CBSE Class 12 Physics 2025: question 23

Hints

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Step-by-step solution

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1. Take a loop of radius R carrying current I, with P on its axis at distance x from the centre. By Biot–Savart law, an element $d\vec{l}$ gives $d\vec{B}=\frac{{\mu }_{0}I}{4\pi }\frac{d\vec{l}\times \vec{r}}{r^3}$, where $r^2=x^2+R^2$.

   ½ mark

2. $d\vec{l}\perp \vec{r}$ for every element, so $|d\vec{l}\times \vec{r}|=rdl$ and $dB=\frac{{\mu }_{0}Idl}{4\pi (x^2+R^2)}$.

   ½ mark

3. Each $d\vec{B}$ has a component along the axis and one perpendicular to it. The perpendicular components from diametrically opposite elements cancel, so only the axial ($x$) components add.

   ½ mark

4. $d{B}_x=dB\cos \theta$ with $\cos \theta =\frac{R}{\sqrt{x^2+R^2}}$, so $d{B}_x=\frac{{\mu }_{0}IRdl}{4\pi (x^2+R^2{)}^{3/2}}$.

   ½ mark

5. Summing over the loop, $\oint dl=2\pi R$, so $\vec{B}=\frac{{\mu }_{0}I{R}^2}{2(x^2+R^2{)}^{3/2}}\hat{i}$, directed along the axis.

   ½ mark

6. At the centre, $x=0$: $\vec{B}=\frac{{\mu }_{0}I}{2R}\hat{i}$.

   ½ mark

Answer: B on axis = μ₀IR²/[2(x²+R²)^{3/2}] along the axis; at the centre B = μ₀I/(2R)

Revise the chapter

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