CBSE Class 12 Physics 2025: question 23(a)

Hints

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Step-by-step solution

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1. (i) Biot–Savart law: $d\vec{B}=\frac{{\mu }_0}{4\pi }\frac{I(d\vec{l}\times \vec{r})}{r^3}$. Here $d\vec{l}$ is the current element and $\vec{r}$ is the position vector of the field point from that element.

   1 mark

2. (ii) Coil A lies in the XY plane, so its field at the centre is along the z-axis: $B_1=\frac{{\mu }_{0}I}{2R}$. Coil B lies in the YZ plane, so its field is along the x-axis: $B_2=\frac{{\mu }_0\sqrt{3}I}{2R}$.

   ½ mark

3. The two fields are perpendicular, so $B=\sqrt{B_1^2+B_2^2}=\frac{{\mu }_{0}I}{2R}\sqrt{1+3}$.

   ½ mark

4. Net field: $B=\frac{{\mu }_{0}I}{R}$.

   ½ mark

5. Direction: $\tan \theta =\frac{B_1}{B_2}=\frac{1}{\sqrt{3}}\Rightarrow \theta =30^{\circ }$. The net field lies in the XZ plane at $30^{\circ }$ to ${\vec{B}}_2$ (the x-axis), i.e. $60^{\circ }$ to ${\vec{B}}_1$ (the z-axis).

   ½ mark

Answer: B = μ₀I/R, in the XZ plane at 30° to the x-axis (field of coil B), i.e. 60° to the z-axis.

Revise the chapter

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