A straight wire 30 cm long and mass 60 mg lies 30° east of north. Earth’s magnetic field is horizontal, due north, and of magnitude 0.8 G. The current needed to float the wire is 10N amperes. Find N. (g = 10 m s⁻².)
Physics · Mixed Chapter
JEE Mains

Question

A straight wire 30 cm long and mass 60 mg lies 30° east of north. Earth’s magnetic field is horizontal, due north, and of magnitude 0.8 G. The current needed to float the wire is 10N amperes. Find N. (g = 10 m s⁻².)

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Weight of the wire: m = 60 mg = 60×10⁻⁶ kg; g = 10 m s⁻². mg = 60×10⁻⁶×10 = 6×10⁻⁴ N. Step 2 — Magnetic force: The wire makes 30° with the horizontal terrestrial field. Magnitude of the force is Fᵦ = ILB sin30°. Here L = 0.30 m and B = 0.8 G = 8×10⁻⁵ T. Step 3 — Vertical equilibrium: ILB sin30° = mg. I = (6×10⁻⁴)/[0.30×8×10⁻⁵×(1/2)] = 50 A. Step 4 — Compare with answer form: I = 10N A ⇒ N = 5. Answer: 5.