Magnetic Effects Of Current Question 21
Question 21 - 01 February - Shift 2
As shown in the figure, a long straight conductor with semicircular arc of radius $\frac{\pi}{10} m$ is carrying current $I=3 A$. The magnitude of the magnetic field. at the center $O$ of the arc is:
(The permeability of the vacuum $=4 \pi \times 10^{-7} NA^{-2}$ )
(1) $6 \mu T$
(2) $1 \mu T$
(3) $4 \mu T$
(4) $3 \mu T$
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Answer: (4)
Solution:
Formula: Magnetic field due to circular loop
$B_C=\frac{\mu_0 I}{4 \pi R}(\pi)(B$ at centre of circular arc)
$ \begin{aligned} & =\frac{\mu_0 I}{4 R}=\frac{4 \pi \times 10^{-7} \times 3}{4 \times \frac{\pi}{10}} \\ & =3 \times 10^{-6} T=3 \mu T \end{aligned} $
