CBSE Class 12 Physics sample paper 2023-24: question 26

Hints

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Step-by-step solution

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1. Region r > a: Take a circular Amperian loop (loop 2) of radius r, concentric with the wire. By symmetry B is tangential and of the same magnitude everywhere on it, length $L=2\pi r$. Whole current I is enclosed.
   Ampere's circuital law: $B(2\pi r)={\mu }_{0}I\Rightarrow B=\frac{{\mu }_{0}I}{2\pi r}$, so $B\propto \frac{1}{r}$.

   1½ marks

2. Region r < a: Take loop 1 of radius r inside the wire. Current is uniform, so the enclosed current is $I_e=I\frac{\pi r^2}{\pi a^2}=\frac{I{r}^2}{a^2}$.
   $B(2\pi r)={\mu }_0\frac{I{r}^2}{a^2}\Rightarrow B=\frac{{\mu }_{0}I}{2\pi a^2}r$, so $B\propto r$.

   1½ marks

Answer: $r>a$: $B={\mu }_{0}I/(2\pi r)$; $r<a$: $B={\mu }_{0}Ir/(2\pi a^2)$

Revise the chapter

• Chapter: Moving Charges and Magnetism (notes, formulas, where marks are lost and every past question)
• Topic: Motion of a Charge in a Magnetic Field