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.

“Be always at war with your vices, at peace with your neighbors, and let each New Year find you a better man.”

– Benjamin Franklin

Happy New Year…

Multiple Choice Questions on Dimensions of Physical Quantities

The world is a dangerous place, not because of those who 

do evil, but because of those who look on and do nothing. 

–Albert Einstein 

Today we will discuss a few questions on units and dimensions:

(1) Van der Waals equation of state for real gases is [P + (a/V²)](V – b) = RT where P is the pressure, V is the volume of one mole of gas, R is the universal gas constant, T is the temperature and a and b are constants. The dimensional formula of a is

(a) [M⁰L⁶T⁰]

(b) [ML^–1T^–2]

(c) [ML⁵T^–2]

(d) [ML^–2T^–2]

(e) [ML³T^–2]

This is a question popular among question setters and has appeared many times in various entrance test papers. Since a/V² appears as added to P, the dimensions of a/V² must be the same as those of P. Therefore, ‘a’ has the dimensions of PV².

Note that pressure P is force per unit ares and hence the dimensional formula for P is [ML^–1T^–2]. The dimensional formula for V² is [L⁶]. Therefore, the dimensional formula for ‘a’ is [ML⁵T^–2].

(2) The dimensional formula for Stefan’s constant is

(a) [MT^–3K^–4]

(b) [ML²T^–2K^–4]

(c) [ML²T^–2]

(d) [MT^–2L⁰]

(e) [MT⁴L⁰]

This question appeared in Kerala Engineering Entrance (KEAM) 2009 question paper. You will find similar questions in other entrance test papers also.

We have E = σT⁴ where E is the energy radiated per second from unit area, σ is Stefan’s constant and T is the temperature of the black body.

Therefore, σ = E/T⁴

Note that energy has dimensional formula [ML²T^–2]. Sine E is the energy radiated per second from unit area, the dimensional formula for E is [MT^–3].

Therefore, the dimensional formula for Stefan’s constant σ is [MT^–3K^–4].

(3) Pick out the correct dimensions in length of the quantity μ₀ε₀ from the following:

(a) 2

(b) 1

(c) – 1

(d) – 2

(e) zero

You know that the speed ‘c’ of electromagnetic waves in free space is given by

c = 1/√(μ₀ε₀)

Therefore μ₀ε₀ = 1/c² and the dimensional formula for μ₀ε₀ is [L^–2T²]. The quantity μ₀ε₀ therefore has – 2 dimensions in length [Option (d)].

(4) Which of the following is dimensionless?

(a) Stress

(b) Gas constant

(c) Frequency of sound

(d) Efficiency of heat engine

(e) Thermal conductivity

The correct option is (d) since the efficiency is the ratio of the output power to the input power.

Applications invited for Kerala Entrance Examinations for Admission to Medical/ Agriculture/ Veterinary/ Engineering/ Architecture Degree Courses 2010 (KEAM 2010)

The Commissioner for Entrance Examinations, Govt. of Kerala, has invited applications for the Entrance Examinations for admission to the following Degree Courses in various Professional Colleges in Kerala for 2010-2011.

(a) Medical (i) MBBS (ii) BDS (iii) BHMS (iv) BAMS (v) BSMS

(b) Agriculture (i) BSc. Hons. (Agriculture) (ii) BFSc. (Fisheries) (iii) BSc. Hons. (Forestry)

(c) Veterinary BVSc. & AH

(d) Engineering B.Tech. [including B.Tech. (Agricultural Engg.)/B.Tech. (Dairy Sc. & Tech.) courses under the Kerala Agricultural University]

(e) Architecture B.Arch.

[Candidates aspiring for B.Arch. course should also submit their application to the CEE. Even though there is no state level Entrance Examination for this purpose, these candidates should write the National Aptitude Test for Architecture and should forward the NATA score and mark list of the qualifying examination to the CEE on or before 05.06.2010].

Dates of Exam:

Engineering Entrance Examination (For Engineering courses except Architecture)

19.04.2010 Monday 10.00 A.M. to 12.30 P.M. Paper-I : Physics & Chemistry.

20.04.2010 Tuesday 10.00 A.M. to 12.30 P.M. Paper-II: Mathematics.

Medical Entrance Examination (For Medical, Agriculture and Veterinary Courses)

21.04.2010 Wednesday 10.00 A.M. to 12.30 P.M. Paper-I : Chemistry & Physics.

22-04.2010 Thursday 10.00 A.M. to 12.30 P.M. Paper-II: Biology.

