# Force acting on the current carrying conductor placed in magnetic field

In this topic we will discuss about Force acting on the current carrying conductor placed in magnetic field, where magnetic field is uniform and current carrying coil is laced at some angle with the magnetic field.

Before to know about this topic students must know these topics –

(i)force on a moving charged particle in uniform magnetic field and . To get the notes on this topic click here-

(ii) Motion of a charged particle in uniform magnetic field . To get the notes on this topic click here-

Force acting on the current carrying conductor placed in magnetic field

Suppose a cylinder of length ‘l’ area of cross-section ‘A’ carrying current ‘I’ , placed in uniform magnetic field ‘B’.

Angle between them is ‘ϴ’ as shown in figure. If it is considered that ‘n’ is the number density of electron,Then total number of electrons present in the conductor is given as,

N = A l n. [ where volume V = Al ]

As we know when a charge moves in a magnetic force , force experienced on it is given by

f =  q v B sinϴ  [ ϴ is the angle between v and B]

so for each electron f=e vd B sinϴ ( where vd is the drift velocity of the electron) ‘,

So, net force on the electrons F = Nf = N e vd B sinϴ =  Aln e vd B sinϴ = n e A vd l B sinϴ

But we know from I = neAvd  ,

So, we can write  F = I l B sinϴ……………eq.

Special case-

(i)  if ϴ=00 or 1800 then ,

Force F = 0 . i.e. a linear conductor carrying current placed parallel or anti parallel to the magnetic field experience no force.

(ii) If ϴ= 900 then, sinϴ=sin90 = 1

Then force will be maximum given by F = BIl .

The direction of the force is given by the Fleming’s left hand rule.