A long wire carries a current of 10 A. A particle of charge q = 2.0 µC travels at a velocity of 5 x 10⁵ ms^{- 1} at a perpendicular distance 20 cm from the wire in a direction opposite to the direction of the current in the wire. Find the magnitude and direction of the force experienced by the particle.

Given I = 10 A,

r = 20 cm= 0.2 m, v = 5 x 10⁵ ms^-1 and q = 2.0 µC = 2.0 x 10^-6 C.

The magnetic field due to current I at the site of the charged particle is

$$\begin{gathered} \mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi r}}=\frac{\left(4\mathrm{\pi }\times {10}^{-7}\right)\times 10}{2\mathrm{\pi }\times 0.2} \\ =1.0\times {10}^{-5} \mathrm{T} \end{gathered}$$

According to Right Hand Thumb rule, the direction of the field is into the page. Hence the charged particle is moving perpendicular to the field, i.e. 0 = 90°. The force experienced by the particle is

$$\begin{gathered} \mathrm{F}=\mathrm{q}.\mathrm{vBsin} \mathrm{\theta } \\ =\left(2.0\times {10}^{-6}\right)\times \left(5\times {10}^5\right)\times \left(1.0\times {10}^{-5}\right)\times \sin {90}^{\circ } \\ =1.0\times {10}^{-5} \mathrm{N} \end{gathered}$$

According to Fleming's Left Hand rule, the direction of the force is lo the right, i.e. perpendicular to the wire and away from it.