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www.GOALias.blogspot.comPhysicsIntegrating from θ = 0 to θ = π/2,EXAMPLE 4.12 EXAMPLE 4.11Iω f∫ð /2ωdω = mB sinθdθ∫0 02ωf/2I =−mBcosθπ0= m B2ωf⎛2mB⎞= ⎜⎝⎟I ⎠1/2⎛2×20⎞= ⎜ 1⎝ 10 − ⎟⎠1/2= 20 s –1 .Example 4.12(a) A current-carrying circular loop lies on a smooth horizontal plane.Can a uniform magnetic field be set up in such a manner thatthe loop turns around itself (i.e., turns about the vertical axis).(b) A current-carrying circular loop is located in a uniform externalmagnetic field. If the loop is free to turn, what is its orientationof stable equilibrium? Show that in this orientation, the flux ofthe total field (external field + field produced by the loop) ismaximum.(c) A loop of irregular shape carrying current is located in an externalmagnetic field. If the wire is flexible, why does it change to acircular shape?Solution(a) No, because that would require τ to be in the vertical direction.But τ = I A × B, and since A of the horizontal loop is in the verticaldirection, τ would be in the plane of the loop for any B.(b) Orientation of stable equilibrium is one where the area vector Aof the loop is in the direction of external magnetic field. In thisorientation, the magnetic field produced by the loop is in the samedirection as external field, both normal to the plane of the loop,thus giving rise to maximum flux of the total field.(c) It assumes circular shape with its plane normal to the field tomaximize flux, since for a given perimeter, a circle encloses greaterarea than any other shape.1604.10.2 Circular current loop as a magnetic dipoleIn this section, we shall consider the elementary magnetic element: thecurrent loop. We shall show that the magnetic field (at large distances)due to current in a circular current loop is very similar in behavior to theelectric field of an electric dipole. In Section 4.6, we have evaluated themagnetic field on the axis of a circular loop, of a radius R, carrying asteady current I. The magnitude of this field is [(Eq. (4.15)],B =2µ IR022 2( x + R )3/2and its direction is along the axis and given by the right-hand thumbrule (Fig. 4.12). Here, x is the distance along the axis from the centre ofthe loop. For x >> R, we may drop the R 2 term in the denominator. Thus,