Multiple Choice Questions Involving Electromagnetic Induction

“There is only one corner of the universe that you can be certain of improving, and that’s your own self.”

– Aldous Huxley

Many questions on electromagnetic induction have been discussed on this site earlier. You can access those questions by clicking on the label, ‘electromagnetic induction’ below this post. Today we shall discuss a few more questions in this section.

(1) An object of mass M (Fig.) carrying a charge +Q is released from rest at the top of a smooth inclined plane of inclination θ. A magnetic field of flux density B acts  throughout the region and is directed normally into the plane of the figure. Which one among the following statements is correct?

(a) The acceleration of the object down the incline depends on Q but does not depend on B.

(b) The acceleration of  the object down the incline depends on B but does not depend on Q.

(c) The acceleration of the object down the incline depends on both Q and B.

(d) The acceleration of the object down the incline does not depend on Q and B.

(e) All the above statements are incorrect.

The component of the gravitational force along the incline is Mgsinθ which can produce in the object an acceleration gsinθ down the incline. When the object moves down along the inclined plane, the magnetic force (Lorentz force) acting on the object is directed perpendicular to the inclined plane such that it tries to press the object against the inclined plane.

[Apply Fleming’s left hand rule to arrive at the direction of the magnetic Lorentz force].

Since the magnetic force on the object is perpendicular to the inclined plane, it has no component along the plane. The acceleration of the object down the incline is gsinθ itself and is independent of Q and B [Option (d)].

(2) In the above question suppose the inclined plane is rough and the object is able to slide down. Which one among the following statements is correct?

(a) The acceleration of the object down the incline depends on Q but does not depend on B.

(b) The acceleration of the object down the incline depends on B but does not depend on Q.

(c) The acceleration of the object down the incline depends on both Q and B.

(d) The acceleration of the object down the incline does not depend on Q and B.

(e) All the above statements are incorrect.

In this case the frictional force acting between the object and the inclined plane reduces the acceleration of the object. The frictional force F depends on the normal force N acting on the object.

[Remember that F = μN where μ is the coefficient of friction]

But the normal force (which is Mgcosθ + magnetic Lorentz force) depends on both Q and B since the magnetic Lorentz force is QvB where v is the velocity of the object.

Therefore, the acceleration of the object down the incline depends on both Q and B [Option (c)].

(3) A plane rectangular coil of length 2L and breadth L has N turns of insulated copper wire in it. Initially (at time t = 0) the plane of the coil is perpendicular to a uniform magnetic field of flux density B . The coil rotates with angular velocity ω such that the axis of rotation of the coil is at right angles to the magnetic field (Fig.). If the voltage induced in the coil is V_maxsinωt, the angular velocity of the coil is

(a) V_max / 2NL²B

(b) 2V_max t / NL²B

(c)V_max/NL²B                                                                                            

(d) NL²B/ V_max

(e) 2NL²B/ V_max

The magnetic flux φ linked with the coil at the instant t is given by      

φ = NAB cos ωt where A is the area of the coil.

The voltage V induced in the coil is is given by

            V = – dφ /dt = NABω sin ωt

It is given in the question that V = V_maxsinωt

Therefore, V_max = NABω

This gives ω = V_max /NAB

Since the area A = 2L², we have ω = V_max / 2NL²B

(4) In the transformer shown in the adjoining figure, the secondary winding has taps A, B, C, D and E. The number of turns between taps A and B is 100. The number of turns between taps B and C is 50, the number of turns between C and D is 40 and that between D and E is 10. The primary has 2000 turns and is connected to 220 volt A. C. mains. Between which taps will you obtain 5.5 volt output? 

(a) Between A and B

(b) Between A and B as well as B and E

(c) Between B and C as well as C and E

(d) Between C and D

(e) Between D and E

Since the primary voltage is 220 V and the number of turns in the primary is 2000, the induced voltage per turn is 220/2000 = 0.11 volt.

Therefore, the number of turns required to produce a voltage of 5.5 volt is (5.5)/(0.11) = 50.

Since there are 50 turns between taps B and C as well as between taps C and E, the correct option is (c).

[You can work this out this way also:

Considering the entire secondary of the transformer, the secondary to primary turns ratio is 200/2000 = 1/10. Therefore, the secondary voltage (between the ends A and E) is 220/10 = 22 V. Since the voltage across the entire secondary containing 200 turns is 22 volt, we will get 5.5 volt across 50 turns (between taps B and C as well as between taps C and E)]

 

Lorentz force – An IIT-JEE 2012 Multiple Correct Answer(s) Type Question

“Making the simple complicated is commonplace; making the complicated simple, awesomely simple, that’s creativity.”

