Two Questions (MCQ) on Magnetic Effect of Electric Current

You will find many questions (with solution) involving magnetic fields on this blog, which can be accessed by trying a search for ‘magnetic field’ using the search box at the top of this page (or, you may click on the label ‘magnetic field’ below this post.

Today we will discuss two multiple choice questions on magnetic field produced by current carrying conductors:

(1) The adjoining figure shows a plane wire loop made of two semicircular portions of radii R₁ and R₂ and two straight portions. The wire loop contains a battery which drives a current I through the loop. What is the magnitude of the magnetic flux density at the common centre ‘O’ of the semicircular portions of this wire loop?

(a) (μ₀I/2π)(1/R₁ + 1/R₂)

(b) (μ₀I/4)(1/R₁ + 1/R₂)

(c) (μ₀I/4)(1/R₂ – 1/R₁)

(d) (μ₀I/4π)(1/R₂ – 1/R₁)

(e) Zero

The magnetic field produced at O by the straight portions of the loop is zero. The semicircular portion of smaller radius R₂ produces a magnetic field μ₀I/4R₂ which is directed normally into the plane of the figure (away from the reader).

[Note that a single turn plane circular coil of radius R produces a magnetic field μ₀I/2R at its centre].

The semicircular portion of larger radius R₁ produces a smaller magnetic field μ₀I/4R₁ which is directed normally outwards (towards the reader).

The resultant magnetic field at the centre O has magnitude (μ₀I/4R₂ – μ₀I/4R₁) = (μ₀I/4)(1/R₂ – 1/R₁), as given in option (c).

[The resultant field is directed normally into the plane of the figure (away from the reader). The above question can be modified to check your understanding of this fact also].

(2) A current I enters a circular coil of radius R, branches into two parts and then recombines as shown in the circuit diagram.

The resultant magnetic field at the centre of the coil is

(a) zero

(b) μ₀I/2R

(c) (¾)(μ₀I/2R)

(d) (¼ )(μ₀I/2R)

(e) (½)(μ₀I/2R)

This question appeared in KEAM 2009 (Engineering) question paper.

The magnetic field produced at the centre by the straight current leads is zero. So it is sufficient to consider the fields produced by the circular portions.

The currents through the branches are inversely proportional to the lengths of the branches while the magnetic fields for a given current are directly proportional to the lengths of the branches. The magnitudes of the fields due to the two branches are therefore equal. Since the fields due to the branches are perpendicular to the plane of the loop and oppositely directed, the net field at the centre is zero. This argument is enough for answering the above question.

If you want to make your argument more rigorous, you will proceed as follows:

Since the longer branch is made of three quarters of the circle and the shorter branch is made of one quarter of the circle, the resistance of the longer branch is 3 times that of the shorter branch. The currents through the shorter and longer branches are therefore given respectively by

I₁ = 3I/4 and

I₂ = I/4

The magnetic field produced at the centre by the shorter branch is directed normally into the plane of the coil (away from the reader) and has magnitude B₁ given by

B₁ = (¼)(μ₀I₁/2R) = 3μ₀I/32R, on substituting for I₁.

The magnetic field produced at the centre by the longer branch is directed normally outwards (towards the reader) and has magnitude B₂ given by

B₂ = (¾)(μ₀I₂/2R) = 3μ₀I/32R, on substituting for I₂.

The fields B₁ and B₂ get canceled and the resultant field at the centre is zero [Option (a)].

[Note that in all situations of branching of current at a circular coil, if the current leads are straight and point to the centre of the coil, the magnetic field at the centre will be zero (the angle shown need not necessarily be 90º)].

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