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Physics · 2025 · 3 marks
Moving Charges and MagnetismMagnetic Field due to a Current Element, Biot-Savart Law3 marksApplyshort answer
CBSE 2025 · Region 4 · Set 1 · Q23
(a) Write vector form of Biot-Savart law. (b) Two insulated long straight wires, each carrying 2 · 0 A current are kept along xx^′ and yy^′ axis as…
(a)
Write vector form of Biot-Savart law.
(b)
Two insulated long straight wires, each carrying $\displaystyle 2 \cdot 0$ A current are kept along $\displaystyle \mathrm{xx}^{\prime}$ and $\displaystyle \mathrm{yy}^{\prime}$ axis as shown in the figure. Find the magnitude and direction of resultant magnetic field at point $\displaystyle \mathrm{P}(4 \mathrm{~m}, 5 \mathrm{~m})$.
Marking-scheme solution
(a)
$\displaystyle d \vec{B}=\frac{\mu_{0}}{4 \pi} \frac{I(d \vec{l} \times \vec{r})}{r^{3}}$
(b)
As $\displaystyle B=\frac{\mu_{0}}{4 \pi} \frac{2 I}{a}$
$\displaystyle \vec{B}_{1}=$ Magnetic field due to current carrying wire along $\displaystyle XX^{\prime}$
$\displaystyle \vec{B}_{1}=\frac{\mu_{0}}{4 \pi} \frac{2 I}{a}=\frac{2 \times 2 \times 10^{-7}}{5}=\left(8 \times 10^{-8} \mathrm{~T}\right)(\hat{k})$
$\displaystyle \vec{B}_{2}=$ Magnetic field due to current carrying wire along $\displaystyle YY^{\prime}$
$\displaystyle \vec{B}_{2}=\frac{\mu_{0}}{4 \pi} \frac{2 I}{a}=\frac{2 \times 2 \times 10^{-7}}{4}=10 \times 10^{-8} \mathrm{~T}(-\hat{k})$
$\displaystyle \vec{B}_{\text {net }}=\vec{B}_{2}+\vec{B}_{1}$
$\displaystyle =2 \times 10^{-8} \mathrm{~T}(-\hat{k})$
Direction of resultant magnetic field along $\displaystyle -Z$ axis.
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CBSE Class 12 Physics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.