– Charles Mingus    

IIT-JEE Multiple Correct Answer(s) Type Questions are designed to have four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. The following question appeared in IIT-JEE 2012 question paper. The question is meant for checking your understanding of Lorentz forces.

Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and magnetic fields E = E₀ ĵ and B = B₀ ĵ. At time t = 0, this charge has velocity v in the x-y plane, making an angle θ with the x-axis. Which of the following option(s) is (are) correct for t > 0?

(A) If θ = 0°, the charge moves in a circular path in the x-z plane.

(B) If θ = 0°, the charge undergoes helical motion with constant pitch along the y-axis.

(C) If θ = 10°, the charge undergoes helical motion with its pitch increasing with time, along the y-axis.

(D) If θ = 90°, the charge undergoes linear but accelerated motion along the y-axis.

Note that the electric and magnetic fields are along the positive y-direction.

If θ = 0, the charged particle is moving at right angles to the electric and magnetic fields (Fig.). The magnetic force (qv×B₀) will make the particle move along a circle and the electric force (qE₀) will push it along the positive y-direction. The path of the particle will therefore be a helix of increasing pitch.

If θ = 10°,the magnetic field will make the particle move along the positive y-direction, following a helical path of constant pitch; but since the electric force (qE₀) pushes it along the positive y-direction, the helical path has increasing pitch.

If θ = 90°, the charged particle is moving parallel to the electric and magnetic fields. The magnetic field has no action on the motion since there is no magnetic force. But the electric force (qE₀) will push the particle along the positive y-direction. The particle is thus in accelerated linear motion. 

Therefore options (C) and (D) are correct.

Multiple Choice Questions on Magnetic Force on Moving Charges

Questions involving magnetic force on moving charges are included in most of the medical, engineering and other degree entrance examinations. Here are some simple questions which may easily tempt you to commit mistakes:

(1) The magnetic Lorentz force equation is F = q v×B. In this equation

(a) F, v and B must be mutually perpendicular.

(b) F must be perpendicular to v but not necessarily to B.

(c) F must be perpendicular to B but not necessarily to v.

(d) v must be perpendicular to both F and B

(e) F must be perpendicular to both B and v.

The equation F = q v×B gives the magnetic force F on a charge ‘q’ when it moves with velocity v in a magnetic field B. The angle between the velocity v and the field B can be any value, but the magnetic force F is at right angles to both v and B. So the correct option is (e).

[The vector product v×B which gives the force F indeed demands that F is at right angles to both v and B].

(2) A charged particle moving in the north east direction at right angles to a magnetic field experiences a force vertically upwards. This charged particle will not experience any force if it moves towards

(a) south west direction

(b) north

(c) east

(d) west

(e) north west

Since the magnetic force is vertical, the magnetic field must be horizontal.

Since the charged particle is moving in the north east direction at right angles to the magnetic field, it follows that the magnetic field must be directed either north west or south east (Fig.). So it will not experience any force if it moves towards north west or south east. The correct option is (e).

[You can use Fleming’s left hand rule to obtain the directions easily. See yourself that if the magnetic field in the above question is along the north west direction, the charge on the particle must be positive to obtain a vertically upward magnetic force. If the magnetic field in the above question is along the south east direction, the charge on the particle must be negative].

(3) A proton traveling vertically downwards experiences a southward force due to a magnetic field directed at right angles to its path.. An electron traveling northward in the same magnetic field will experience a magnetic force directed

(a) downwards

(b) upwards

(c) towards east

(d) towards west

(e) towards south east

Since the proton (which is positively charged) experiences a southward force while traveling vertically downwards, the perpendicular magnetic field must be acting towards the east.

[As required by Fleming’s left hand rule, hold the fore-finger, middle finger and thumb of your left hand in mutually perpendicular directions, with the middle finger pointing downwards (in the present case) and the thumb pointing southwards. The fore-finger then points towards the east].

If the proton were to move northward in this magnetic field, it would experience a downward magnetic force. Since the electron is negatively charged, it will experience an upward magnetic force [Option (b)]..

A Question on Lorentz Force

The following MCQ appeared in Kerala Engineering Entrance 2007 question paper. I have seen this question on many occasions with more or less the same options. Here is the question:

A uniform electric field and a uniform magnetic field exist in a region in the same direction. An electron is projected with a velocity pointed in the same direction. Then the electron will

(a) be deflected to the left without increase in speed

(b) be deflected to the right without increase in speed

(c) not be deflected but its speed will decrease

(d) not be deflected but its speed will increase

(e) be deflected to the right with increase in speed

This is a simple question and the most suitable option is (c) since the magnetic field (being parallel to the electron) does not exert any force, but the electric field decelerates the electron (since the charge on the electron is negative).

