# Torque on a current carrying coil placed in magnetic field .

In this topic we will find the expression for the torque on a current carrying coil placed in magnetic field. But before to derive the expression for ‘Torque on a current carrying coil placed in magnetic field’ , students must know the expression for the force experienced on the current carrying coil placed in uniform magnetic field. To get the notes on this topic click here-

Torque on a current carrying coil placed in a magnetic field-

Suppose a rectangular coil PQRS carrying current I is placed in a uniform magnetic field B as shown in figure ( a) .

Let,  PQ = RS = l and QP=SP=b , suppose I is the current flowing through the coil.

Let, ϴ is the angle between the plane of the coil with the magnetic field.

As we know force on a current carrying coil is given as F = BIL sinϴ ,

Then we can find force on PQ is given as , F1 = BIl sinϴ [ but ϴ=900]

Therefore,  F1 = BIl sin900  out side the plane ( using Fleming’s left hand rule)

Similarly, we can find force on SR is given as , F3 = BIl sinϴ [ but ϴ=900]

Therefore,  F3 = BIl sin900   Inside the plane ( using Fleming’s left hand rule).

Again,  Force on QR is given as , F2 = BIb sinϴ

Therefore,  F2 = BIb sinϴ  directed downward ( using Fleming’s left hand rule),

And again, Then we can find force on SP is given as , F4 = BIb sinϴ

Therefore,  F4 = BIb sinϴ  directed upward ( using Fleming’s left hand rule)

It is clear that F2 and F4 will cancel out each other,

But due to pair of force F1 and F3 these force form a couple , and cause of turning effect,

So, torque Ʈ = either force x perpendicular distance

= F1 bcosϴ = BIlb cos(90-α)=BIl sinα ;

But here lx b = area ‘A’, ; So we can write , Ʈ= BIA sinϴ,

If there are N number of turns in the coil then we can write net torque , Ʈ= BI NA sinϴ.