## Thursday, November 16, 2006

### Additional Multiple Choice Questions on Electromagnetic Induction

In continuation of the post on 5th October 2006, let us discuss some more questions on electromagnetic induction. Generally, students are enthusiastic in working out problems in electromagnetic induction.
The following question is one that can be included under ‘different type’:
A straight copper rod of length ‘L’ is rotating (with angular velocity ‘ω’) about an axis perpendicular to the rod and passing through a point O on the rod, at a distance L/4 from one end of the rod. A uniform magnetic field ‘B’ parallel to the axis of rotation exists in the entire region where the rod rotates. The potential difference developed between the ends of the rod is
(a) ½ BωL2 (b) BωL2 (c) ¼ BωL2 (d) ¾ BωL2 (e) 2BωL2
The easiest method of solving this problem is to find out the velocity (v) of the centre of the rod and calculate the potential difference developed between the ends of the rod using the expression, V = BLv, for the motional emf (V) in a straight conductor of length ‘L’ moving in a magnetic field with velocity ‘v’. The value of ‘v’ is the velocity of the centre of the conductor which is ωL/4 since the centre is at a distance L/4 from the axis of rotation. So the answer is BL(ωL/4) = ¼ BωL2 .
If this method does not look sufficiently appealing to you, you may use the expression for the voltage induced across a conductor of length ‘L’ rotating about a perpendicular axis through one end: V = ½ BωL2 . In the present case, you have to imagine two conductors of lengths ¾ L and ¼ L respectively. The potentials at their ends with respect to the point O (through which the axis of rotation passes) are ½ Bω(9/16)L2 and ½ Bω(1/16)L2 . The potential difference between the ends is therefore (9/32)BωL2 – (1/32) BωL2 = ¼ BωL2 .
Here is another simple question:
When the number of turns in a coil is made 2.5 times the original number without changing the area, the self inductance of the coil becomes
(a) 2.5 times (b) 5 times (c) 6.25 times (d) 1.25 times (e) 10 times
You should remember that the self inductance of a coil is equal to the magnetic flux linked with the coil when unit current flows in it. The flux linked with the coil is directly proportional to the number of turns in the coil. It is directly proportional to the magnetic field (produced by the coil) also, which again is directly proportional to the number of turns. Hence the self inductance is directly proportional to the square of the number of turns.
The self inductance of the coil mentioned in the question will therefore become 6.25 times the initial value.
The following MCQ also is simple but there is chance of picking out a wrong answer.
A plane rectangular coil is rotating with angular velocity ‘ω’ in a uniform magnetic field. If the axis of rotation is parallel to the plane of the coil and perpendicular to the magnetic field, what is the phase difference between the magnetic flux linkage and the induced emf in the coil?
(a) π (b) zero (c) π/4 (d) π/2 (e) ω/π
If you take the time to be zero when the plane of the coil is perpendicular to the magnetic field B, the flux (Φ) at any instant can be written as Φ = BAcosωt where A is the area of the coil. The induced emf, V = -dΦ/dt = BAω sinωt. Since the flux is a sine function and the induced voltage is a cosine function, the phase difference between the flux linkage and the induced emf is π/2.
[If you take the time to be zero at some instant when the plane of the coil is parallel to the magnetic field, Φ = BAsinωt and V = -BAω cosωt. Again, one is a sine function and the other is a cosine function, indicating a phase difference of π/2].
Consider now the following MCQ which appeared in JIPMER 2000 test paper:
The induced emf in a coil is proportional to
(a) magnetic flux through the coil (b) area of the coil (c) rate of change of magnetic flux through the coil (d) product of magnetic flux and area of the coil.
The correct option is (c). Note that there has to be a time rate of change of magnetic flux for obtaining an induced voltage.
The following MCQ pertaining to motional emf is typical and similar questions often find place in Medical and Engineering Entrance test papers:
A 10m long iron rod, while remaining in the east-west horizontal direction, is falling with a speed of 5m/s. If the horizontal component of earth’s magnetic field is 0.3×10-4 Wb/m2, the emf developed between the ends of the rod is
(a) 0.15V (b) 1.5V (c) 0.15mV (d) 1.5mV (e) 3mV
Since the rod is lying along the east-west direction, it will cut the horizontal magnetic field lines of the earth. The motional emf developed between the ends of the rod is V = BLv = 0.3×10-4×10×5 = 1.5×10-3V = 1.5mV.
Let us now consider the following multiple choice question which appeared in I.I.T. Screening 2002 test paper:
A short circuited coil is placed in a time varying magnetic field. Electric power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be
(a) halved (b) same (c) doubled (d) quadrupled
The resistance of the coil will become 16 times the initial value since there is a four fold increase due to the four fold increase of length and another four fold increase due to the four fold decrease in the area of cross section (when the radius is reduced to half the initial value). Note that the resistance is given by R = ρL/A where ‘ρ’ is the resistivity, L is the length and A is the area of cross section.
The induced voltage in the coil is increased to four times since the number of turns is made four times.
We have, power P = V2/R. So, when the value of V is quadrupled and the value of R is made 16 fold, the value of P remains unchanged and hence (b) is the correct option.
You should become capable of working out things in your mind, without wasting time in writing mathematical steps, when you deal with questions of this type.