Today we will discuss two questions on magnetic fields due to current carrying wires. These questions appeared in EAMCET (Agriculture-Medicine) 2010 question paper:

(1) A wire loop PQRS is constructed by joining two semicircular coils of radii *r*_{1} and *r*_{2} respectively as shown in the figure. If the current flowing in the loop is *i*, the magnetic induction at the point O is

(1) (μ_{0}*i*/4)(1/*r*_{1 }– 1/*r*_{2})

(2) (μ_{0}*i*/4)(1/*r*_{1 }+ 1/*r*_{2})

(3) (μ_{0}*i*/2)(1/*r*_{1 }– 1/*r*_{2})

(4) (μ_{0}*i*/2)(1/*r*_{1 }+ 1/*r*_{2})

The straight portions PQ and RS do not produce any magnetic field at O. The *semicircular* portions of radii *r*_{1} and *r*_{2 }produce magnetic fields μ_{0}*i*/4*r*_{1} and μ_{0}*i*/4*r*_{2} respectively at the centre O.

[Remember that the magnetic flux density (magnetic induction) at the centre of a circular loop of radius *r* is μ_{0}*i*/2*r* and hence the magnetic flux density due to a *semicircular *loop is μ_{0}*i*/4*r*].

The field due to the portion of smaller radius *r*_{1} has greater magnitude μ_{0}*i*/4*r*_{1} and is directed perpendicular to the plane of the loop, *towards the reader*. The field due to the portion of larger radius *r*_{2} has smaller magnitude μ_{0}*i*/4*r*_{2} and is directed perpendicular to the plane of the loop, *away from the reader*.

The resultant flux density at O is therefore μ_{0}*i*/4*r*_{1}– μ_{0}*i*/4*r*_{2}, which is equal to (μ_{0}*i*/4)(1/*r*_{1 }– 1/*r*_{2}), as given in option (1).

[The resultant field at O is directed normal to the plane of the loop, *towards the reader*. This point too can be incorporated in the question to make it a little more difficult].

(2) A wire of length 6.28 m is bent into a circular coil of 2 turns. If a current of 0.5 A exists in the coil, the magnetic moment of the coil is, in Am^{2}

(1) π/4

(2) ¼

(3) π

(4) 4π

The magnetic moment *m *of a plane circular coil of area *A* with *n *turns carrying current *i* is given by

*m = niA *

The radius *r *of the coil is given by

2×2π*r* = 6.28

Therefore, *r = *6.28/4π = ½ and area *A =* π*r*^{2} = π/4* *

Substituting, *m =* 2×0.5 × π/4 = π/4.

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