CBSE Class 12 Physics 2024: question 18

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. Let the conductor have length $L$, cross-section $A$ and $n$ mobile carriers per unit volume, each of charge $q$.

   Total number of mobile carriers $=nAL$.

   Force on one carrier moving with drift velocity ${\vec{v}}_d$ is $q({\vec{v}}_d\times \vec{B})$. So the total force is
   $\vec{F}=nALq({\vec{v}}_d\times \vec{B})$ ...(1)

   ½ mark

2. The current through the conductor is $I=nqA{v}_d$ ...(2)

   ½ mark

3. Define $\vec{L}$ as a vector of magnitude $L$ along the direction of the current. From (1) and (2), $nqA{v}_d\vec{L}=I\vec{L}$, so
   $$\vec{F}=I(\vec{L}\times \vec{B})$$

   ½ mark

4. Zig-zag conductor: Yes, the result still holds.

   Divide the conductor into small straight elements $d\vec{l}$ and add their forces vectorially: $\vec{F}=I\int d\vec{l}\times \vec{B}$.

   For a uniform $\vec{B}$ this becomes $I{\vec{L}}_{\text{eff}}\times \vec{B}$, where ${\vec{L}}_{\text{eff}}$ is the vector joining the two ends of the conductor.

   ½ mark

Answer: $\vec{F}=I(\vec{L}\times \vec{B})$; yes, it is valid for a zig-zag conductor (sum over small straight elements $d\vec{l}$)

Revise the chapter

• Chapter: Moving Charges and Magnetism (notes, formulas, where marks are lost and every past question)
• Topic: Torque on a Current Loop and Magnetic Dipole