Answer:
the new current on the wire is 3.64 A.
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;
[tex]F = BIL(sin \theta)\\\\[/tex]
F ∝ I
[tex]\frac{F_1}{I_1} = \frac{F_2}{I_2} \\\\I_2 = \frac{F_2I_1}{F_1} \\\\I_2 = \frac{0.063\ \times\ 1.5 }{0.026} \\\\I_2 = 3.64 \ A[/tex]
Therefore, the new current on the wire is 3.64 A.
