Question

An electron is moving with a velocity of 3 x 10^8 m/s perpendicular to magnetic field of 0.5T, then the force experienced by the electron

Answer 1

Magnetic Force on a moving charge is equal to

F = q(v×B)

where v × B is cross product of velocity and magnetic field.

so v×B = vBsinθ

now Acc. to ques. θ = 90°

hence vBsin90° => 3×10^(8) × 0.5 × 1 => 1.5×10^(8)

now as we know charge on electron is q = 1.6 × 10^ (-19) coulomb

So Mag. Force will be => 1.6×10^(-19) × 1.5×10^(8)

= 2.4×10^(-11)