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.

Multiple Choice Questions (MCQ) on electromagnetic induction

Questions based on electromagnetic induction are simple and interesting. Usually you will get questions from this section in any Medical and Engineering Entrance Test. Consider the following two questions:
(1) Lenz’s law is a consequence of the law of conservation of
(a) linear momentum (b) charge (c) angular momentum (d) energy (e) none of the above
According to Lenz’s law, the induced current will always oppose the change which causes it. Indeed, the induced current has to oppose the change because, otherwise the change which causes the current will persist and the current will continue to flow once it is started. You will then have a supply of energy (in the form of electric current) without any external agency doing any work. This will then violate the law of conservation of energy, which is impossible. So, Lenses law holds good in accordance with the law of conservation of energy. Option (d) is the correct answer.
(2) The electrical entity inductance can be compared to the mechanical entity
(a) energy (b) impulse (c) momentum (d) torque (e) inertia
The correct option is (e). Inertia in Mechanics is the property by which a body (or, a mechanical system) tries to oppose any change in its state of rest or of uniform motion. Inductance is the property by which an electric circuit tries to oppose any change of current flowing in it. So, inductance and inertia are comparable.
The above two questions high light two basic points. Now consider the following:
A bar magnet is released into a copper ring which is directly below it. What about the acceleration of the magnet? Greater than ‘g’ or equal to ‘g’ or less than ‘g’?
The acceleration is less than ‘g’ since the falling magnet will generate an induced current in the copper ring and the induced current will oppose the motion of the magnet.
Now, consider the following M.C.Q.:
A jet plane is flying horizontally at a speed of 1800 km/hour. What is the potential difference developed between the tips of its wings if the wing span is 25m? Earth’s magnetic field at the location is 0.4 gauss and the angle of dip is 30˚.
(a) 25mV (b) 250mV (c) 500mV (d) 2.5V (e) 5V
The motional emf developed between the tips of the wings is given by V = B_vLv where B_v is the vertical component of the earth’s magnetic flux density, L is the distance between the tips of the wings (wing span) and ‘v’ is the velocity. [Note that this emf is produced because of the cutting of the vertical field lines and this is why we use the vertical component of the field]. We have B_v = Bsin30 = 0.4×10^-4×½ = 0.2×10^-4 tesla. Also, L = 25m and v = 500m/s. The emf then works out to be 0.25V = 250mV. [Note that gauss is the cgs unit of magnetic flux densitywhich is often used. One tesla = 10⁴ gauss].
Let us consider another question involving Faraday’s disc, the first electric generator:
A circular copper disc 10cm in diameter rotates 1800 times per minute about a central axis at right angles to the plane of the disc. A uniform magnetic field of 1 tesla is applied perpendicular to the plane of the disc. The voltage induced between the centre and the edge of the disc is
(a) 0.235V (b) 0. 47V (c) 2.35V (d)4.7V (e) zero
The motional emf induced when a conductor moves perpendicular to a magnetic field is the product of the area swept per second (by the conductor) and the magnetic field. Therefore, in the present case, induced voltage = area swept per secong by a radius × B = n πr²B = (1800/60) × π ×(0.05)² ×1= 0.2356V. The correct option therefore is (a).
Let us modify this question as follows:
A copper rod 10cm long rotates 1800 times per minute about an axis passing through one end at right angles to the rod. A uniform magnetic field of 1 tesla is applied perpendicular to the plane of rotation. The voltage induced between the ends of the rod is
(a) 0.235V (b) 0. 942V (c) 2.35V (d)4.7V (e) zero
This mcq is similar to the previous one. The voltage induced will be four times the previous value since the area swept is four times. The correct option is (b).
Now let us modify the above ‘rod problem’ as follows:
An aluminium rod 10 cm long rotates 1800 times per minute about an axis passing through its centre at right angles to it. A uniform magnetic field of 1 tesla is applied perpendicular to the plane of rotation. The voltage induced between the ends of the rod is
(a) 0.235V (b) 0. 47V (c) 2.35V (d)4.7V (e) zero
This is a simple question but there is chance of committing a mistake! The correct option is neither (a) nor (b). The potential at the ends (with respect to the mid point of the rod) will be the same (0.235V) so that the potential difference between the ends will be zero [Option (e)].
You should note that the motional emf is generated because of the shifting of mobile charge carriers due to the Lorentz force.
The following two questions (mcq) appeared in the Kerala Engineering Entrance test paper of 2006:
(1) A copper disc of radius 0.1m is rotated about its centre with 20 revolutions per second in a uniform magnetic field of 0.1T with its plane perpendicular to the field. The emf induced across the radius of the disc is
(a) π /20 volt (b) π /10 volt (c) 20π milli volt (d) 10π milli volt (e) 2π milli volt
The emf induced = n πr²B = 20π(0.1)² ×0.1 = 0.02 π volt = 20π milli volt [Option (c)].
(2) A varying magnetic flux linking a coil is given by Φ = Xt². If at time t=3s, the emf induced is 9V, then the value of X is
(a) 0.66 Wb.s^-2 (b) 1.5 Wb.s^-2 (c) -0.66 Wb.s^-2 (d) -1.5 Wb.s^-2 (e) -0.33 Wb.s^-2
The induced emf = -dΦ/dt. Therefore, -2Xt = 9 when t=3s so that -6X = 9, from which X = -1.5. The correct option is (d).