Genius is one percent inspiration and ninety-nine percent perspiration.
– Thomas Alva Edison
Today I give you afree response practice question involving magnetic force on moving charges. The question is meant for AP Physics C aspirants. You may work out the question yourself and compare your answer with the model answer given below the question. Here is the question:
A proton (mass 1.67×10–27 kg, charge 1.6×10–19 C) atrest at the origin is acted on by a uniform electric field of intensity 900 V/m, acting along the positive X-direction in the region from x = 0 to x = 8.35 cm. A uniform magnetic field is applied perpendicular to the direction of motion of the proton in the region from x = 8.35 cm to x = 16.7 cm so that the proton is finally brought to a detector located at x = 0, y = – 8.35 cm, z = 0. Now answer the following questions:
(a) In which direction should the magnetic field be applied. Pick out the answer by putting a tick (√) mark against the correct option among the following:
(i) Along the negative Y-direction _____
(ii) Along the negative Z-direction _____
(iii) Along the positive Z-direction _____
Justify your answer.
(b) Calculate the velocity with which the proton enters the magnetic field.
(c) Calculate the magnetic flux density needed to bring the proton exactly at the detector at the point (0, – 8.35 cm, 0).
(d) The magnetic field is gradually decreased from the value asked for in part (c). Calculate the value of the magnetic field which will allow the proton to just cross the magnetic field.
(a) The magnetic field is to be applied along the positive Z-direction. [Option (iii)].
At the moment the proton enters the magnetic field, its velocity is along the positive X-direction. The magnetic force on the electron at that moment should be downwards (along the negative Y-direction) so that it suffers a negative Y-displacement by the time it reaches the detector. By Fleming’s left hand rule (motor rule), this can be achieved only if the magnetic field is directed along the positive Z-direction.
(b) In the adjoining figure O is the origin from where the proton is released and P is the position of the detector. It is accelerated along the positive X-direction by the electric field E (let us say). The work done by the electric field on the proton in moving through a distance d along the electric field is qEd where q is the charge on the proton. This is equal to the kinetic energy of the proton at the moment it reaches the magnetic field. Therefore, we have
qEd= ½ mv2 where m is the mass of the proton and v is its velocity.
Therefore, v = √(2qEd/m)
Or, v = √[(2×1.6×10–19×900×8.35×10–2)/(1.67×10–27)] = 1.2×105 ms–1, directed along the positive X-direction.
(c) In the magnetic field the proton moves along a circular path of radius r obtained by equating the magnetic force on the proton to the centripetal force:
qvB = mv2/r where B is the magnetic flux density.
Therefore, r = mv/qB and the magnetic flux density needed to bring the proton exactly at the detector is given by
B = mv/qr
As the proton has to reach the detector at the point (0, – 8.35 cm, 0), the diameter of its semicircular path (as demanded by geometry) in the magnetic field is 8.35 cm so that the radius of the path is r = 4.175 cm.
Therefore, B = (1.67×10–27×1.2×105)/(1.6×10–19×4.175×10–2) = 3×10–2 tesla.
(d) If the magnetic field is decreased from the above value, the radius of its path is increased. The limiting value of the magnetic field upto which the proton is prevented from crossing the field is that which will make the radius of the path equal to the extent of the magnetic field (r = 8.35 cm). The path of the proton in this case is indicated by the dotted semicircle in the figure.
Since the proton will cross the magnetic field on decreasing the field by a very small amount from this limiting value, it is the magnetic field which will allow the proton to just cross the magnetic field. This field (B1 )is given by
B1 = mv/qr where r = 8.35 cm = 0.0835 m
The value of B1 is half the value obtained in part (c) since the radius is doubled.
As promised in the last post we will now discuss some questions (MCQ) on wave motion. Here are the questions:
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(1) When a tuning fork of frequency f is excited and held near one end of a straight pipe of length L open at both ends, the air column in the pipe vibrates in its fundamental mode and is in resonance with the tuning fork. The pipe is now kept vertical in a jar containing water so that half the length of the pipe is inside water. What should be the frequency of the tuning fork to be used to make the air column vibrate in its fundamental mode in resonance with the tuning fork now?
(a) f
(b) 2 f
(c) f/2
(d) f/4
(e) 4 f
In the fundamental mode the air column in the open pipe (pipe open at both ends) vibrates with consecutive antinodes at its ends so that the length L of the pipe is equal to λ/2 where λ is the wave length of sound in air. Therefore, λ = 2L
The air column in the closed pipe (pipe closed at one end) on the other hand vibrates in its fundamental mode, with a node at the closed end (at the water surface inside the pipe) and the neighbouring antinode at the open end so that L/2 = λ/4. Again we obtain λ = 2L. The frequencies in the two cases are same so that the correct option is (a).
