Answer:
the new current on the wire is 3.64 A.
Step-by-step explanation:
Given;
first force on the wire, F₁ = 0.026 N
second force on the wire, F₂ = 0.063 N
first current on the wire, I₁ = 1.5 A
second current on the wire, I₂ = ?
The force on a current carrying conductor placed in a magnetic field is given as;
![F = BIL(sin \theta)\\\\](https://img.qammunity.org/2022/formulas/physics/college/hmx797y5hcgn1fzc3j5bh3zgrlh4izf7kt.png)
F ∝ I
![(F_1)/(I_1) = (F_2)/(I_2) \\\\I_2 = (F_2I_1)/(F_1) \\\\I_2 = (0.063\ *\ 1.5 )/(0.026) \\\\I_2 = 3.64 \ A](https://img.qammunity.org/2022/formulas/physics/college/puwgp7f4a9v4n3jvy5sgn1zlj2zogc2020.png)
Therefore, the new current on the wire is 3.64 A.