Question

A long straight wire AB carries a current of 5 A. P is a proton travelling with a velocity of 2 × 10⁶ m/s, parallel to the wire, 0.2 m from it and in a direction opposite to the current, as shown in Figure below. Calculate the force which magnetic field of the current carrying conductor AB exerts on the proton.

Answer 1

The magnetic field due to the current-carrying wire is perpendicular to the plane of the paper, in a downward direction.

i.e., `vec B = -(mu_0 I)/(2 pi d)  vec k`

Force `vec F = q  vec v xx vec B`

= `e(- v vec j) xx (-(mu_0 I)/(2 pi d) vec k)`

= `(mu_0 ev I)/(2 pi d) vec i`

Given that d = 0.2 m, ν = 2 × 10⁶ m/s, I = 5 A

∴ F = `(mu_0 e xx 2 xx 10^6 xx 5 xx 10^5)/(2 pi xx 0.2) vec i`

= `(2 xx 10^-7 xx 1.6 xx 10^-19 xx 2 xx 10^6 xx 5 xx 10^5)/0.2`

= 16 × 10−19 N