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QPages 230-234: Q16: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q16 Solution: Prove that the force acting on a current-carrying wire, joining …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often apply $F=iBL\sin\theta$ using the wire's total length, treating $L$ as a scalar magnitude instead of the end‑to‑end displacement vector $\vec{L}=\vec{r}_b-\vec{r}_a$. The correct approach is to evaluate the vector integral $\vec F= i\int d\vec l\times\vec B$, which for a uniform $\vec B$ reduces to $i(\vec r_b-\vec r_a)\times\vec B$, showing the force depends only on the endpoints, not on the wire's shape.

Prove that the force acting on a current-carrying wire, joining two fixed points a and b in a uniform magnetic field, is independent of the shape of the wire.

Step-by-Step Verified Solution

**Answer:** The force is independent of the shape of the wire. **Step-by-Step Explanation:** The force on a wire carrying current

i

in a uniform magnetic field

\vec{B}

is

\vec{F} = i \vec{L} \times \vec{B}

, where

\vec{L}

is the vector from point a to point b. This depends only on the endpoints, not on the shape of the wire.

All Questions in this Chapter

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