A straight copper wire of 100 g and length 2.0 m carries a current of 1.5 A through it. The minimum magnetic field that needs to be applied to keep the wire suspended in mid-air is about

Correct option is D

Correct answer is D
Explanation:
Given:
Mass of the wire, m = 100 g = 100/1000 = 0.1 kg
Length of the wire, L = 2.0 m
Current through the wire, I = 1.5 A
Acceleration due to gravity, g = 9.8 m/s2
The gravitational force is:
Fg = mg ⇒ Fg = 0.1 × 9.8 = 0.98 N
The magnetic force is:
FB = I × L × B
For equilibrium: FB = Fg
I × L × B = mg ⇒ B = mg / (I × L)
Substitute values:
B = (0.1 × 9.8) / (1.5 × 2.0) ⇒ B = 0.98 / 3.0 ⇒ B ≈ 0.33 T