KEAM (Medical) 2007 Questions on Electromagnetic Induction

The following questions which appeared in Kerala Medical Entrance 2007 question paper are simple, as is usually the case as far as the questions on electromagnetic induction are concerned:

(1) A wire of length 50 cm moves with a velocity of 300 m/minute perpendicular to a magnetic field. If the emf induced in the wire is 2V, the magnitude of the magnetic field in tesla is

(a) 2 (b) 5 (c) 0.4 (d) 2.5 (e) 0.8

The emf (V) induced in a wire of length ‘L’ metre moving with velocity ‘v’ metre/second perpendicular to a magnetic field ‘B’ tesla is given by

V = BLv from which B = V/Lv

Be careful to convert the given velocity in to meter/second.

Thus we have B = 2/(0.5×5) = 0.8 tesla

(2) Whenever a magnet is moved either towards or away from a conducting coil, an emf is induced, the magnitude of which is independent of

(a) the strength of the magnetic field

(b) the speed with which the magnet is moved

(c) the number of turns in the coil

(d) the resistance of the coil

(e) the area of cross section of the coil

The magnetic flux linked with a circuit (and the emf induced) is independent of the resistance of the circuit. So the correct option is (d)

However, remember that the induced charge flowing in a circuit depends on the resistance as induced charge = (change of flux)/resistance

You will find more multiple choice questions with solution at physicsplus

A Multiple Choice Question on Lenz’s Law

Two plane coils A and B are arranged coaxially as shown. Using an external battery, an increasing or decreasing clockwise current (as judged by viewing along the axis in the direction marked by the arrow) can be made to flow through coil A. The directions of the induced current through coil B when the current in coil A is (i) increased and (ii) decreased are

(a) clockwise in cases (i) and (ii)

(b) anticlockwise in cases (i) and (ii)

(c) anticlockwise in case (i) and clockwise in case (ii)

(d) clockwise in case (i) and anticlockwise in case (ii)

(e) unpredictable

This is a question based on Lenz’s law in electromagnetic induction, according to which the induced current has to oppose the magnetic flux change which produces the induced current.

When the clockwise current in coil A increases as in case (i), there is an increasing magnetic flux through coil B. The direction of the magnetic field lines (due to the current in coil A) through coil B is along the direction of view marked by the arrow. The induced current in coil B should therefore produce a magnetic field in the opposite direction. Evidently, this will be the case if the sense of induced current in coil B is anticlockwise.

When the current in coil A is decreasing as in case (ii), the flux through coil B is decresing. The direction of the magnetic field lines (due to the current in coil A) is same as in case (i). But the induced current in coil B has to oppose the decrease of the flux. This can be done by producing a magnetic field in the same direction as that produced by coil A. Evidently, the sense of the induced current through coil B must be clockwise for this. So, the correct option is (c).

[Instead of the above lengthy argument, you can have a relatively shorter argument like this:

Coil A has to move away from coil B when the flux linked with coil B increases because of the increasing current in coil A. The required repulsive force between the coils can be produced only if they carry unlike currents. So, the induced current in B in case (i) is anticlockwise. In case (ii) the decrement in the flux linked with coil B can be opposed if coil B moves towards coil A. The attractive force required for this can be produced if the coils carry like currents (currents in the same direction). So, the induced current in B is clockwise in case (ii)].

IIT-JEE 2007 Assertion-Reason Type MCQ on Kinetic Energy of Gas Molecules

Assertion-Reason Type Multiple Choice Questions are often interesting to answer, as is the case with the following one which appeared in IIT-JEE 2007 question paper:

STATEMENT-1

The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and volume

because

STATEMENT-2

The molecules of a gas collide with each other and the velocities of the molecules change due to the collision

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.

(C) Statement-1 is True, Statement-2 is False

(D) Statement-1 is False, Statement-2 is True

Statement-1 is true. To prove this you may use the expression for the translational kinetic energy (E) of a gas molecule, given by

E = (3/2) kT

where ‘k’ is Boltzman’s constant and ‘T’ is the absolute (Kelvin) temperature.

[Remember that the translational kinetic energy of any type of molecule is (3/2)kT since the energy per degree of freedom is ½ kT and in our three dimensional space, there can be three translational degrees of freedom only]

The total translational kinetic energy of all the molecules of the given mass of gas is (3/2) kT×nV where ‘V’ is the volume of the gas and ‘n’ is the number of molecules per unit volume.

But, the pressure (P) of the gas is given by

P = nkT .

Therefore, total translational kinetic energy = (3/2)PV = 1.5 PV.

Statement -2 is true. In fact this follows from the postulates of kinetic theory. But, the pressure exerted by a gas is not because of the collision between the gas molecules, but is because of the collision of gas molecules with the walls of the container. So, statement-2 is not a correct explanation for statement-1.

The correct option therefore is (B).

KEAM 2007(Medical) Question on Electric Dipole

The following MCQ which appeared in Kerala Medical Entrance 2007 question paper is straight forward, requiring your knowledge of the expression for torque on a dipole:

An electric dipole consists of two opposite charges each 0.05 μC separated by 30 mm. The dipole is placed in an uniform external electric field of 10⁶ NC^–1. The maximum torque exerted by the field on the dipole is

(a) 6×10^–3 Nm (b) 3×10^–3 Nm (c) 15×10^–3 Nm

(d) 1.5×10^–3 Nm (e) 9×10^–3 Nm

The torque on a dipole of moment p in an electric field E is the vector product p×E. The magnitude of the torque (τ) is pEsinθ where θ is the angle between dipole moment vector and the electric field vector. The maximum value of the torque is pE (when θ = 90º).

Note that p = q×2a where ‘q’ is the charge and 2a is the separation between the charges.

Therefore, maximum torque = (0.05×10^–6×30×10^–3) ×10⁶ = 1.5×10^–3 Nm.

Electrostatic Potential- An IIT-JEE 2007 MCQ

The following MCQ which appeared in IIT-JEE 2007 question paper is meant for checking whether you have the correct idea about electric potential and potential difference:

Positive and negative point charges of equal magnitude are kept at (0,0, α/2) and(0,0,–α/2), respectively. The work done by the electric field when another positive point charge is moved from (–α,0,0) to (0,α,0) is

(a) positive

(b) negative

(c) zero

(d) depends on the path connecting the initial and final positions

You should have a three dimensional mental view of the Cartesian coordinate system to work out this problem in no time. Even without drawing a figure, some of you will be able to work this out. But you may draw a quick figure to help you.

In the figure shown, the positions of the equal positive and negative charges and the movement of the third positive charge from P to Q are indicated. The path shown is straight, but it can be any curve, since the work done is independent of the path in an electrostatic field and is dependent on the initial and final positions only.

Since P and Q are equidistant from the positive and negative charges, the resultant potentials at P as well as Q due to these charges is zero. Therefore, no work need be done for moving the third charge from P to Q [Option (c)].