Electromagnetic Induction- Two Multiple Choice Questions

You should be able to find the answers for the following questions without difficulty. For each question you should take about one minute at the most (if you have understood basic principles in this section thoroughly).

(1) A long straight conductor carries a current that increases at a constant rate. The current induced in the circular conducting loop placed near the straight conductor (fig.) is

(a) zero

(b) constant and in the clockwise direction

(c) constant and in the anticlockwise direction

(d) increasing and in the clockwise direction

(e) increasing and in the anticlockwise direction

A current flows in the circular loop because of the increasing magnetic flux linked with it. This increase in the magnetic flux is to be opposed in accordance with Lenz’s law. The magnetic field lines produced by the straight conductor are directed normally into the plane of the circular coil (away from the reader). Therefore, the direction of the magnetic field lines produced by the induced current in the circular coil must be directed normal to the plane of the circular coil and towards the reader. The induced current should therefore flow in the anticlockwise sense.

The magnitude of the induced current is constant since the rate of increase of flux is constant. The correct option is (c).

(2) In the circuit shown a coil of inductance 0.5 H and negligible resistance is connected to a 12 V battery of negligible internal resistance using a resistive network. If I₁ is the current delivered by the battery immediately after closing the switch and I₂ is the current (delivered by the battery) long time after closing the switch,

(a) I₁ = 0.5 A and I₂ = 0.6 A

(b) I₁ = 0.6 A and I₂ = 0.6 A

(c) I₁ = 0.5 A and I₂ = 0.5 A

(d) I₁ = 0 A and I₂ = 0.5 A

(e) I₁ = 0.6 A and I₂ = 0.5 A

The current through the inductance cannot rise abruptly on closing the switch (because of the time constant L/R of the LR circuit). At the moment of switching the circuit on, the circuit behaves as though the inductance branch is absent. The initial current is therefore given by

I₁ = 12 V/ (16+8) Ω = 0.5 A

The current through the inductance rises exponentially with time t from zero to the final steady value I [in accordance with the equation I = I₀(1– e^–Rt/L) with usual notations]. After a long time, the current through the inductance becomes steady and there is no voltage drop across the inductance. The inductance is in effect short circuited and the final steady current is given by

I₂ = 12 v/(16+4) Ω = 0.6 A

[Note that the parallel combined value of 8 Ω and 8 Ω is 4 Ω]

The correct option is (a).

You will find many useful posts on electromagnetic induction on this site. You can access them by clicking on the label ‘electromagnetic induction’ below this post.

Joint Entrance Examination for Admission to IITs and other Institutions- (IIT-JEE 2009)

The Joint Entrance (2009) Examination for Admission to IITs and other Institutions (IIT-JEE 2009) will be held on April 12, 2009 (Sunday) as per the following schedule:

09:00 – 12:00 hrs Paper – 1
14:00 – 17:00 hrs Paper – 2
Application materials of the Joint Entrance Examination 2009 will be issued with effect from 19th November, 2008. On-line submission of Application will also commence on the same day.
Application materials can be purchased from designated branches of Banks and all IITs between 19.11.2008 and 24.12.2008 by paying Rs. 500/- in case of SC/ST/PD/Female candidates, Rs.1000/- in case of male general/OBC candidates and US$ 100 in the case of candidates appearing in Dubai centre.
The last date for postal request of application materials is 16-12-2008.
The last date for receipt of the completed application at the IITs is 24th December, 2008.
Online Submission of Application: This facility will be available between 19.11.2008 and 24.12.2008 between 8:00 hrs and 17:00 hrs. through the JEE websites of the different IITs.
The JEE websites of the different IITs are given below:

IIT Bombay: http://www.jee.iitb.ac.in
IIT Delhi: http://jee.iitd.ac.in
IIT Guwahati: http://www.iitg.ac.in/jee
IIT Kanpur: http://www.iitk.ac.in/jee
IIT Kharagpur: http://www.iitkgp.ernet.in/jee
IIT Madras: http://jee.iitm.ac.in
IIT Roorkee: http://www.iitr.ac.in/jee
Visit one of these web sites (http://jee.iitm.ac.in) for more details.

AIEEE 2008 Question Involving Kirchoff’s Laws

Occasionally you will find questions requiring the application of Kirchoff’s laws. Here is a question that appeared in AIEEE 2008 question paper which I give here for clearing your doubts regarding the direction of the current you will have to mark in closed loops in direct current networks:

A 5 V battery with internal resistance 2 Ω and a 2 V battery with internal resistance 1 Ω are connected to a 10 Ω resistor as shown in the figure.

The current in the 10 Ω resistor is

(a) 0.27 A, P₁ to P₂

(b) 0.27 A, P₂ to P₁

(c) 0.03 A, P₁ to P₂

(d) 0.03 A, P₂ to P₁

The circuit is redrawn, indicating the currents in the three branches. We have marked the directions of the currents I₁ and I₂ in the directions we normally expect the 5 V and the 2 V batteries to drive their currents. [Note that there can be situations in which the direction we mark is wrong.

There is nothing to be worried about such situations since you will obtain a negative current as the answer and you will understand that the real direction is opposite to what you have marked in the circuit].

Applying Kirchoff’s voltage law (loop law) to the loops ABP₂P₁ and P₁P₂CD we have respectively,

5 = 2×I₁ + 10×(I₁ – I₂) and

2 = 1×I₂ –10×(I₁ – I₂)

The above equations can be rewritten as

12 I₁ – 10 I₂ = 5 and

–10 I₁ + 11 I₂ = 2

These equations can be easily solved to give I₁ = 2.34 A (nearly) and I₂ = 2.31 A (nearly) so that the current (I₁ – I₂) = 0.03 A.

Since the currents I₁ and I₂ are obtained as positive, the directions we marked are correct and the current flowing in the 10 Ω resistor is 0.03 A, flowing from P₂ to P₁ [Option (4)].

[Suppose we had marked the current I₂ as flowing in the opposite direction. The current flowing in the 10 Ω resistor will then be (I₁ + I₂). We will then obtain I₁ = 2.34 A and I₂ = – 2.31 A, the negative sign indicating that the real direction of I₂ is opposite to what we marked. The current flowing through the 10 Ω resistor will again be obtained correctly as (I₁ + I₂) = 2.34 A– 2.31 A = 0.03 A].