AP Physics C – Magnetic Fields – Answer to Free Response Practice Question on Magnetic Force on Moving Charges

A free response practice question on magnetic force on moving charges was posted on 8^th July 2013. As promised I give below a model answer for your benefit, along with the question:

The adjoining figure represents a cathode ray tube in a cathode ray oscilloscope. The voltage applied (with respect to the cathode) to the final accelerating anode is V and the distance between the anode and the fluorescent screen of the cathode ray tube is L. The mass of the electron is m and the magnitude of its charge is e.  The cathode ray tube is so oriented that in the undeflected condition the electron beam is horizontal and is along the magnetic meridian, with the electrons traveling from south to north. The horizontal component of the earth’s magnetic field at the place is B_h. Now, answer the following questions:

(a) Calculate the velocity with which the electrons strike the fluorescent screen, assuming that the space between the accelerating anode and the fluorescent screen is devoid of electric fields and the electrons start from the cathode with negligible speed.

(b) The cathode ray tube is rotated through 90º about a verti8cal axis so that the electrons travel from east to west. In what direction will the spot of light (produced by the impact of the electron beam) on the fluorescent screen be deflected? Select the correct option from the following, by putting a tick mark (√) against one of the following:

Upwards………

Downwards………

Leftwards………

Rightwards……..

Justify your answer.

(c) If the cathode ray tube is rotated through 90º in the opposite direction, will the direction of deflection of the spot of light on the screen change? Justify your answer.

(d) Derive an expression for the specific charge (charge to mass ratio) of the electron in terms of the given data and the shift s of the electron spot on the screen, on rotating the cathode ray tube through 90º. Assume that the shift s is small compared to the distance between the accelerating anode and the fluorescent screen.

Answer:

(a) The electrons acquire kinetic energy under the action of the accelerating potential V applied on the accelerating anode. If the final velocity acquired by the electrons is v, we have

             ½ mv² = eV

Therefore v = √(2eV/m)

This is the velocity with which the electrons strike the fluorescent screen since their velocity after leaving the aperture of the accelerating anode remains unchanged in the field free space between the anode and the screen.

(b) Upwards.

Earth’s magnetic field is directed from south to north. Electrons traveling from east to west make a conventional current flowing from west to east (since the electrons are negatively charged).

[The current produced by electrons traveling from east to west has the same direction as the current produced by positive charges traveling from west to east]

Therefore, in order to apply Fleming’s left hand rule for fining the direction of the magnetic force on the electron beam, we have to hold the forefinger along the south to north direction and the middle finger along the west to east direction. The thumb will then point upwards.

(c) If the cathode ray tube is rotated through 90º in the opposite direction, the electron beam will be deflected downwards since the electrons  travel from west to east.

(d) When the electrons travel parallel to the horizontal component (B_h) of earth’s magnetic field, there is no magnetic force on them. This follows from the expression for magnetic force F:

             F =evB_h sinθ where θ is the angle between the magnetic field and the direction of the conventional current. Since θ = 180º, F = 0.

When the cathode ray tube is rotated through 90º, the electrons travel at right angles to the horizontal component (B_h)  of earth’s magnetic field. In this case the magnetic force is maximum (F_max) and is given by

             F_max = evB_h. 

Since the deflection (s) produced is small (Fig.), we may assume that this force acts vertically throughout the path of the electrons.

The above vertical force produces a vertical acceleration and the vertical displacement s suffered by the electrons is given by the equation of uniformly accelerated motion,

             s = ut + ½ at² ……….. (i)

where u is the initial vertical velocity (at the instant of leaving the accelerating anode), which is zero, t is the time taken by the electron to travel from the accelerating anode to the fluorescent screen and a is the vertical acceleration which is given by

             a = F_max/m = evB_h/m  

The time taken by the electron to travel the distance L between the anode and screen is given by

             t = L/v

Substituting for u, a and t in Eq. (i) we have

             s = 0 + ½ (evB_h/m) (L/v )² = ½ (e/m) (B_hL²/v)

The speed v of the electrons, as shown in part (a) is √(2eV/m)

Substituting for v we have

             s = ½ (e/m) (B_hL²/√(2eV/m)

This gives e/m = 8s²V/ B_h²L⁴

AP Physics C – Magnetic Fields – Free Response Practice Question on Magnetic Force on Moving Charges

“One of the deep secrets of life is that all that is really worth the doing is what we do for others.”