Now, suppose the question setter had changed the last option as “not deflected but its speed will first decrease and will then increase”. Then, this last option would be the correct option since the electron would be decelerated first and then would be accelerated in the opposite direction. [Remember speed is a scalar quantity and the modified option takes advantage of this].

Magnetic Force on Moving Charges

"You may never know what results come of your actions, but if you do nothing, there will be no results."

– Mahatma Gandhi

 
Most of you might be remembering the expression for the magnetic force ‘F’ on a charge ‘q’ moving with velocity ‘v’ in a magnetic field of flux density ‘B’:
F = qvB sinθ where θ is the angle between v and B. Retain the order qvB. It helps you to remember this expression in its vector form:
F = qv×B

The vector form gives you the magnitude (qvB sinθ) of the force and its direction which is along the direction of the cross product vector v×B. You should remember that the direction of v is the direction of motion of a positive charge. If you have a negatively charged particle such as an electron, you should reverse the direction of v to get the direction of the force.
If there is an electric field (E) also in the region, the net force on the charge is given by
F = q(v×B + E).
This is the Lorentz force equation.

Let us consider the following MCQ which appeared in the IIT-JEE Screening 2003 question paper:

For a positively charged particle moving in XY plane initially along the X-axis, there is a sudden change in its path due to the presence of electric and/or magnetic fields beyond P. The curved path is shown in the XY plane and is found to be non-circular.Which one of the following
combinations is possible? (i,j,k are unit
vectors along X,Y,Z directions)
(a) E = 0; B =bi + ck 
(b) E = ai; B =ck + ai
(c) E = 0; B =cj + bk(d) E = ai; B =ck + bj
As the curved path is confined to the XY plane,
the component of magnetic field effective in
deflecting the particle is the Z-component only.
The other component should be the X-component since the X-component cannot deflect the charged particle proceeding along the X-direction. There has to be an electric field since the path of the particle is non-circular. All these conditions are satisfied by option (b).
Now consider the following multiple choice questions:

(1)An electron is projected horizontally from south to north in a uniform horizontal magnetic field acting from west to east. The direction along which it will be deflected by this magnetic field is
(a) vertically upwards 
(b) vertically downwards 
(c) northwards 
(d) southwards 
(e) eastwards
Since the electron is negatively charged, the direction of its velocity is to be reversed for finding the direction of the cross product vector v×B. The correct option therefore is (a). You can use Fleming’s left hand rule (motor rule) also for finding the direction of the magnetic force. But, when you apply the rule, remember that the direction of the conventional current is that of positive charge.
(2) An electron enters a uniform magnetic field of flux density B=2i+6j-√8 k with a velocity V=i+3j-√2 k where i,j and k are unit vectors along the X, Y and Z directions respectively. Then
(a) both speed and path will change 
(b) speed alone will change 
(c) the path will become circular 
(d) the path will become helical (e) neither speed nor path will change
The vectors B and V are parallel as the components of B are twice the components of V. So the magnetic force on the electron is zero and the correct option is (e).
Generally, in problems of the above type, you will have to find the angle between the vectors V and B using cosθ = V.B/ VB. In the present case we have, cosθ = (2+18+4)/ √(12×48) = 1 so that θ= zero. If in a similar problem, the value of θ works out to be 90˚(if cosθ works out to be zero), the correct option would be (c). If the value of θ were neither zero nor 90˚, the path would be helical and the correct option would be (d).

(3) Doubly ionized atoms X and Y of two different elements are accelerated through the same potential difference. On entering a uniform magnetic field, they describe circular paths of radii R₁ and R₂. The masses of X and Y are in the ratio
(a) R₁/R₂ (b) √(R₁/R₂) (c) √(R₂/R₁) (d) (R₂/R₁)² (e) (R₁/R₂)²
Since both atoms are doubly ionized, they have the same charge ‘q’ and since they are accelerated by the same potential difference ‘V’, they have the same kinetic energy, qV. If m₁ and m₂ are the masses and v₁ and v₂ are the velocities of X and Y respectively, we have, ½ m₁v₁² = ½ m₂v₂² from which m₁/m₂ = v₂²/v₁² = [qBR₂/m₂]²/ [qBR₁/m₁]², on substituting for the velocities from the centripetal force equation, mv²/R = qvB.
Therefore, m₁/m₂ = (R₁/R₂)², given by option (e).