CBSE 2025 · Region 5 · Set 3 · Q18 · 2 marks
A current of $\displaystyle 5$ A is passing along +X direction through a wire lying along X -axis. Find the magnetic field $\displaystyle \overrightarrow{\mathrm{B}}$ at a point $\displaystyle \overrightarrow{\mathrm{r}}=(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}) \mathrm{m}$ due to $\displaystyle 1$ cm element of the wire, centered at origin.
Marking-scheme solution
Using Biot-Savart's law:
$\displaystyle \vec{B}=\frac{\mu_{0}}{4 \pi} \frac{I d \vec{l} \times \vec{r}}{r^{3}}$
$\displaystyle =\frac{\left(10^{-7}\right)\left[5\left(10^{-2}\right) \hat{i} \times(3 \hat{i}+4 \hat{j})\right]}{5^{3}}$
$\displaystyle =1.6 \times 10^{-10} \hat{k} \mathrm{~T}$
Moving Charges and MagnetismMagnetic Field due to a Current Element, Biot-Savart LawApplyvery_short_answermedium
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CBSE Class 12 Physics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.