CBSE Class 12 Physics sample paper 2023-24: question 28(b)

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. Field of conductor 1 (current $I_1$) at conductor 2, distance a away: $B_1=\frac{{\mu }_0{I}_1}{2\pi a}$

   ½ mark

2. Force on length $l$ of conductor 2: $F_2=I_{2}l{B}_1\sin 90^{\circ }=\frac{{\mu }_0{I}_1{I}_{2}l}{2\pi a}$, so $\frac{F_2}{l}=\frac{{\mu }_0{I}_1{I}_2}{2\pi a}$, directed towards conductor 1.

   ½ mark

3. Similarly, the field of conductor 2 at conductor 1: $B_2=\frac{{\mu }_0{I}_2}{2\pi a}$

   ½ mark

4. Force per unit length on conductor 1: $\frac{F_1}{l}=\frac{{\mu }_0{I}_1{I}_2}{2\pi a}$, directed towards conductor 2. Equal and opposite forces; for currents in the same direction the force is attractive.

   ½ mark

5. Putting $I_1=I_2=1$ A, $a=1$ m: $\frac{F}{l}=\frac{4\pi \times 10^{-7}\times 1\times 1}{2\pi \times 1}=2\times 10^{-7}$ N/m.
   One ampere is that steady current which, flowing in each of two infinitely long, straight, parallel conductors placed 1 m apart in vacuum, makes each conductor experience a force of $2\times 10^{-7}$ N per metre of its length.

   1 mark

Answer: $F/l={\mu }_0{I}_1{I}_2/(2\pi a)$, attractive; 1 A gives $2\times 10^{-7}$ N/m at 1 m separation

Revise the chapter

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