CBSE Class 12 Physics: Magnetic Force on a Current-Carrying Conductor

Step 1 (Governing Law & Definition): [0.5 Mark]

The magnetic force $\vec{F_m}$ on a single charge $q$ moving with velocity $\vec{v_d}$ (drift velocity) in a magnetic field $\vec{B}$ is given by the Lorentz force expression:

$$\vec{F_m} = q(\vec{v_d} \times \vec{B})$$

For a conductor of length $L$ and cross-sectional area $A$, the total number of free electrons is $N = nAL$, where $n$ is the number density of electrons.

Step 2 (Mathematical Derivation): [1.0 Mark]

The total force $\vec{F}$ on the conductor is the sum of forces on all free electrons:

$$\vec{F} = N \cdot q(\vec{v_d} \times \vec{B}) = (nAL) \cdot (-e)(\vec{v_d} \times \vec{B})$$

Since the current $I$ is related to drift velocity by $I = nAev_d$, we can write the current vector as $I\vec{L} = -nAev_d\vec{L}$. Substituting this into the force equation:

$$\vec{F} = (nAev_d) (\vec{L} \times \vec{B})$$ $$\vec{F} = I(\vec{L} \times \vec{B})$$

Step 3 (Final Conclusion & Justification): [0.5 Mark]

Final Expression: $\vec{F} = I(\vec{L} \times \vec{B})$

Justification for Zig-Zag Conductor: Yes, it is valid. A zig-zag conductor can be considered as a series of infinitesimal straight segments $d\vec{l}$. The total force is the vector sum of forces on these segments: $\vec{F} = \int I(d\vec{l} \times \vec{B}) = I(\int d\vec{l}) \times \vec{B}$. Since $\int d\vec{l}$ is the displacement vector between the two ends of the conductor, the force depends only on the net displacement, not the path taken.