CBSE 2026 · Region 5 · Set 1 · Q23 · 3 marks
Write the expression for the magnetic field due to a current element in vector form. Consider a $\displaystyle 1$ cm segment of a wire, centered at the origin, carrying a current of $\displaystyle 10$ A in positive x-direction. Calculate the magnetic field $\displaystyle \overrightarrow{\mathrm{B}}$ at a point ( $\displaystyle 1 \mathrm{~m}, 1 \mathrm{~m}, 0$ ).
Marking-scheme solution
Magnetic field due to a current element ( $\displaystyle \mathbf{d} \boldsymbol{l}$ ) at a distance $\displaystyle \mathbf{r}$,
\[\begin{aligned}
& d \vec{\mathrm{B}}=\frac{\mu_{0}}{4 \pi} \frac{I(\overrightarrow{d l} \times \vec{\mathrm{r}})}{\mathrm{r}^{3}} \\
& \mathrm{~dB}=\frac{\mu_{0} i d l \sin \theta}{4 \pi \mathrm{r}^{2}} \\
& \quad \mathrm{r}=\sqrt{2} \mathrm{~m} \\
& \mathrm{~dB}=10^{-7} \times \frac{10 \times 10^{-2} \sin 45^{\circ}}{2} \\
& \quad \mathrm{~dB}=3.53 \times 10^{-9} \mathrm{~T}
\end{aligned}
\]
Direction : along +ve Z-axis
Moving Charges and MagnetismMagnetic Field due to a Current Element, Biot-Savart LawApplyshort_answermedium
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CBSE Class 12 Physics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.