A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is held in mid-air, its weight balanced by a uniform horizontal magnetic field. Take g = 9.8 m s⁻². The magnitude of the field is:

1. For mid-air suspension the magnetic force balances gravity: $mg = IlB$. 2. Rearranging: $B = \dfrac{mg}{Il}$. 3. Substitute $m = 0.2$ kg, $g = 9.8$, $I = 2$ A, $l = 1.5$ m: $B = \dfrac{0.2 \times 9.8}{2 \times 1.5}$. 4. $B = \dfrac{1.96}{3} = 0.65$ T. _Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.137_