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 = BvLv where Bv 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 Bv = 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).
(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 = BvLv where Bv 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 Bv = 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).
In the copper rod question
ReplyDeletei tnink the area swept would be four tmes as radius is doubled........
You are absolutely correct anonymous!
ReplyDeleteI have already corrected it. Thank you very much.
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ReplyDeleteThis site is helpful for me.....
ReplyDelete