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)]..

IIT JEE 2010 Question (MCQ) on Magnetic Force on Current Carrying Conductors

“I object to violence because when it appears to do good, the good is only temporary; the evil it does is permanent.”

– Mahatma Gandhi

The following question which appeared in IIT JEE 2010 question paper is worth noting:

A thin flexible wire of length L is connected to two adjacent fixed points and carries a current I in the clockwise direction, as shown in the figure.

When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is

(a) iBL

(b) iBL /π

(c) iBL /2π

(d) iBL /4π

Considering an elemental length dL of the wire, the magnetic force F acting on the element is given by

F = idLB = iRdθB, as is clear from the figure.

The tension T developed in the wire is resolved into rectangular components T sin(dθ/2) and T cos(dθ/2) as indicated in the figure.

The magnetic force F is related to the tension T as

F = iRdθB = 2T sin(dθ/2)

Since dθ is a very small angle, sin(dθ/2) ≈ dθ/2

Therefore, the above equation gives T = iRB

But R = L /2π

Therefore, T = iBL /2π

Now, consider the following MCQ which is quite simple contrary to the impression of some of you:

Two straight, infinitely long parallel wires P and Q carry equal currents (I) in opposite directions (fig.). A square loop of side a carrying current i is arranged symmetrically in between the straight wires so that the plane of the loop is in the plane containing the wires P and Q. What is the torque acting on the current loop?

(a) Zero

(b) μ₀Iia²/2πd

(c) (μ₀Iia²/2π)[1/(d+a)]

(d) (μ₀Iia²/2π)[1/d + 1/(d+a)]

(e) (μ₀Iia²/2π)[1/d – 1/(d+a)]

The resultant magnetic forces acting on the four sides of the current loop act at their centres. The forces on the opposite sides of the loop are oppositely directed as you can easily verify using Fleming’s left hand rule. Since the lines of action of the resultant forces on opposite sides of the loop coincide, there is no lever arm to produce a torque and hence the torque is zero.

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).