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QPages 230-234: Q29: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q29 Solution: A current is passed through a silver strip of width and area of …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often insert the electron charge $-e$ directly into $i=n e A v_d$, which flips the sign of the drift velocity; instead use the magnitude $e$ for the formula and remember that electron drift is opposite to the conventional current, so the Hall electric field direction must be determined using the negative charge sign after computing $v_d$.

A current

i

is passed through a silver strip of width

d

and area of cross-section

A

. The number of free electrons per unit volume is

n

. a) Find the drift velocity

v

of the electrons. b) If a magnetic field

B

exists in the region, as shown in the figure, what is the average magnetic force on the free electrons? c) Due to the magnetic force, the free electrons get accumulated on one side of the conductor along its length. This produces a transverse electric field in the conductor, which opposes the magnetic force on the electrons. Find the magnitude of the electric field which will stop further accumulation of electrons. d) What will be the potential difference developed across the width of the conductor due to the electron-accumulation? The appearance of a transverse emf, when a current-carrying wire is placed in a magnetic field, is called Hall effect.

Step-by-Step Verified Solution

**Answer:** (a)

v = \frac{i}{n e A}

(b)

F = e v B

(c)

E = v B

(d)

V = v B d

. **Step-by-Step Explanation:** (a) Drift velocity

v = \frac{i}{n e A}

. (b) Magnetic force on each electron:

F = e v B

. (c) Transverse electric field at equilibrium:

E = v B

. (d) Hall voltage:

V = E d = v B d

.

All Questions in this Chapter

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