\right)$
• D. ${\sin }^{-1}\left(1/2\right)$

Answer 1

The correct option is D ${\sin }^{-1}\left(1/2\right)$

Given,
length of the conductor $l=1\text{mm}$

Distance of the point $P$ from the conductor, $r=10\text{m}$

Since, $l>>r$ so, we can use Biot-Savart law, to calculate the magnetic field at point $P$,

$B=\frac{{\mu }_0}{4\pi }\frac{il\sin \theta }{r^2}$

$\therefore \sin \theta =\frac{B{r}^2}{{10}^{-7}\times i\times l}$

$=\frac{(5\times {10}^{-12})\times \left({10}^2\right)}{{10}^{-7}\times 10\times \left({10}^{-3}\right)}=\frac{1}{2}$

$\Rightarrow \theta ={\sin }^{-1}\left(1/2\right)$