CBSE Class 12 Physics 2024: question 24(b)

Hints

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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. (i) Ampere's circuital law: The line integral of the magnetic field around any closed loop in vacuum equals ${\mu }_0$ times the net current threading the loop:
   $$\oint \vec{B}\cdot d\vec{l}={\mu }_{0}I$$

   The integral is taken along the closed loop (Amperian loop) that bounds the surface. $I$ is the net current passing through that surface, with its sign given by the right-hand rule.

   1 mark

2. (ii) Field of a long straight wire: $B=\frac{{\mu }_{0}I}{2\pi r}$. The midpoint is at $r=0.1$ m from each wire.

   $B_1=\frac{4\pi \times 10^{-7}\times 5}{2\pi \times 0.1}=1\times 10^{-5}$ T (due to the 5 A wire)

   ½ mark

3. $B_2=\frac{4\pi \times 10^{-7}\times 10}{2\pi \times 0.1}=2\times 10^{-5}$ T (due to the 10 A wire)

   ½ mark

4. For currents in the same direction, the two fields at the midpoint are opposite to each other. So
   $B=B_2-B_1=1\times 10^{-5}$ T

   ½ mark

5. The net field is $10^{-5}$ T. It is perpendicular to the plane of the wires, in the direction of the field due to the 10 A wire (given by the right-hand rule for that wire).

   ½ mark

Answer: (ii) B = 1 × 10⁻⁵ T, perpendicular to the plane of the wires, along the field of the 10 A wire

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