A conducting strip carries current I = 4.8 A in a perpendicular magnetic field B = 1 T. Width = 1 mm and length = 1 cm, as shown. Free-electron density is n = 10²⁹ m⁻³. Find the Hall potential in the power-of-ten form specified in the original image and determine x. [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

A conducting strip carries current I = 4.8 A in a perpendicular magnetic field B = 1 T. Width = 1 mm and length = 1 cm, as shown. Free-electron density is n = 10²⁹ m⁻³. Find the Hall potential in the power-of-ten form specified in the original image and determine x. [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Equate magnetic and electric forces on a carrier: eEₕ = evB, so Eₕ = vB. Step 2 — Current satisfies I = neAv, and Hall voltage across the strip gives Vₕ = IB/(ne t), with thickness t = 1 cm (the dimension perpendicular to both B and Hall field in this drawing). Step 3 — Vₕ = 4.8/(10²⁹ × 1.6 × 10⁻¹⁹ × 0.01) = 3 × 10⁻⁸ V. Step 4 — Hence x = 3 in x × 10⁻⁸ V.