TORQUE EXPERIENCED BY A CURRENT LOOP IN A UNIFORM MAGNETIC FIELD

Torque on a current loop in a uniform magnetic field:

A rectangular coil PQRS suspended in a uniform magnetic field B. The axis of the rectangular coil is perpendicular to the field. 

Let Ｉ = current flowing through the coil PQRS

     a, b = sides of the coil PQRS

       A  = ab = area of the coil

       θ = angle between the direction of B and normal to the plane of the coil.

 

Direction of the area and B makes an angle θ

all the forces acting on the sides of the rectangular coil PQRS

According to Fleming's left hand rule, 

i) the magnetic force on the side QR is F₁

 and it is acting upward.

F₁ = Ｉ(a xB) = ＩaB

ii) the magnetic force on the side SP is F'₁ and it is acting downward. 

F'₁ = Ｉ(a xB) = ＩaB

So, net force along vertical direction is zero as 

F₁ and F'₁ are equal and opposite as both are acting along the axis of the coil.

iii) the magnetic force on the side SR is F and it is coming out of the board.

F = Ｉ(b xB) = ＩbB

iv) the magnetic force on the side QP is F' and it is going into the board.

F' = Ｉ(b xB) = ＩbB

coil as seen from the top : m is the direction of the magnetic moment as well as coil area (perpendicular to the plane of the coil).

Therefore F and F' will produce a torque τ

We know, 

τ = one of the force x perpendicular distance between them 

τ = F a sin θ = ＩbBa sin θ = ＩBA sin θ

[ ∵ ab = A]

If the rectangular loop has N turns, the torque increases N times ie.,

τ = NＩBA sin θ

But there is one physical quantity called "magnetic moment" or m = NＩA

τ = m B sin θ = m x B

The direction of the torque τ is such that it rotates the loop clockwise about the axis of the loop. 

The torque will be zero when θ = 0 ie., When the plane of the loop is perpendicular to the magnetic field. 

The torque will be maximum when θ = π/2 and τ = NＩBA ie., when the plane of the loop is parallel to the magnetic field.