Find the force acting on a current carrying conductor placed in an uniform magnetic field.

Explanation

Consider a straight conductor PQ of length l, area of cross-section A, carrying current I, placed in uniform magnetic field `vec B.` Let the conductor be placed along X-axis and let `theta` be the angle between axis of the conductor and magnetic field. Let l be the current that is following through the conductor from end P to Q. Therefore, electorns are flowing frm the end Q and P with velocity `vec V _(d).`
Let -e be the charge on each electron.
Force acting on one electron moving with velocity `vecV _(d)` in the magnetic field `vec B` is given by
`vec f = - e ( vec V _(d) xx vec B)`
Let n = number f electrons per unit volume.
Therefore, total number of electrons in the conductor = n Al .
Total force (F) acing on nAl electrons moving with velocity `vecV _(d)` in mangetic field
`vec B`is `vec F =- n Al e (vec V_(d) xx vec B)`
`implies F =- n Al e (V _(d) B sin theta)`
`implies F =- (n Ae V _(d)) (lB sin theta)`
But we know that, `I =- n A e V _(d)`
`implies F = I ( lB sin theta)`
vector form, `vecF = I ( vecl xx vecB)`
The direction of `vecF` is given by right hand screq rule of Fleming .s left hand rule. The direction of `vecF` is perpendicular to the plance containing `vecl and vecB.`