– Lewis Carroll

 

The cathode ray oscilloscope is one of the most versatile instruments which is used as a test instrument in a variety of forms in almost all fields of scientific activity. Most of you may be familiar with this impressive instrument which is essentially a highly evacuated glass bulb containing an electron gun and a fluorescent screen, with provisions for deflecting the electron beam, usually by applying voltages across deflecting plates. I give below a free response question related to the cathode ray tube:

The adjoining figure represents a cathode ray tube in a cathode ray oscilloscope. The voltage applied (with respect to the cathode) to the final accelerating anode is V and the distance between the anode and the fluorescent screen of the cathode ray tube is L. The mass of the electron is m and the magnitude of its charge is e.  The cathode ray tube is so oriented that in the undeflected condition the electron beam is horizontal and is along the magnetic meridian, with the electrons traveling from south to north. The horizontal component of the earth’s magnetic field at the place is B_h. Now, answer the following questions:

(a) Calculate the velocity with which the electrons strike the fluorescent screen, assuming that the space between the accelerating anode and the fluorescent screen is devoid of electric fields and the electrons start from the cathode with negligible speed.

(b) The cathode ray tube is rotated through 90º about a verti8cal axis so that the electrons travel from east to west. In what direction will the spot of light (produced by the impact of the electron beam) on the fluorescent screen be deflected? Select the correct option from the following, by putting a tick mark (√) against one of the following:

Upwards………

Downwards………

Leftwards………

Rightwards……..

Justify your answer.

(c) If the cathode ray tube is rotated through 90º in the opposite direction, will the direction of deflection of the spot of light on the screen change? Justify your answer.

(d) Derive an expression for the specific charge (charge to mass ratio) of the electron in terms of the given data and the shift s of the electron spot on the screen, on rotating the cathode ray tube through 90º. Assume that the shift s is small compared to the distance between the accelerating anode and the fluorescent screen.

Try to answer this question. You are required to answer it within 15 minutes and it carries 15 points.

I’ll be back soon with a model answer for your benefit.

AP Physics B & C - Multiple Choice Practice Questions on Magnetic Fields

“The best way to find yourself is to lose yourself in the service of others."

– Mahatma Gandhi

Questions on magnetic fields due to current carrying wires and the magnetic force on moving charges and current carrying wires are generally interesting. Such questions have been discussed on many occasions on this site. You can access them by clicking on the label ‘magnetic field’ below this post. Note that the percentage goal from this section for AP Physics C Electricity and Magnetism Exam is 20% while that for AP Physics B Exam is 4% (for the May 2012 Exam)

Today we will discuss a few more multiple choice practice questions in this section

(1) An electron moves with speed v in a direction perpendicular to a constant magnetic field B. Which one of the following graphs best represents the radius r of the path of the electron as a function of its speed

The radius r of the circular path followed by the electron in the perpendicular magnetic field B is given by

r = mv/qB where m is the mass of the electron and q is its charge.

[The above expression is obtained by equating the centripetal force to the magnetic force on the electron:

mv²/r = qvB]

In a constant magnetic field the radius r is directly proportional to the speed v and the graph shown in option (b) is the answer.

(2) Two vertical metal plates A and B are arranged in an evacuated chamber with their planes parallel to each other and they carry charges + Q and – Q as indicated in the adjoining figure. A uniform magnetic field B exists in the region between the plates. An electron entering with horizontal velocity ‘v’ rightwards in the central region between the plates is found to proceed undeviated throughout the region. What is the direction of the magnetic field? (Ignore the effect of gravity)

(a) horizontally leftwards

(b) horizontally towards plate A

(c) horizontally towards plate B

(d) vertically downwards

(e) vertically upwards

Since the electron is negatively charged, the electric field in the region between the plates will exert a horizontal force on the electron. This force is directed towards the positive plate A. Since the electron proceeds undeviated throughout the region between the plates, the magnetic force on the electron must be equal and opposite to the electric force. This means that the magnetic force on the electron is directed horizontally towards the plate B

The rightward moving electron is equivalent to a leftward current. Therefore, on applying Fleming’s left hand rule, we find that the magnetic field must act vertically upwards.

[To apply Fleming’s left hand rule, hold the fore finger, middle finger and thumb of your left hand along mutually perpendicular directions such that the middle finger points along the direction of the current (horizontally leftwards in the present case) and the thumb points along the direction of the force (horizontally towards plate B in the present case). Your fore finger will then point vertically upwards].

(3) If the magnitude of the electric field in the above question is E and the mass and speed of the electron are respectively m and v, what is the magnitude of the magnetic field in between the plates?

(a) E/v

(b) Ev/m

(c) E/mv

(d) v/E

(e) E

Since the electron proceeds undeviated under the action of crossed electric and magnetic fields, we can equate the magnitudes of the magnetic and electric forces. Therefore, we have

Bev = Ee

This gives B = E/v.