(2) A string of length L metre has mass M kg. It is kept strtched under a tension T newton. If a transverse jerk is given at one end of this string how long does it take for the disturbance to reach the other end?
(a) √(T/LM)
(b) L√(T/M)
(c) L√(M/T)
(d) √(LT/M)
(e) √(LM/T)
The time taken (t) is given by
t = L/v where v is the velocity of the disturbance.
But v = √(T/m) where T is the tension and m is the linear density (mass per unit length) of the string. Since m = M/Lwe obtain v = √(TL/M)
Therefore, t = L/√(TL/M)= √(LM/T)
(3) Transverse waves travel along a stretched wire of uniform cross section area A with a speed of 100 ms–1. If the wire were of cross section area A/2 and stretched under the same tension, the speed of the transverse waves would be
(a) 100 ms–1
(b) 200 ms–1
(c) 50 ms–1
(d) 100√2 ms–1
(e) 100/√2 ms–1
Speed (v) of transverse waves in a stretched string is given by
v = √(T/m)where T is the tension in the string and ‘m’ is the linear density (mass per unit length) of the string.
Since m = Aρ where A is the cross section are andρ is the density of the material of the wire, the linear density is directly proportional to the cross section area. When the cross section area is reduced to half the initial value, the linear density is reduced to half the initial value.
With the first wire we have
100 = √(T/m)
With the wire of half the area of cross section the velocity v’ is given by
v’ = √(2T/m) on replacing m in the above expression with m/2
Therefore v’ =100√2 ms–1
(4) When two tuning forks are sounded together 4 beats are heard per second. One tuning fork is of frequency 346 Hz. When its prong is loaded with a little wax, the number of beats is increased to 6 per second. The frequency of the other fork is
(a)352 Hz
(b) 340 Hz
(c)342 Hz
(d) 346 Hz
(e) 350 Hz
The frequency of the unknown fork must be either 342 Hz or 350 Hz since 4 beats are produced initially. When the fork of frequency 346 Hz is loaded with wax, its frequency is reduced. The number of beats then increased since its frequency is lower than that of the unknown fork.
The frequency of the unknown fork must therefore be 350 Hz.
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(5) In the adjoining figures (a), (b), (c), (d) and (e) S is a source of sound of real frequency n and L is a listener. The speeds of the source and the listener are equal (V) in all cases except in (e) where the source is at rest and the listener is moving with speed V (as in the other cases). There is no wind. Pick out the wrong statement from the following:
(a) In the case shown in figure (a) the apparent frequency as heard by the listener is equal to the real frequency of the source.
(b) In the case shown in fig (b) the apparent frequency as heard by the listener is the greatest.
(c) In the case shown in fig (c) the apparent frequency as heard by the listener is the least.
(d) In the case shown in fig (d) the apparent frequency as heard by the listener is less than the real frequency of the source.
(e) In the case shown in fig (e) the apparent frequency as heard by the listener is greater than the real frequency of the source.
When the source moves towards the listener and/or the listener moves towards the source the apparent frequency is increased. When the source moves away from the listener and/or the listener moves away from the source the apparent frequency is decreased. If there is no relative motion between the source and the listener there is no change in the frequency. So statement (d) is incorrect. (Since there is no relative motion between the source and the listener there is no change in the frequency. Note that they are moving with the same velocity).
Option (d) is the answer.
(6) Pure sound notesfrom two sources make the molecules of air at a location vibrate simple harmonically in accordance with the equations
y1 = 0.008 sin (604 π t) and
y2 = 0.007 sin (610 π t) respectively.
The number of beats heard by a person at the location will be
(a) 1
(b) 2
(c) 3
(d) 4
(e) 6
The expressions are of the simplest form of simple harmonic motion,
y = a sin ωt where y is the displacement at the instant t, a is the amplitude, and ω is the angular frequency. The angular frequency ω is related to the linear frequency f as ω = 2πf.
The linear frequencies of the two sounds are therefore 302 Hz and 305 Hz. [2πf1 = 604 π and 2πf2 = 610 π]
The number of beats heard is 305 – 302 = 3 [Option (c)].
“The weak can
never forgive. Forgiveness is the attribute of the strong.”
– Mahatma Gandhi
You need to remember the following points to make you strong in answering multiple choice questions involving wave motion including sound:
(1). We come across three types of waves. They are:
(i) Mechanical waves
(ii) Electromagnetic waves and
(iii) Matter waves
Sound waves, surface waves in water and the waves in a stretched string are examples of mechanical waves. They require a material medium for their propagation.
