Question 92: Magnetism and Matter

Question

As shown in the figure, a long straight conductor with semicircular arc of radius $\frac{\pi}{10}$m is carrying current $\mathrm{I=3A}$. The magnitude of the magnetic field, at the center O of the arc is :
(The permeability of the vacuum $=4\pi	imes10^{-7}~\mathrm{NA}^{-2}$)
![](https://app-content.cdn.examgoal.net/fly/@width/image/1ldodban9/90760155-e6e9-488e-8dc2-07b3df0bf6bf/4e45dd50-a3ab-11ed-bed3-417993225459/file-1ldodbana.png?format=png)

Accepted Answer

Magnetic field due to two straight wire will be zero as point O is on the axis of the wire. Magnetic field at $B$ will be only due to the semicircular arc.
$
 \begin{aligned}
	ext { So, } B & =\left(\frac{\mu_0 i}{4 r}ight)=\frac{\left(4 \pi 	imes 10^{-7}ight) 	imes 3}{4 	imes\left(\frac{\pi}{10}ight)} \\
& =3 	imes 10^{-6} \mathrm{~T} \\
& =3 \mu \mathrm{T}
\end{aligned}
$
![](https://app-content.cdn.examgoal.net/fly/@width/image/1ldodban9/90760155-e6e9-488e-8dc2-07b3df0bf6bf/4e45dd50-a3ab-11ed-bed3-417993225459/file-1ldodbana.png?format=png)

Answer

$4\mu\mathrm{T}$

Answer

$3\mu\mathrm{T}$

Answer

$6\mu\mathrm{T}$

Answer

$1\mu\mathrm{T}$

Magnetism and Matter

As shown in the figure, a long straight conductor with semicircular arc of radius \frac{\pi}{10}m is carrying current \mathrm{I=3A}. The magnitude of the magnetic field, at the center O of the arc is : (The permeability of the vacuum =4\pi\times10^{-7}~\mathrm{NA}^{-2})

Choose the correct answer: