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QPages 230-234: Q28: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q28 Solution: A conducting wire of length , lying normal to a magnetic field…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often forget that the electron charge is negative, so the magnetic force is $\mathbf{F}= -e\,\mathbf{v}\times\mathbf{B}$, opposite to the $v\times B$ direction used for positive charges. Consequently the induced electric field inside the wire points opposite to the magnetic force on electrons, and the sign of the resulting potential difference must reflect the electron’s negative charge.

A conducting wire of length ll, lying normal to a magnetic field BB, moves with a velocity vv, as shown in the figure. a) Find the average magnetic force on a free electron of the wire. b) Due to this magnetic force, electrons concentrate at one end, resulting in an electric field inside the wire. The redistribution stops when the electric force on the free electrons balances the magnetic force. Find the electric field developed inside the wire when the redistribution stops. c) What potential difference is developed between the ends of the wire?

Step-by-Step Verified Solution

**Answer:** (a) F=evBF = e v B (b) E=vBE = v B (c) V=vBlV = v B l. **Step-by-Step Explanation:** (a) The magnetic force on an electron is F=evBF = e v B. (b) At equilibrium, the electric force balances the magnetic force: eE=evB⇒E=vBeE = e v B \Rightarrow E = v B. (c) The potential difference is V=El=vBlV = E l = v B l.

All Questions in this Chapter

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