Electromagnetic waves can propagate even through through empty space. They are propagated with different velocities through different media. Visible light, ultraviolet rays, X-rays, infrared rays, microwaves and radio waves are examples of electromagnetic waves. All electromagnetic waves propagate through free space with the same velocity equal to 3×108ms–1. This is the maximum velocity that can be attained by any particle.
The concept of matter waves is more abstract and is used in explaining the dual nature of matter (de Broglie's theory).
In this post we will concentrate in the study of mechanical waves only.
(2). In a transverse wave the particles of the medium oscillate perpendicular to the direction of propagation of the wave. In a longitudinal wave the particles of the medium oscillate parallel to the direction of propagation of the wave.
Ripples on the surface of water and the waves in a stretched wire when you pluck or strike it are examples of transverse waves. Sound waves are longitudinal waves.
(3). The velocity (v) of a wave is related to its frequency (n) and wave length (λ) as
v = nλ
(4). Speed (v)of transverse waves in a stretched string is given by
v =√(T/m) where T is the tension in the string and ‘m’ is the linear density (mass per unit length) of the string
(5) Frequency (n) of vibration of a stretched string is given by
n = (1/2ℓ)√(T/m) where ℓ is the length of the string.
Note that this is the frequency in the fundamental mode. This frequency is called the fundamental frequency or the frequency of the first harmonic. A string can vibrate in various modes and generally, the frequency is given by
n = (s/2ℓ)√(T/m) where s = 1,2,3,….etc. In the fundamental mode, s = 1.
In the next mode s = 2 and the frequency of vibration is twice the frequency in the fundamental mode. The note produced by the string in this case is called the first overtone or the second harmonic. When s = 3 we obtain the second overtone or the third harmonic.
The possible frequencies of vibration of a given length of a string with a given tension are integral multiples of the fundamental frequency.
(6) Speed of sound (v) in a medium is generally given by v =√(E/ρ) where E is the modulus of elasticity and ρ is the density of the medium.
(i) Newton-Laplace equation for the velocity (v) of sound in a gas is
v = √(γP/ρ) where γ is the ratio of specific heats of the gas, P is the pressure and ρ is the density of the gas.
(ii) Velocity of sound in a solid rod is given by
v =√(Y/ρ) where Y is the young’s modulus and ρ is the density of the solid.
(7) Equation of a progressive wave:
The simplest equation for a wave is that which represents a wave proceeding in a particular direction (let us say, the x-direction) with the particles of the medium vibrating simple harmonically. Unlike in the case of the equation of a simple harmonic motion, the equation for a wave contains ‘x’ in addition to ‘t’ since the equation basically shows the variation of the displacement ‘y’ of any particle of the medium with space and time.
You will encounter the wave equation in various forms. For a progressive wave proceeding along the positive X-direction, the wave equation is
y = A sin [ω(t–x/v) + φ]
where A is the amplitude of the wave, ω is the angular frquency, v is the velocity (of the wave) and φ is the initial phase of the particle of the medium at the origin.
This has some similarity to the general equation of a simple harmonic motion [y = A sin (ωt + φ)] with an initial phaseφ. But the term ωt is replaced by ω(t–x/v) since the state of vibration of the particle of the medium at the origin at any instant t is attained by a particle of the medium at distance x only after a time x/v. The wave equation is thus an equation that gives the state of vibration of the particle of the medium at any location x at any time t.
If the initial phase of the particle at the origin (φ) is taken as zero, the above equation has the following forms:
(a) y = A sin ω(t–x/v)
(b) Since ω = 2π/T where T is the period of the wave (which is the period of vibration of the particles of the medium), the wave equation can be written also as
y = A sin [(2π/T)(t – x/v)]
(c) Since T = λ/v, where λ is the wave length, the wave equation can be written also as
y = A sin [(2π/λ)(vt–x)]
(d) Another form of the wave equation obtained from the above is
y = A sin [2π(t/T – x/ λ)]
(e) The equation y = A sin (ωt – kx) also can be used to represent a progressive wave proceeding along the positive x-direction. This follows from the form shown at (a), where k = ω/v which is equal to2π/λ.
It will be useful to remember that the velocity of the wave, v = Coefficient of t /Coefficient of x
(8) Equation of a progressive wave proceeding in thenegative X-direction is
y = A sin ω(t + x/v)
Note that the negative sign of the term containing x in the case of the equation for a wave proceeding along the positive X-direction is replaced with positive sign. This has to be so in order to ensure the phase lead of the particle at distance x compared to the particle at the origin at any instant t. You can replace the positive sign of the x term in all the wave equations given above to obtain the equation of a wave proceeding in the negative x-direction.
(9) Equation of a stationary wave is y = 2A cos(2πx/λ) sin(2πvt/λ) if the stationary wave is formed by the superposition of a wave with the same wave reflected at a free boundary of the medium (such as the free end of a string or the open end of a pipe).