[The mass of the electron given in the question just serves the purpose of a distraction]

The following questions are specifically meant for AP Physics C aspirants:

(4) A charged particle entering at right angles to a uniform magnetic field has its trajectory as shown in the figure, during a small time interval. The magnetic field is directed normally into the plane of the trajectory and the speed of the particle is decreasing continuously. Which one among the following statements is correct?

(a) The particle is negatively charged and is moving in the clockwise sense

(b) The particle is negatively charged and is moving in the anticlockwise sense

(c) The particle is positively charged and is moving in the clockwise sense

(d) The particle is positively charged and is moving in the clockwise sense

(e) Nothing can be conclusively mentioned about the motion of the particle since the nature of its charge is not given

The particle must be moving in the clockwise sense since the radius of its initial path has to be greater because of the greater initial speed.

[We have radius r = mv/qB so that r α v. Qualitatively speaking, a particle with larger velocity cannot be easily affected by a force].

On applying Fleming’s left hand rule (motor rule) it is seen that the particle is negatively charged.

[When you hold the fore finger of your left hand along the direction of the magnetic field (normally into the plane of the figure) and the middle finger along the initial direction of the particle, you will find that the thumb is pointing radially outwards and not inwards as required for the curved trajectory given in the figure. This indicates that the direction of the conventional current is opposite to the direction of motion of the particle. Therefore the particle must be negatively charged.

(5) A straight wire of length 0.2 m is oriented along the z-axis. It carries a current of 5 A flowing along the positive z-direction. The magnetic force acting on this wire when it is placed in a magnetic field of flux density 0.4 j – 0.5 k tesla is (i, j and k are unit vectors along the x, y and z-directions respectively)

(a) – 0.4 j N

(b) 0.4 k N

(c) 0.9 i N

(d) – 0.4 i N

(e) – 0.9 i N

The magnetic force F on the wire is given by

F = I L×B = 5[(0.2 k)×(0.4 j – 0.5 k)] = – 0.4 i newton.

[The magnetic force on the wire acts along the negative x-direction].

Now see these questions with solution.

Multiple Choice Practice Questions on Magnetic Fields for AP Physics B & C

Equations to be remembered in the section ‘magnetic fields’ were discussed in the post dated 1^st February 2008, followed by some multiple choice practice questions in the post dated 2^nd February 2008. A free response practice question for AP Physics C aspirants was discussed in the post dated 9^th February 2008, followed by another free response practice question of common interest to AP Physics B as well as AP Physics C aspirants. You can access all these posts by clicking on the label ‘magnetic field’ below this post.

Today we will discuss a few more multiple choice questions (for practice) on magnetic fields:

(1) Electrons and protons of the same momentum enter normally into a uniform magnetic field of flux density B. Mass of the electron is m_e and the mass of the proton is m_p. If R_e and R_p represent the radii of the paths of the electron and the proton respectively, then the ratio R_e/R_p is

(a) m_p/m_e

(b) m_e/m_p

(c) (m_p/m_e)^1/2

(d) (m_e/m_p)^1/2

(e) 1

In the case of a particle of mass m and charge q projected with speed v perpendicular to a magnetic field of flux density B, the magnetic force of magnitude qvB supplies the centripetal force required for the circular motion of the particle within the field so that we have

qvB = mv²/R where R is the radius of the path.

Therefore, R = mv/qB

The electron and proton have the same quantity of charge (even though the signs are opposite). The momentum mv is given as the same in the question. Therefore they move along circular paths of the same radius so that the ratio R_e/R_p = 1 [Option (e)].

(2) An particle of mass m and charge q is projected perpendicular to crossed static electric and magnetic fields. The magnetic field has magnitude k tesla while the electric field has magnitude 2k newton per coulomb. The kinetic energy of the electron is found to increase from 2.5 electron volt to 4 electron volt. How much energy is supplied (to the particle) by the magnetic field?

(a) 4 eV

(b) 1.5 eV

(c) 1.0 eV

(d) 0.5 eV

(e) zero

Even though there is a magnetic force on a charged particle moving through a magnetic field, the direction of this force is always perpendicular to the direction of the displacement of the particle. Therefore the work done by the magnetic field on the charged particle is zero. In other words, the energy supplied (to the particle) by the magnetic field is zero. [Option (e)].

(3) The attractive force between two identical coaxial circular coils separated by a small distance d (fig.) and carrying the same current I is F. If the separation is halved and the current in both coils reversed and doubled, the force becomes

(a) – F

(b) F

(c) 4F

(d) 8F

(e) – 8F

Two identical coils separated by a small distance can be treated as two parallel straight wires so that the magnetic force between them is directly proportional to the product of the currents in them and inversely proportional to the separation between them.

