A straight current carrying copper wire with 28 Ampere current is held normal to the magnetic field such that copper wire is balanced against gravity acting on it. Find the value of magnetic field if mass per unit length of wire is (46.6 gram/meter).

$1.6 × 10^{-2} T$

The correct answer is Option (2) → $1.6 × 10^{-2} T$

The magnetic force $(F_{mag})$ on a straight current carrying -

$F_{mag}=I×L×B$

The gravitational force $(F_{grav})$ is -

$⇒I.L.B=m.g$

$⇒B=\frac{mg}{IL}=\frac{40.6×10^{-3}×9.8}{28}$

$≃0.0162T$