“You must not lose faith in humanity. Humanity is like an ocean; if a few drops of the ocean are dirty, the ocean does not become dirty.”
– Mahatma Gandhi
Questions involving electromagnetic induction have been discussed on several occasions on this site. You can access all of them either by trying a search for ‘electromagnetic induction’ using the search box provided on this page or by clicking on the label ‘electromagnetic induction’ below this post. Questions on electromagnetic induction are generally interesting and so the aspirants of AP Physics B as well as AP Physics C will definitely ‘enjoy’ them. The following questions are meant for gauging the depth of your knowledge and understanding in this area:
(1) A particle of mass m and charge –q moves with uniform speed from point A to point B along a straight path AOB in the positive y-direction. A plane conducting circular loop is arranged near the path AOB (as shown in the figure) so that the loop and the path AOB are in the same plane. Pick out the correct statement regarding the current induced in the loop during the interval the charged particle moves from A to B:
(a) The induced current increases and flows clockwise in the loop
(b) The induced current increases and flows anticlockwise in the loop
(c) The induced current increases, reaches a maximum and then decreases; but it flows clockwise in the loop
(d) The induced current increases, reaches a maximum and then decreases; but it flows anticlockwise in the loop
(e) The induced current changes its direction
This question may confuse many among you quite a lot. The particle has a negative charge and that too may enhance your doubts.
Think of the moving charge as a moving current element. Since the charge is negative and is moving along the positive y-direction, the direction of the conventional current is along the negative y-direction. At the circular loop this current element gives rise to a magnetic field perpendicular to the plane of the loop, directed towards the reader. When the charged particle is at A, the magnetic flux through the loop is small and when it is nearest to the circular loop (at O), the magnetic flux through the loop is maximum. The increasing magnetic flux induces a current in the loop. By Lenz’s law the induced current must flow in the clockwise sense (so that the magnetic flux produced by the induced current opposes the increasing magnetic flux produced by the moving charge).
When the charge moves from O to B, the magnetic flux linked with the loop decreases. The current induced in the loop now changes its direction since the magnetic flux produced by the induced current has to oppose the decreasing magnetic flux produced by the moving charge. The correct option is (e).
[The last option can be further refined and stated as follows:
(e) The induced current flows in the clockwise sense and anticlockwise sense respectively when the charged particle traces the paths AO and OB].
(2) A rectangular conducting loop PQRS (Fig.) is located in a magnetic field B which acts perpendicularly into the plane of the loop. The magnetic field is constant and it exists over a wide region, well beyond the extent of the loop. The loop is moved, within the magnetic field, with uniform velocity v in a direction mutually perpendicular to the direction of the magnetic field and the side QR. Pick out the correct statement regarding the induced current (if any):
(a) The induced current in the side PQ flows from P to Q.
(b) The induced current in the side RS flows from S to R.
(c) There is no induced current in the loop since the voltage induced in the side PS as well as QR is zero.
(d) There is no induced current in the loop since no voltage is induced in any of the sides of the loop.
(e) There is no induced current in the loop since the voltages induced in the sides PS and QR are equal and they act in opposition in the loop.
No current is induced in the loop since the magnetic flux linked with the loop is the same in all positions of the loop. So there is no change of flux to produce an induced emf in the loop. Even though the last three options indicate that the induced current in the loop is zero, the reason for this is correct in the case of option (e) only. Equal motional emfs are generated in the sides PS and QR and these act in the loop as two voltage sources in opposition.
(3) A copper rod moves at a constant velocity in a direction perpendicular to its length. A constant magnetic field mutually perpendicular to the rod and its direction of motion exists in the region. Pick out the correct statement from the following:
(a) The electric potential at the centre of the rod is maximum when the rod moves.
(b) The electric potential at the centre of the rod is minimum when the rod moves.
(c) The electric potential is the same at all points of the rod when it moves.
(d) An electric field is produced in the rod when it moves.
(e) The electric potential is zero at the centre of the rod and increases towards the ends
When the rod moves at right angles to the magnetic field, the free electrons in the rod experience magnetic (Lorentz) force and shift towards one end. Thus there is an accumulation of negative charges at one end of the rod and consequently there exists an electric field in the rod [Option (d)].
The following question is specifically for AP Physics C aspirants: 
(4) The series combination of an inductance L and a resistance R shown in the adjoining figure is part of an electric circuit which drives a current i through the combination. When the current through the combination is 6 mA and is decreasing at the rate of 10 mA s–1, the potential difference between points A and B is 6 mV. When the current through the combination is 6 mA and is increasing at the rate of 10 mA s–1, the potential difference between points A and B is 12 mV. The values of the inductance L and resistance R are respectively
(a) 4 H and 6 Ω.
(b) 3 H and 6 Ω
(c) 0.5 H and 3.5 Ω
(d) 0.3 H and 1.5 Ω
(e) 1.5 H and 1.5 Ω
It is the potential difference between the points A and B that drives a current through L and R. When the current in the circuit decreases, the inductor has to try to oppose the decrease by inducing a voltage in it to aid the 6 mV across the points A and B. Therefore Kirchhoff’s voltage equation for this case is
– L(di/dt) + iR = P.D. between A and B
[We could have better written this as (P.D. between A and B) + L(di/dt) = iR]
Substituting known values, we have
– L×10×10–3 + 6×10–3 R = 6×10–3
Or, – L×10 + 6R = 6 -----------(i)
When the current in the circuit increases, the inductor has to try to oppose the increase by inducing a voltage in it to oppose the 12 mV across the points A and B. Therefore Kirchhoff’s voltage equation for this case is
L(di/dt) + iR = P.D. between A and B
[We could have better written this as (P.D. between A and B) – L(di/dt) = iR]
Substituting known values, we have
L×10 + 6R = 12 -----------(ii)
Adding Eq(i) and Eq(ii) we have 12 R = 18 from which R = 1.5 Ω.
On substituting the value of R in Eq(ii) we obtain L = 0.3 H [Option (d)].
No comments:
Post a Comment