If the reflection is at a rigid boundary (such as the fixed end of a string or the closed end of a pipe), the equation for the stationary wave formed is
y = – 2A sin(2πx/λ) cos(2πvt/λ).
The negative sign and the inter change of the sine term and the cosine term in this expression (when compared to the expression for the stationary wave formed by reflection at a free boundary) occurred because of the phase change of π suffered due to the reflection at the rigid boundary. The important thing to note is that the amplitude has a space variation between the zero value (at nodes) and a maximum vlue 2A (at the anti nodes). Further, the distance between consecutive nodes or consecutive anti nodes is λ/2.
(10) When two sound waves of nearly equal frequencies and amplitudes, traveling in the same direction, are superimposed, the frequency of the resultant sound heard by a listener is the average of the two frequencies. The intensity of the sound will waver (waxing and waning) at a frequency equal to the difference between the two frequencies (beat frequency). If the beats are to be clearly heard, the difference between the frequencies of the individual waves should not be more than 8 or so.
If n1 and n2 are the individual frequencies (n1> n2), beat frequency = n1– n2
(11)A closed pipe or closed organ pipe in acoustics (branch of physics dealing with sound) means a tube closed at one end. An open pipe or open organ pipe means a tube open at both ends. When a standing wave (stationary wave) is formed in an organ pipe, the closed end will be a node and the open end will be an antinode. This is why the length of the pipe in the fundamental mode is equal to λ/4 (which is the distance between neighbouring node and antinode) in a closed pipe. In an open pipe, in the fundamental mode, the length of the pipe is equal to λ/2 since the consecutive antinodes are located at the ends of the tube.
Note that a closed pipe can produce odd harmonics only where as an open pipe can produce all harmonics. In other words, the frequencies of vibration of the air column in a closed pipe are in the ratio 1: 3 : 5 : 7 : etc., while those in an open pipe are in the ratio 1 : 2 : 3 : 4 : 5 : etc.
(12) The phenomenon by which the frequency of a wave as measured by an observer is changed because of the motion of the source, observer and the medium is called Doppler Effect.
Let the source S of sound (fig.) move with velocity vS, the listener move with velocity vL and the wind blow with velocity w, all in the same direction as shown in the figure.
Wind→w S●→vS L●→vL
The apparent frequency (n’) of sound is then given by
n’ = n(v + w – vL)/ (v + w – vS)
where n is the real frequency of the sound and v is the velocity of sound. It may be noted that the above relation has been derived on the assumption that the source is moving towards the listener, the listener is moving away from the source and the wind is blowing from the source to the listener.
Even if you don’t remember the above expression, you will be able to answer many questions if you note that the apparent frequency increases if the source moves towards the listener or the listener moves towards the source. If they move away from each other, the apparent frequency decreases.
If the wind blows from source to the listener the wind velocity w gets added to the velocity v of sound. If the wind is blowing from the listener to the source, the wind velocity w gets subtracted from the velocity v of sound. If the air is stll (w = 0), the apparent frequency (n’) of sound is given by
n’ = n(v – vL)/ (v – vS)
Light and other electromagnetic radiations do not require a medium for their propagation. Hence the velocity w will be absent in the expression for the apparent frequency. Further, the speed of sound v is to be replaced by the speed of light c and the apparent frequency in the case of Doppler Effect in light is given by
n’ = n(c– vL)/ (c – vS)
Doppler effect in light is used to estimate the recessional velocities of stars and galaxies by measuring the red shift of the spectral lines. When a star moves away from the observer, the frequency of light emitted by the star is decreased and the the wave length is increased. The expansion of the universe is proved from this red shift. The shift in frequency is (n – n’) which is related to the recessional velocity vS as
vS = c(n – n’)/ n’
This can be written in terms of the wave length shift λ’ – λ as
vS = c (λ’–λ)/ λ
Or, vS = cz where z = (λ’–λ)/ λ
z = (λ’– λ)/ λ is called Doppler shift or spectral shift.
In the case of a star moving towards the observer, the apparent frequency is increased and the wave length is decreased giving rise to a blue shift as in the case of one member of a binary star system. The velocity of approach of the star is given by
vS = c(n’ – n)/ n’
Or, vS = c (λ– λ’)/ λ
[Doppler shift of the microwaves reflected by aircraft and automobiles is used for measuring their velocities].
[It won’t be possible for you to remember all the expressions you came across in this post. Many questions at your level can be answered even if you do not remember them. But you will definitely get many useful ideas by examining the expressions given above].
In the next post we will discuss questions in this section. Meanwhile find some useful questions with answers at the following locations:
