IIT JEE 2012 Questions on Magnetic Field due to Infinitely Long Current Carrying Cylinders

“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 two questions pertaining to the magnetic field produced by infinitely long current carrying cylinders. The first question is the usual single correct answer type multiple choice question. The second one is an integer answer type question, the answer to which is a single digit integer ranging from 0 to 9. Here are the questions with their solution.   

(1) An infinitely long conducting cylinder with inner radius R/2 and outer radius R carries a uniform current density along its length. The magnitude of the magnetic field B‌ as a function of the radial distance r from the axis is best represented by

‌‌Case (i): r < R/2

Using Ampere’s circuital law we find that the magnetic fiel at points at distance r < R/2 is zero.

[We have ∫B.dℓ = μ₀I.

Since the current passing through the surface enclosed by the path of integration is zero, the value of B is zero at distance r < R/2]

‌‌Case (ii)

For distance r lying between R/2 and R (or, R/2 ≤ r < R) we have

            ∫B.dℓ = μ₀[πr²J – π(R/2)²J] where J is the current density

‌[πr²J – π(R/2)²J is the current passing through the surface enclosed by the path of integration in this case]

Or, B×2πr = μ₀[πr²J – π(R/2)²J]

This gives B = (μ₀J/2)[r – (R²/4r)]

Case (iii)

For r ≥ R we have

            B.dℓ = μ₀[πR²J – π(R/2)²J]

Or, B×2πr = μ₀[πR²J – π(R/2)²J]

This gives B = (μ₀J/2r)[R² – (R²/4r)] = 3μ₀JR²/8r

The graph shown in option (D) indicates the above three cases correctly.

(2) A cylindrical cavity of diameter ‘a’ exists inside a cylinder of diameter 2a as shown in the figure. Both the cylinder and the cavity are infinitely long. A uniform current density J flows along the length. If the magnitude of the magnetic field at the point P is given by N μ₀ aJ/12, then the value of N is

            If the cylinder has no cavity, the magnetic flux density B₁ at the point P is given by

            ∫B₁.dℓ = μ₀πa²J

Therefore, B₁×2πa = μ₀πa²J

Or, B₁ = μ₀aJ/2

Since there exists a cavity, the current is reduced by π(a/2)²J and the magnetic field is reduced by B₂.

The field B₂ is given by

            ∫B₂.dℓ = μ₀π(a/2)²J

Therefore, B₂×2π(3a/2) = μ₀π(a/2)²J

[The circular path of integration in this case has radius a+(a/2) = 3a/2]

Or, B₂ = μ₀aJ/12

The magnitude B of the magnetic field at the point P due to the actual conductor with the cavity is given by

            B = B₁ – B₂ =  μ₀aJ/2– μ₀aJ/12

Or, B = 5μ₀aJ/12

In the question the above field is given as Nμ₀aJ/12

            Therefore N = 5.

AIPMT Main Exam – Two Questions Involving Discharge of a Capacitor Through an Inductor

"Live as if you were to die tomorrow. Learn as if you were to live forever."

– Mahatma Gandhi

Questions appearing in AIPMT Main (Physics) question paper are generally not as simple as those appearing in AIPMT Preliminary question paper. Those who get qualified (by passing the preliminary examination) should bear this in mind and prepare accordingly. I give below two questions involving the discharge of a charged capacitor through an inductor:

(1) A condenser of capacity C is charged to a potential difference of V₁. The plates of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to V₂ is

(1) [C(V₁ – V₂)²/L]^1/2

(2) C(V₁² – V₂²)/L

(3) C(V₁² + V₂²)/L

(4) [C(V₁² – V₂²)/L]^1/2

The above question appeared in AIPMT Main 2010 question paper.

            The initial energy of the charged capacitor is ½ CV₁². When the potential difference across the capacitor reduces to V₂  the energy is ½ CV₂². Therefore, the energy lost by the capacitor is ½ C(V₁² – V₂²). This amount of energy is gained by the inductor.

If I represents the discharge current at the instant when the potential difference across the capacitor reduces to V₂ we have

             ½ C(V₁² – V₂²) = ½ LI²

[Remember that the energy of an inductor carrying a current I is ½ LI² and the energy is store in the magnetic field]

From the above equation we obtain

             I = [C(V₁² – V₂²)/L]^1/2

(2) Capacitance of 6 μF is charged by 6 V battery. Now it is connected with inductor of 5 mH. Find the current in the inductor when 1/3rd of total energy is magnetic.

This question was asked as half the part of a free response (and not a multiple choice) question in the AIPMT Main 2007 question paper.

[Current practice is to ask objective type (multiple choice) questions].

            The initial energy E₁ of the charged capacitor is given by

            E₁ = ½ CV² = ½ ×(6×10^–6)×6²

Since one-third of total energy is mentioned as magnetic in the question, we understand that the energy of the inductor carrying the discharge current I is E₁/3.

Therefore we have

            E₁/3 = ½ LI²  

Or, (1/3) ×[½ ×(6×10^–6)×6²] = ½ ×(5×10^–3)I²  

This gives I = 0.12 A

KEAM (Engineering) 2010 Questions (MCQ) on Electronics

"You may never know what results come of your actions, but if you do nothing, there will be no results."

–Mahatma Gandhi

Questions on electronics will be generally interesting to most of you. Today we will discuss questions in this section which appeared in Kerala engineering entrance (KEAM - Engineering) 2010 question paper. Here are the questions with their solution:

 

(1) A full wave rectifier with an a.c. input is shown:

The output  voltage across R_L is represented as

  

The rectified output voltage will be a direct voltage but there will be very large amount of ripples. The capacitor C acts as a filter to remove the ripples; but there will still be a small amount of  ripples in the output. Therefore the correct option is (e).

(2) In the given circuit the current through the battery is

(a) 0.5 A

(b) 1 A

(c) 1.5 A

(d) 2 A

(e) 2.5 A

Since the diode D₁ is reverse biased, no current will flow through the D₁ branch. Diodes D₂ and D₃ are forward biased and hence the battery drives currents through the 20 Ω resistor and the series combination of the two 5 Ω resistors.

The current driven through the 20 Ω resistor is 10 V/20 Ω = 0.5 A.

The current driven through the 10 Ω resistor is 10 V/10 Ω = 1 A.

Therefore, total current through the battery is 0.5 A + 1 A = 1.5 A

(3) The collector supply voltage is 6 V and the voltage drop across a resistor of 600 Ω in the collector circuit is 0.6 V, in a transistor connected in common emitter mode. If the current gain is 20, the base current is

(a) 0.25 mA

(b) 0.05 mA

(c) 0.12 mA

(d) 0.02 mA

(e) 0.07 mA

We have I_CR_C = 0.6 V where I_C is the collector current and R_C is the resistance in the collector circuit.

Therefore,  I_C×600 Ω = 0.6 V from which I_C = 0.6/600 A = 10^–3 A = 1 mA.

Since the current gain β is given by

            β = I_C/I_B where I_B is the base current, we have

            I_B = I_C/β = 1 mA/20 = 0.05 mA.

(4) A pure semiconductor has equal electron and hole concentration of 10¹⁶ m^–3. Doping by indium increases n_h to 5×10²² m^–3. Then the value of n_e in the doped semiconductor is

(a) 10⁶ m^–3

(b) 10²² m^–3

(c) 2×10⁶ m^–3

(d) 10¹⁹ m^–3

(e) 2×10⁹ m^–3

According to the law of mass action we have

            n_i² = n_en_h where n_i is the electron concentration as well as the hole concentration in the intrinsic (pure) semiconductor, n_e is the electron concentration in the doped semiconductor and n_h is the hole concentration in the doped semiconductor.

Therefore n_e = n_i²/n_h = (10¹⁶)²/(5×10²²) = 2×10⁹ m^–3

Multiple Choice Questions on Dimensions of Physical Quantities [Including KEAM (Engineering) 2011 question]

I believe in standardizing automobiles, not human beings

– Albert Einstein

In most of the entrance examination question papers you will find at least one question on dimensions of physical quantities. Here are a few typical multiple choice questions in this section:

(1) Which one among the following quantities is a dimensional constant?

(a) Dielectric constant of water

(b) Speed of light in free space

(c) Viscosity of water

(d) Ratio of specific heats of a diatomic gas

(e) Reynolds number

Options (a) and (c) are not constants since the dielectric constant and viscosity depend on other parameters. Options (d) and (e) are dimensionless numbers.

[Reynolds number (used in assessing the turbulence of fluids) is the ratio of inertial force to force of viscosity]

The correct option is the speed of light in free space which is a fundamental constant with dimensions LT^–1.

(2) If F denotes force and t time, then in the equation F = at^–1 + bt², the dimensions of a and b respectively are

(a) LT^–4 and LT^–1

(b) LT^–1 and LT^–4

(c) MLT^–4 and MLT^–1

(d) MLT^–1 and MLT^–4

(e) MLT^–3 and MLT^–2

The above question appeared in Kerala Engineering Entrance ((KEAM) 2011 question paper.

The dimensions of at^–1 and bt² have to be that of force which is MLT^–2. Therefore the dimensions of a must be MLT^–1 and the dimensions of b must be MLT^–4.

The correct option is (d).

The following question appeared in Karnataka CET 2004 question paper:

(3) The physical quantity having the same dimensions as Planck’s constant h is

(1) Boltzmann constant

(2) force

(3) linear momentum

(4) angular momentum

Most of you will remember that the unit of h is joule second. Therefore h has the dimensions of the product of work and time which is ML²T^–2×T = ML²T^–1.

Angular momentum is the moment of linear momentum and has dimensions L×MLT^–1 = ML²T^–1.

The correct option is (4).

The following question appeared in EAMCET 2008 Engineering Entrance Exam question paper. You will answer it in no time if you remember that the dimensions of Planck’s constant are those of angular momentum.

(4) The energy (E), angular momentum (L) and universal gravitational constant (G) are chosen as fundamental quantities. The dimensions of universal gravitational constant in the dimensional formula for Planck’s constant (h) is

(a) zero

(b) – 1

(c) 5/3

(d) 1

Since the dimensions of Planck’s constant are those of angular momentum, it follows that the dimensions of universal gravitational constant in the dimensional formula for Planck’s constant (h) is zero.

[In terms of E, L and G which are assumed as fundamental quantities in the above question, Planck’s constant (h) has zero dimension in E, one dimension in L and zero dimension in G].

Electrostatics - Multiple Choice Questions on Electric Field and Potential

Today we will discuss a few simple, but interesting, multiple choice questions on electrostatics. Here are the questions:

(1) ABC is an equilateral triangle. When positive charges Q and 2Q are placed at points A and B, the electric potential at the mid point (O) of AB is found to be 180 V. What is the electric potential at the vertex C of the triangle under this condition?

(a) 60 V

(b) 90 V

(c) 120 V

(d) 180 V

(e) 240 V

The distance of the vertex C of the triangle from the charges is twice the distance from the point O. Since the electric potential due to a point charge is inversely proportional to the distance, the potential at the vertex C must be half the value at C. So the answer is 90 V.

[Mathematically, you will write (with usual notations)

(1/4πε₀)(3Q/r) = 180 and

(1/4πε₀)(3Q/2r) = x

This gives x = 90 V]

(2) Equal and opposite point charges + Q and – Q are fixed at the corners A and B of an equilateral triangle (Fig.). If a negative point charge (– q) is placed at the vertex C of the triangle, the net force on this charge is directed

(a) towards right

(b) towards left

(c) towards A

(d) towards B

(e) towards the mid point of side AB

The force on the charge – q due to the charge + Q is attractive and is directed along CA. The force due to the charge – Q is repulsive and is directed along BC. Since these forces have the same magnitude, their resultant acts parallel to BA. The net force on the charge – q is thus directed leftwards [Option (b)].

(3) A cube of side 4 cm has a constant electric potential of 2 volt on its surface. If there are no charges inside the cube, the potential at a distance of 1 cm from the centre of the cube is

(a) zero

(b) 0.25 V

(c) 0.5 V

(d) 1 V

(e) 2 V

Since the surface of the cube is an equipotential surface and there are no charges inside the cube, the electric field inside the cube must be zero. This means that the electric potential everywhere inside the cube is 2 V itself [Option (e)].

(4) A spherical conductor of radius R has a central cavity (Fig.) with a negative point charge (–q) located at the centre of the cavity (without touching the surface of the cavity). The electric fiel at a point P distant r from the centre of the sphere is

(a) q/4πε₀r² directed towards the sphere

(b) q/4πε₀r² directed away from the sphere

(c) q/4πε₀r² directed upwards

(d) q/4πε₀r² directed downwards

(e) zero

The negative charge at the centre of the cavity will induce positive charge q on the surface of the cavity and negative charge –q on the outer surface of the sphere. The situation is similar to that of a spherical conductor carrying a charge –q so that the magnitude E of the electric field at the point P is given by

E = q/4πε₀r²

Since the field is due to negative charge, the direction of the field is towards the centre of the sphere.

IIT-JEE Multiple Correct Answer Type Questions on Thermodynamics

"Hate the sin, love the sinner."
–Mahatma Gandhi

Today we will discuss two multiple correct answer type questions on thermodynamics. These were included in the IIT-JEE 2009 question paper:

(1) The figure shows the P–V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then,

(A) the process during the path A → B is isothermal

(B) heat flows out of the gas during the path B → C → D

(C) work done during the path A → B → C is zero

(D) positive work is done by the gas in the cycle ABCDA.

Statement (A) is incorrect since a curve representing an isothermal in a PV diagram must be a hyperbola and not an arc of a circle.

We have PV = nRT with usual notations. The value of V is smaller at D compared to that at B where as the value of P is the same. The product PV = nRT is therefore smaller at D. This means that the temperature at D is less than that at B and hence statement (B) is correct.

During the path AB the gas expands, doing work. During the path BC the gas is compressed and work is done on the gas. But these works are unequal since the area under AB is greater than the area under BC. Therefore, statement (c) is incorrect.

The cycle of operations ABCDA is clockwise and the cyclic curve encloses an area. Therefore, positive work is done by the gas and statement (D) is correct.

So the correct options are (B) and (D).

(2) C_V and C_P denote the molar specific heat capacities of a gas at constant volume and constant pressure respectively. Then

(A) C_P – C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas

(B) C_P + C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas

(C) C_P/C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas

(D) C_P.C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas.

In the case of ideal gases C_P – C_V = R, the universal gas constant. Therefore statement (A) is wrong.

C_P and C_V are larger for a diatomic ideal gas compared to a mono-atonic ideal gas. Therefore, C_P + C_V is larger for a diatomic ideal gas. Statement (B) is correct.

[For diatomic ideal gas C_P = 7R/2 and C_V = 5R/2. For mono-atomic ideal gas C_P = 5R/2 and C_V = 3R/2].

Obviously statement (C) is wrong.

Statement (D) is correct since both C_P and C_V are larger for a diatomic ideal gas than for a mono-atomic ideal gas.

Therefore, options (B) and (D) are correct.

You will find a useful post on thermodynamics here.