Karnataka CET 2010 Questions (MCQ) on Magnetic Field due to Current Carrying Conductors

Questions on magnetic field due to current carrying conductors are generally interesting. Four questions were included from this section in the Karnataka CET 2010 question paper. Here are those questions with solution:

(1) Two thick wires and two thin wires, all of same material and same length, form a square in three different ways P, Q and R as shown in the figure:

With correct connections shown, the magnetic field due to the current flow, at the centre of the loop will be zero in the case of ……

(a) Q and R only

(b) P only

(c) P and Q only

(d) P and R only

In all cases the current in the upper branch produces a magnetic field acting normally into the plane of the figure where as the current in the lower branch produces a magnetic field acting normally outwards (towards the reader). The two fields are thus in opposition. But the currents in the two branches are equal in the case of P and R since the resistances are equal. Therefore, the magnetic field due to the current flow, at the centre of the loop will be zero in the case of P and R only [Option (d)].

(2) Magnetic field at the centre of a circular coil of radius R due to a current I flowing through it is B. The magnetic field at a point along the axis at distance R from the centre is……

(a) B/2

(b) B/4

(c) B/√8

(d) √8 B

The magnitude of the magnetic field (B_axis) on the axis of a plane circular coil at a distance x from the centre of the coil is given by

B_axis = (μ₀nR²I) /2(R²+x²)^3/2 where R is the radius of the coil and I is the current in the coil.

Therefore, the magnetic field (B₁) at a point along the axis at distance R from the centre is given by

B₁ = (μ₀nR²I) /2(R²+R²)^3/2 = (μ₀nR²I) /(2×2√2 ×R³)

Or, B₁ = (μ₀nI) /(2R×2√2)

But the magnetic field at the centre of the coil is given by

B = (μ₀nI) /2R

Therefore, we have B₁ = B/2√2 = B/√8 as given in option (c).

(3) A current I is flowing through a loop. The direction of the current and the shape of the loop are as shown in the figure. The magnetic field at the centre of the loop is μ₀I/R times….. (MA = R, MB = 2R, ÐDMA = 90º)

(a) 5/16, but out of the plane of the paper

(b) 5/16, but into the plane of the paper

(c) 7/16, but out of the plane of the paper

(d) 7/16, but into the plane of the paper

The magnetic field at the centre of a single turn plane circular coil carrying current I is given by

B = (μ₀I) /2r where r is the radius

The loop given in the question consists of three-fourths (DIA) of a circular loop of radius R, one fourth (BIC) of a circular loop of radius 2R and two straight wires AB and CD. The straight wires do not produce any magnetic field at the common centre M of the circular arcs. The circular arcs DIA and BIC produce magnetic fields acting normally into the plane of the loop so that they add up to produce the resultant field at M.

Therefore, magnetic field at M = [(¾)× μ₀I /2R] + [(¼ )× μ₀I /2(2R)]

Or, magnetic field at M = (μ₀I/R) [(3/8) + (1/16)] = (μ₀I/R)(7/16)

The correct option is (d).

(4) PQ and RS are long parallel conductors separated by certain distance. M is the mid point between them (see the figure). The net magnetic field at M is B. Now, the current 2A is switched off. The field at M now becomes……

(a) 2 B

(b) B

(c) B/2

(d) 3 B

If the magnitude of the magnetic field produced by the conductor RS (which carries current 1 A) is B, the magnitude of the magnetic field produced by the conductor PQ (which carries current 2 A) must be 2B. But these fields are oppositely directed. That’s why the resultant field at M has magnitude B when both conductors carry currents. When the current 2A is switched off, the field at M becomes B.

[If one of the options were – B, you would pick it out as the correct one].

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.

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)].

Now, find an interesting question in this section here.

Magnetic Fields Produced by Current Carrying Conductors

Let us consider the following MCQ which appeared in AIEEE 2004 question paper:
The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 μT. What will be its value at the centro of the loop?
(a) 250 μT (b) 150 μT (c) 125 μT (d) 75 μT
The magnetic flux density(B) at a point on the axis at distance ‘x’ from the centre for a single turn loop is given by
B = μ₀ r²I / [2(r²+ x²)^3/2] where μ₀ is the permeabiloty of free space, ‘r’ is the radius of the loop and ‘I’ is the current.
The magnetic field(B') at the centre of the loop is given by
B' = μ₀ I/2r
Dividing, B/B' = r³/(r²+ x²)^3/2 from which B' = B[(r²+ x²)^3/2] /r³.
You may substitute distances in cm itself since their units will get cancelled in the ratio. On substituting B in μT itself, you will get the answer in μT.
Therefore, B' = 54×[(9+16)^3/2]/27 = 54×125/27 = 250 μT.
The following multiple choice question appeared in Kerala Medical Entrance 2003 test paper:

