Question

A long straight conductor kept along X′X axis, carries a steady current I along + x direction. At an instant t, a particle of mass m and charge q at point (x, y) moves with a velocity \(\vec v\)along + y direction. Find the magnitude and direction of the force on the particle due to the conductor.

Answer 1

Magnetic field due to the current carrying long wire is \(\text {B} = \frac{\mu _0 I}{2 \pi r}\)

Where r is perpendicular distance from wire

Now from Lorentz force

\(\vec F_m = q (\vec v \times \vec B ) = q \left[v \cdot \hat j \times \frac{\mu _0 I}{2 \pi y} (\hat k)\right]\)

\(\vec {F}_ m = \frac{\mu _0 I (qv)}{2 \pi y} \hat i\)