Torque On A Current Loop Assignment Help | Torque On A Current Loop Homework Help

Torque on a Current Loop in a uniform Magnetic Field

Consider a rectangular loop, height l and with b, carrying a current i. The axis of the loop is perpendicular to the uniform magnetic field B. The normal to the plane of the loop makes an angle θ with the direction to magnetic induction B.

The forces acting on bottom and top sides 1 and 3 and F1 and F3. They are equal, opposite and act along the same time.
        |F1| = - |F3| = |ib x B|
        = I b B sin (90 – θ) = ibB cos θ

By the vector product rule, this force is upward on the top side (3) and the downward on the bottom side (1). The net force on the rigid loop from these elements is thus zero, as is the torque.

Forces acting on the vertical sides 2 and 4 are F2 and F4.
They are equal and opposite but do not act along the same line.
The common magnitude of F2 and F4 is ilb.

These two forces produce a torque.
τ = ilB b sin θ – iAB sin θ
where A = lb = area of the loop.
If there are N turns on the lop.
τ = NiAB sin θ.
We define the magnetic dipole moment of a plane current loop as

 | μ | = μ = Area x Current = NiA
The direction of μ is given by the right handed s screw rule.
. :    τ = μ x B = μ B sin θ

The moving coil galvanometer is based on this concept.

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