JEE MAIN - Physics (2023 - 1st February Evening Shift - No. 14)
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 $$=4\pi\times10^{-7}~\mathrm{NA}^{-2}$$)
_1st_February_Evening_Shift_en_14_1.png)
$$4\mu\mathrm{T}$$
$$3\mu\mathrm{T}$$
$$6\mu\mathrm{T}$$
$$1\mu\mathrm{T}$$
Explanation
_1st_February_Evening_Shift_en_14_2.png)
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.
$$ \begin{aligned} \text { So, } B & =\left(\frac{\mu_0 i}{4 r}\right)=\frac{\left(4 \pi \times 10^{-7}\right) \times 3}{4 \times\left(\frac{\pi}{10}\right)} \\\\ & =3 \times 10^{-6} \mathrm{~T} \\\\ & =3 \mu \mathrm{T} \end{aligned} $$
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