If we did all things we are capable of, we would literally astound ourselves.

– Thomas A. Edison

Showing posts with label electromagnetic induction. Show all posts
Showing posts with label electromagnetic induction. Show all posts

Wednesday, July 09, 2008

Kerala Engineering Entrance 2008 Questions on Electromagnetic Induction

Questions involving electromagnetic induction at your level are usually simple and interesting. Here are the two questions which appeared in Kerala Engineering Entrance 2008 question paper:

(1) If the self inductance of 500 turn coil is 125 mH, then the self inductance of similar coil of 800 turns is

(a) 48.8 mH

(b) 200 mH

(c) 187.5 mH

(d) 320 mH

(e) 78.1 mH

Self inductance is the total magnetic flux linked with the coil when unit current flows through the coil. Since the total flux and the magnetic field are directly proportional to the number of turns in the coil, the self inductance is directly proportional to the square of the number of turns in the coil. Therefore we have

5002/8002 = 125/L where L is the self inductance of the 800 turn coil.

This yields L = 320 mH.

(2) The flux linked with a circuit is given by Φ= t3 + 3t 7. The graph between time (xaxis) and induced emf (y–axis) will be a

(a) straight line through the origin

(b) straight line with positive intercept

(c) straight line with negative intercept

(d) parabola through the origin

(e) parabola not through the origin

The induced emf, V = – dΦ/dt = – 3t2 – 3

If V is plotted against t, a parabola which does not pass through the origin will be obtained. So the correct option is (e).

Saturday, July 21, 2007

Two Multiple Choice Questions on Electromagnetic Induction

(1) The initial rate of increase of current when a battery of emf 6V is connected in series with an an inductance of 2H and a resistance of 12 Ω is

(a) 0.5 A s–1 (b) 1 A s–1 (c) 3 A s–1 (d) 3.5 A s–1 (e) 4.5 A s–1

Initially the current in the circuit is zero but the rate of increase of current is maximum. There will be no voltage drop (which is equal to R×I) across the resistance and the entire voltage will appear across the inductance. The magnitude of the voltage across the inductance will be L(dI/dt) so that initially we have

L(dI/dt) = V, from which dI/dt = V/L = 6/2 = 3 A s–1.

[You can calculate the rate of increase of current at instant using the expression for the current in an LR circuit during the rise, given by I = I0(1– e–Rt/L) where ‘e’ is the base of the natural logarithm, ‘I’ is the current at the instant ‘t’ and I0 is the final (maximum) current equal to V/R.

Differentiating the above equation, dI/dt = (I0R/L)e–Rt/L.

Initially, since t = 0, dI/dt = I0R/L = (V/R)×R/L = (6/12)×12/2 = 3 As–1

The rate of decrease of current at any instant ‘t’ when the current in an LR circuit is switched off is – (I0R/L)e–Rt/L since the current during the decay is given by I = I0 e–Rt/L]

(2) The emf induced in a coil at time t = 0.5 s when the magnetic flux (Φ) linked with the coil changes as Φ = t2 – 4t + 2 weber is

(a) 3V (b) 4V (c) 5V (d) 6.5V (e) 8V

The emf (V) induced is given V = – dΦ/dt = –d[t2 – 4t + 2]/dt = –2t + 4.

On substituting for t (=0.5 s), V = 3 volt

Friday, July 13, 2007

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

Friday, July 06, 2007

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)].