An infinitely long straight conductor carries a current of 5A as...

Magnetic Field due to a current carrying wire at $20cm$ in terms of current $I$ is given by,

$B = \dfrac{{\mu}_{0}I}{2\pid}$

$B = \dfrac{2 \times {10}^{- 7} \times 5}{20 \times {10}^{- 2}}$

$B = 0 . 5 \times {10}^{- 5}T$

Now,

Force on charge particle $= qvB$

${F}_{m} = eVB$

${F}_{m} = 1 . 6 \times {10}^{- 19} \times {10}^{5} \times 0 . 5 \times {10}^{- 5}$

$= 8 \times {10}^{- 20}N$