CBSE Class 12 Physics 2023: question 32(a)

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Step-by-step solution

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1. (i) Biot–Savart law: the magnetic field $d\vec{B}$ at a point due to a current element $Id\vec{l}$ is proportional to I, to the length dl and to $\sin \theta$ (θ between $d\vec{l}$ and $\vec{r}$), and inversely proportional to the square of the distance r: $d\vec{B}=\frac{{\mu }_0}{4\pi }\frac{Id\vec{l}\times \vec{r}}{r^3}$, $dB=\frac{{\mu }_0}{4\pi }\frac{Idl\sin \theta }{r^2}$.

   1 mark

2. For a circular loop of radius a, every element $d\vec{l}$ is perpendicular to the radius joining it to the centre, so $\theta =90^{\circ }$, $\sin \theta =1$ and $r=a$: $dB=\frac{{\mu }_0}{4\pi }\frac{Idl}{a^2}$, and all $d\vec{B}$ point in the same direction (perpendicular to the plane of the loop).

   ½ mark

3. $B=\int dB=\frac{{\mu }_{0}I}{4\pi a^2}{\int }_0^{2\pi a}dl=\frac{{\mu }_{0}I}{4\pi a^2}(2\pi a)$

   1 mark

4. $B=\frac{{\mu }_{0}I}{2a}$, directed along the axis of the loop (given by the right-hand thumb rule).

   ½ mark

5. Field lines (described): closed loops encircling the wire on each side; near the centre they become nearly straight and parallel to the axis, passing through the loop. For anticlockwise current (seen from the front) the field at centre points out of the loop towards the viewer; the loop behaves like a magnetic dipole with one face N and other S.

   ½ mark

6. (ii) Current due to revolving electron: $i=\frac{e}{T}=\frac{ev}{2\pi r}$

   ½ mark

7. $i=\frac{1.6\times 10^{-19}\times 10^7}{2\pi \times 10^{-10}}=\frac{0.8\times 10^{-2}}{\pi }$ A

   ½ mark

8. $i\approx 2.55\times 10^{-3}$ A $=2.55$ mA

   ½ mark

Answer: (i) $B=\frac{{\mu }_{0}I}{2a}$ along the axis; (ii) $i\approx 2.55$ mA

Revise the chapter

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