Sale of Application Form through selected post offices: From  07-12-2009 to 06-01-2010

To see the list of selected post offices, visit http://www.cee-kerala.org/
Last Date for Receipt of Application by CEE: 06-01-2010 (Wednesday), before 5 P.M.

Facility for online submission of application in the case of non-reservation category and Non-Keralite category is available.

Find details at http://www.cee-kerala.org/

In addition to the multiple choice questions (of the type expected to appear in KEAM 2010) on this site, you will find many similar useful questions with solution at the site  http://physicsplus.blogspot.com.  If you want to see earlier KEAM questions only, type in ‘Kerala’ in the search box at the top left of this page and click on the search button or press the enter key.

EAMCET 2009 (Medical) Questions (MCQ) on Rotational Motion

The following two questions were included from rotational motion in the EAMCET 2009 (Medical) question paper. (The first question has appeared in many entrance exam question papers. It was included in the EAMCET 2009 Engineering question paper also). Here are the questions with solution:

(1) A rod of length ‘l’ is held vertically stationary with its lower end located at a position P on the horizontal plane. When the rod is released to topple about P, the velocity of the upper end of the rod with which it hits the ground is

(1) √(g/l)

(2) √(3gl)

(3) 3√(g/l)

(4) √(3g/l)

When the rod falls its gravitational potential energy mgl/2 gets converted into rotational kinetic energy of the rod. (Note that ‘m’ is the mass of the rod and initially the centre of gravity of the rod is at a height l/2 with respect to the horizontal plane).

Therefore we can write

             ½ Iω² = mgl/2 where I is the moment of inertia of the rod about an axis passing through the end (at P) of the rod and perpendicular to the length of the rod and ‘ω’ is the angular velocity of the rod when it hits the horizontal plane.

Here I = ml²/3.

[Usually you will remember the moment of inertia of a rod about a normal axis through its middle as ml²/12. The moment of inertia about a normal axis through one end is obtained by applying the parallel axis theorem: I = ml²/12 + m(l/2)² = ml²/3].

Substituting for I we have

             ½ (ml²/3)ω² = mgl/2

Since ω = v/l where ‘v’ is the velocity with which the rod hits the ground, we have

             ½ (ml²/3)(v/l)² = mgl/2

This gives v = √(3gl)

(2) A rigid uniform rod of mass M and length ‘L’ is resting on a smooth horizontal table. Two marbles each of mass ‘m’ and traveling with uniform speed ‘v’ collide with the two ends of the rod simultaneously and inelastically as shown. The marbles get stuck to the rod after the collision and continue to move with the rod. If m = M/6 and v = L ms^–1, then the time taken by the rod to rotate through π/2 is

(1) 1 sec

(2) 2π sec

(3) π sec

(4) π/2 sec

Because of the collision, the rod will rotate about a normal axis through its middle with an angular velocity ω given by

             Iω = mvL/2 + mvL/2 where ‘I’ is the moment of inertia of the rod carrying the masses m and m at its ends.

[Note that we have equated the final angular momentum of the system (containing the rod and the masses) to the initial angular momentum. Before the collision the two masses have angular momentum about the central axis. These are shown on the right hand side of the above equation].

Since v = L the above equation gets modified as

             Iω = mL²

After the collision, the rod and the masses move together and the total angular momentum is given by

             Iω = [(ML²/12) + 2m(L/2)²] ω

[The first term within the square bracket above is the moment of inertia of the rod and the second term is the moment of inertia of the two masses].

From the above two equations, we have

             mL² = [(ML²/12) + mL²/2 ] ω 

Since m = M/6 the above equation becomes

             M/6 = [(M/12) + (M/12)] ω = (M/6) ω

Therefore ω = 1 radian /sec and the time taken by the rod to rotate through π/2 radian is π/2 sec.

You will find many questions on rotational motion on this site. You can access all of them by clicking on the label ‘rotation’

You will find many useful questions with solution in this section at physicsplus and at AP Physics Resources.