[Recall that the force per unit length between two straight parallel current carrying wires separated by a distance d (in free space or air) is μ₀I₁I₂ /2πd where μ₀ is the permeability of free space].

When the separation is halved the magnetic force is doubled. When the current in both coils is doubled, the magnetic force is quadrupled. The magnitude of the force thus becomes 8F. The force will be attractive itself on reversing the current in both coils. The correct option is (d).

[Remember that ‘like currents attract and unlike currents repel’. If the current in one of the coils were reversed in the above situation, the answer would be – 8F since the force becomes repulsive].

(4) A charged particle travels along positive x-direction with constant speed v through a region of space where both electric and magnetic fields exist. If E and B are respectively the magnitudes of the electric and magnetic fields and the electric field is directed along the negative y-direction, it follows that

(a) E = vB and the magnetic field is along the positive z-direction

(b) E = vB and the magnetic field is along the negative z-direction

(c) E = vB and the magnetic field is along the positive y-direction

(d) B = vE and the magnetic field is along the negative z-direction

(e) B = vE and the magnetic field is along the positive z-direction

The electric force on the charge q must be equal and opposite to the magnetic force and hence we have

Eq = qvB from which E = vB.

[The last three options are therefore eliminated].

Since the electric field is along the negative y-direction, the electric force on a positive charge (let us say) is along the negative y-direction and it follows that the magnetic force on the positive charge must be along the positive y-direction. Using Fleming’s left hand rule you will easily find that the magnetic field must be along the negative z-direction [Option (b)].

[You will get the same result if the particle has charge negative].

The following questions are for AP Physics C aspirants (But AP Physics B aspirants also can ‘enjoy’ them):

(5) A plane square loop PQRS of side ‘a’ made of thin copper wire has ‘n’ turns and it carries a direct current ‘I’ ampere in the direction shown in the adjoining figure. This wire loop is placed in a magnetic field of flux density ‘B’ tesla, which is directed perpendicularly in to the plane of the loop. What is the torque acting on the loop?

(a) IaB

(b) nIaB

(c) Ia²B

(d) nIa²B

(e) zero

The magnetic force on the side PQ of the loop is nIaB and is directed upwards through the mid point of PQ in accordance with Fleming’s left hand rule (motor rule). The magnetic force on the side RS of the loop is nIaB itself; but it is directed downwards through the mid point of RS. These two forces cannot produce a torque since they have the same line of action.

The magnetic force on the side QR of the loop is nIaB and is directed leftwards through the mid point of QR. The magnetic force on the side SP of the loop is nIaB itself; but it is directed rightwards through the mid point of SP. These two forces also cannot produce a torque since they have the same line of action.

Thus the net torque acting on the loop is zero [Option (e)].

[You should note that if the loop is placed in the magnetic field with its plane parallel to the direction of the magnetic field, the torque acting on the loop will be maximum which is equal to nIa²B].

Generally the torque acting on the current loop is nIAB sin θ where θ is the angle between the magnetic field vector B and the area vector A of the coil. The magnitude of the area vector is ℓb for a rectangular coil of length ℓ and breadth b. The direction of the area vector is perpendicular to the plane of the coil and is determined by the direction of flow of the current in the coil (direction of advance of a right hand screw turned in the sense of flow of the current), as indicated in the figure in which a plane circular coil is shown.

The expression for the torque τ in vector form will be convenient:

τ = nI A×B

When A and B are parallel or anti-parallel the torque is zero as we have in the above question].

(6) A plane circular coil of radius R has N turns and it carries a current I. It is placed in a uniform magnetic field of flux density B such that the plane of the coil is parallel to the magnetic field. What is the magnitude of the net magnetic force on the coil?

(a) nIRB

(b) nI πRB

(c) nI πR²B

(d) 2nI πRB

(e) zero

In the previous question we found that the magnetic forces acting on the horizontal sides of the square loop are equal in magnitude and opposite in direction. Similarly the magnetic forces acting on the vertical sides of the square loop also are equal in magnitude and opposite in direction. The net force acting on the square loop will therefore be zero. The same result holds in the case of a rectangular loop also.

The plane circular loop can be imagined to be made of a large number of horizontal and vertical segments (fig.). The net magnetic force on the horizontal segments as well as the vertical segments can easily be seen to be zero so that the net magnetic force on the entire loop is zero [Option (e)].

[You should note that the net magnetic force on a plane loop of any shape, when placed in a uniform magnetic field is zero. But there will be a torque on the current loop unless the magnetic field is perpendicular to the plane of the loop. If the current loop is placed in a non-uniform magnetic field, there will be a net force and a torque. These statements are true for a bar magnet placed in a magnetic field, for obvious reasons].