Answer:
Current in the wire is given as
[tex]i = 1.17 \times 10^{-7} A[/tex]
Explanation:
magnetic field due to long current carrying wire is given as
[tex]B = \frac{\mu_0 i}{2\pi r}[/tex]
so we have magnetic force on moving charge is given as
[tex]F = qvB[/tex]
so we have
[tex]F = (1\times 10^{-6})(1.5 \times 10^6)(\frac{\mu_0 i}{2\pi (0.35)})[/tex]
so we have
[tex]1\times 10^{-6} = 1.5 \times \frac{2 i}{0.35}[/tex]
[tex]i = 1.17 \times 10^{-7} A[/tex]
