If we did all things we are capable of, we would literally astound ourselves.

– Thomas A. Edison

Monday, October 04, 2010

EAMCET (Medical) 2010 Questions on Magnetic Fields Due to Current Carrying Conductors

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 r1 and r2 respectively as shown in the figure. If the current flowing in the loop is i, the magnetic induction at the point O is

(1) (μ0i/4)(1/r1 1/r2)

(2) (μ0i/4)(1/r1 + 1/r2)

(3) (μ0i/2)(1/r1 1/r2)

(4) (μ0i/2)(1/r1 + 1/r2)

The straight portions PQ and RS do not produce any magnetic field at O. The semicircular portions of radii r1 and r2 produce magnetic fields μ0i/4r1 and μ0i/4r2 respectively at the centre O.

[Remember that the magnetic flux density (magnetic induction) at the centre of a circular loop of radius r is μ0i/2r and hence the magnetic flux density due to a semicircular loop is μ0i/4r].

The field due to the portion of smaller radius r1 has greater magnitude μ0i/4r1 and is directed perpendicular to the plane of the loop, towards the reader. The field due to the portion of larger radius r2 has smaller magnitude μ0i/4r2 and is directed perpendicular to the plane of the loop, away from the reader.

The resultant flux density at O is therefore μ0i/4r1 μ0i/4r2, which is equal to (μ0i/4)(1/r1 1/r2), 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 Am2

(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

2×2πr = 6.28

Therefore, r = 6.28/4π = ½ and area A = πr2 = π/4

Substituting, m = 2×0.5 × π/4 = π/4.

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