A wire of certain length carries a steady current. It is first bent to form a circular coil of one turn. The same wire is next bent to form a circular coil of three turns. The ratio of magnetic induction at the centre of the coil in the two cases is
(a) 9:1 (b) 1:9 (c) 1:3 (d) 3:1 (e) 1:1

Since the magnetic induction (flux density) at the centre of a circular coil is given by
B = μ₀ nI/(2r) where μ₀ is the permeability of free space, ‘n’ is the number of turns in the coil, I is the current and ‘r’ is the radius of the coil, we have, B α n/r for a given current.

The ratio of magnetic inductions at the centre is therefore B/B' = (n/n') ×(r'/r) = (1/3) ×(1/3) = 1/9, since 2πr = 3×2πr' so that r'/r = 1/3. The correct option therefore is (b).
Here is a simple question (but be careful):

An infinitely long aluminium pipe of radius ‘r’ carries a current. The magnetic flux density out side the pipe at a distance 3r/2 from the axis is 0.04 T. The magnetic flux density inside the pipe at a distance r/2 from the axis will be
(a) 0.01 T (b) 0.02 T (c) 0.04 T (d) zero (e) infinite
Don’t be distracted by the value of the field out side. You should remember that the magnetic flux density at any point inside a long current carrying pipe is zero. The correct option therefore is (d).

Questions involving Magnetic Fields

The following M.C.Q. which appeared in the All India Pre –Medical/Dental Entrance-2005 (C.B.S.E) test paper will be interesting to you:
An electron moves in a circular orbit with a uniform speed ‘v’. It produces a magnetic field ‘B’ at the centre of the circle. The radius of the circle is proportional to
(a) √(B/v) (b) B/v (c) √(v/B) (d) v/B
The electron, moving along the circular path is equivalent to a current I = e/T = ev/(2πr) where ‘e’ is the electronic charge, T is the period, ‘v’ is the speed and ‘r’ is the radius of the circular motion. The magnetic field at the centre of the circle, B = μ₀I/2r = μ₀ev/4πr² so that r α √(v/B). The correct option is (c).
Now, consider the following simple question. (Be careful however, so that you won’t pick out a wrong answer):
A long straight conductor P carries a current ‘I’ while another parallel long straight conductor Q carries a current 2I in the same direction. The magnetic field midway between them is B. If the current 2I in the conductor Q is switched off, the magnetic field midway between the conductors will be
(a) B/3 (b) B (c) –B (d) –B/2 (e) 2B
The magnetic fields produced by the conductors are in opposition in the region between them. The field produced by Q is twice the field produced by P. If the field produced by Q is 2B, the field produced by P is –B. This is indeed the case, since the net field is B when both currents are switched on. When the current 2I in Q is switched off, the field due to Q alone is present. The answer therefore is –B [Option (c)].
The following question also is simple. It is designed to check your understanding of certain basic things:
A straight conductor of length ‘L’ carrying a current ‘I’ is bent in the form of a semicircle. The magnetic flux density (in tesla) at the centre of the semicircle is approximately
(a) 10^-6I/L (b) 10-7I/L (c) 10^-6π²I/L (d) 10⁷π²I/L (e) π²I/L
The magnetic field at the centre of a single turn circular current carrying coil is μ₀I/2r and hence the magnetic field at the centre of a semicircle is μ₀I/4r. The radius of the semicircle obtained by bending the straight conductor of length L is, r = L/π. On substituting this value, the magnetic field at the centre of the semicircle is μ₀πI/ 4L. Since μ₀ = 4π ×10^-7, the field is 10^-7π² I/L. But π² is approximately equal to 10 so that the answer is 10^-6I/L [Option (a)].
Let us now consider the following M.C.Q. which appeared in the Karnataka Common Entrance test paper of 2005:
The electrons in the beam of a television tube move horizontally from south to north. The vertical component of the earth’s magnetic field points down. The electron is deflected towards
(a) west (b) no deflection (c) east (d) north to south.
It will be convenient to use Fleming’s left hand rule here. You have to hold the fore-finger, middle finger and thumb of your left hand along mutually perpendicular directions so that the fore-finger points along the magnetic field and the middle finger along the direction of the conventional current (and hence opposite to the direction of the electron). The thumb will then give you the direction of deflection, which you can easily obtain as along the east [Option (c)]. You will find more multiple choice questions (with solution) at physicsplus: Magnetic Force on Moving Charges