CBSE Class 12 Physics 2026: question 23

Hints

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Step-by-step solution

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1. Biot–Savart law: $d\vec{B}=\frac{{\mu }_0}{4\pi }\frac{Id\vec{l}\times \vec{r}}{r^3}$

   1 mark

2. Here $d\vec{l}=0.01\hat{i}$ m, $\vec{r}=(\hat{i}+\hat{j})$ m, $r=\sqrt{2}$ m, $r^3=2\sqrt{2}$ m³.
   $d\vec{l}\times \vec{r}=0.01(\hat{i}\times \hat{i}+\hat{i}\times \hat{j})=0.01\hat{k}$

   1 mark

3. $\vec{B}=10^{-7}\times \frac{10\times 0.01}{2\sqrt{2}}\hat{k}=\frac{10^{-8}}{2.83}\hat{k}\approx 3.5\times 10^{-9}\hat{k}$ T

   1 mark

Answer: B ≈ 3.5 × 10^-9 T along +z (k̂)

Revise the chapter

• Chapter: Moving Charges and Magnetism (notes, formulas, where marks are lost and every past question)