A 2A current carrying straight metal wire of resistance 1 Ω, resistivity 2 × 10⁻⁶ Ω m, area of cross-section 10 mm² and mass 500 g is suspended horizontally in mid air by applying a uniform magnetic field B⃗. The magnitude of B is ________ × 10⁻¹ T (given, g=10 m / s²).

Solution

Physics sir · 3 steps

The magnetic push on the wire must hold up its weight, so set BIL = mg and find L from the resistance.

1. Length from resistance. Use R = ρ L / A with A = 10 mm² = 10 × 10⁻⁶ m² = 10⁻⁵ m². Then L = RA/ρ = (1 × 10⁻⁵)/(2 × 10⁻⁶) = 5 m.
2. Balance the forces. For the wire to float, the magnetic force must point straight up and equal the weight. Take B⃗ horizontal and perpendicular to the wire, so the force is BIL.

   BIL = mg ⇒ B = mg/IL = (0.5 × 10)/(2 × 5)

3. So. B = 5/10 = 0.5 T = 5 × 10⁻¹ T, so the blank is 5.

The common slip: 0.5. Reporting B in tesla gives 0.5, but the blank is in units of 10⁻¹ T, so the entry is 5.

The idea to keep. A wire hanging in mid air means the magnetic force BIL exactly cancels mg. If L is not given, look for R, ρ and A and use L = RA/ρ.

Answer from the NTA final answer key. Solution by PrepWiser.