Derive the expression for torque on a
current carrying loop
Answers
Consider a rectangular coil PQRS suspended in a uniform magnetic field B→, with its axis perpendicular to the field.
Figure (a) A rectangular loop PQRS in a uniform magnetic field B→.
(b) Top view of the loop, magnetic dipole moment m→ is shown.
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.
According to Fleming's left hand rule, the magnetic forces on sides PS and QR are equal, opposite and collinear (along the axis of the loop), so their resultant is zero.
The side PQ experiences a normal inward force equal to IbB while the side RS experiences an equal normal outward force. These two forces form a couple which exerts a torque given by
τ= Force × perpendicular distance
=IbB×asinθ
=IBAsinθ
If the rectangular loop has N turns, the torque increases N times i.e.,
τ= NIBAsinθ
But NIA=m, the magnetic moment of the loop, so
τ=mBsinθ
In vector notation, the torque τ→ is given by
τ→=m→×B→
The direction of the torque τ is such that it rotates the loop clockwise about the axis of suspension.
Answer:
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