Magnetic Field
Exercises, Q41 to Q50
41. Fe⁺ ions are accelerated through a potential difference of 500 V and are injected normally into a homogeneous magnetic field B of strength 20·0 mT. Find the radius of the circular paths followed by the isotopes with mass numbers 57 and 58. Take the mass of an ion = A(1·6x10⁻²⁷) kg where A is the mass number.
ANSWER: Let the mass of an ion of Fe =m, since it is singly charged, charge on it q =e. If the velocity of the ion after traveling a potential difference V =v, then the kinetic energy of the ion,
½mv² =qV =eV
→v² =2eV/m
→v =√(2eV/m)
Since the ions are injected normally into the uniform magnetic field, they will describe circular paths inside it. The radius of the circle will be given as,
r =mv/qB =mv/eB
→r =(m/eB)*√(2eV/m)
=√(2mV/eB²)
Putting the values,
r=√(2A*1.6x10⁻²⁷*500/1.6x10⁻¹⁹*0.02²)
=√(A*25x10⁻³)
=0.5√(A/10) m
=50√(A/10) cm
For the isotope 57,
r =50√(57/10) cm =119 cm
For the isotope 58,
r =50√(58/10) cm =120 cm.
42. A narrow beam of singly charged potassium ions of kinetic energy 32 keV is injected into a region of width 1.00 cm having a magnetic field strength of 0.500 T as shown in the figure (34-E16). The ions are collected at a screen 95.5 cm away from the field region. If the beam contains isotopes of atomic weights 39 and 41, find the separation between the points where these isotopes strike the screen. Take the mass of a potassium ion = A(1.6x10⁻²⁷) kg where A is the mass number. 
The figure for Q-42
ANSWER: The angle of deviation of the ions will be the same as the angle subtended by the circular arc (path) inside the magnetic region at the center of the circle. Let the angle of deviation =δ. 
Diagram for Q-42
From the figure, sin δ =d/r
DH ≈d/2 =1/2 cm =0.50 cm
DF =DH+HF =0.50 +95.50 cm
=96 cm.
After deflection, the ion strikes at point E on the screen. F is the point on the screen where the ion would have stricken without deflection. From triangle EFD,
EF =DF*tan δ
= DF*sin δ/cos δ
=DF*(d/r)/√{1-(d/r)²}
=DF*d/√(r²-d²) ---------- (i)
We know DF = 96 cm, d =1·0 cm, q =e. We need to find r. We know,
Kinetic energy U =½mv²
→v =√(2U/m)
Also, r =mv/qB =mv/eB
=(m/eB)√(2U/m)
=√(2Um/e²B²)
=√{2*32x10³x1.6x10⁻¹⁹xA*1.6x10⁻²⁷/(1.6x10⁻¹⁹)²*0·5²}
=0.16√(A/10)
For isotope of A.W. 39,
r =0.16√(39/10) m =0.316 m
=31.6 cm
For the isotope of A.W. 41,
r =0.16√(41/10) m =0.324 m
=32·4 cm.
So from (i), the distance EF for the heavier isotope
=96*1/√(32.4²-1²) cm
=2.96 cm
For the lighter isotope, the distance EF,
=96*1/√(31·6²-1²)
=3.04 cm
Hence the separation between the points where these isotopes strike the screen,
=3·04 -2·96 cm
=0·08 cm
=0·80 mm
43. Figure (34-E17) shows a convex lens of focal length 12 cm lying in a uniform magnetic field B of magnitude 1.2 T parallel to its principal axis. A particle having a charge of 2.0x10⁻³ C and mass 2.0x10⁻⁵ kg is projected perpendicular to the plane of the diagram with a speed of 4.8 m/s. The particle moves along a circle with its center on the principal axis at a distance of 18 cm from the lens. Show that the image of the particle goes along a circle and find the radius of that circle. 
The figure for Q-43
ANSWER: The radius of the circle in which the particle moves,
r =mv/qB
=2x10⁻⁵*4·8/(2x10⁻³*1·2)
=4/100 m = 4 cm
Given that the center of this circle lies on the principal axis. At any instant, let us draw the image of the radius joining the particle and the center (assuming it as an object). 
Diagram for Q-43
Since the focal length of the convex lens =12 cm and the object is at 18 cm, the object is placed between focus and the center of curvature. The image will be inverted, beyond the center of curvature on the other side and it will be a real image. Since the particle distance from its center Q remains the same in the plane of the circle, the image distance P'Q' will also remain the same from the principal axis. Only the image will be on the other side of the principal axis. So the image will also move in a circle with its center Q'.
To find the radius of that circle we need to know the distance of the image radius OQ' =v from the lens. u = -18 cm, f =12 cm. From the lens formula,
1/v -1/u =1/f
→1/v -1/(-18) =1/12
→1/v =1/12 -1/18 =1/36
→v =36 cm.
From the triangle PQO and P'Q'O,
P'Q' =OQ'*PQ/OQ
=36*4/18 cm
=8 cm.
44. Electrons emitted with negligible speed from an electron gun are accelerated through a potential difference V along the X-axis. 
The figure for Q-44
These electrons emerge from a narrow hole into a uniform magnetic field B directed along this axis. However, some of the electrons emerging from the hole make slightly divergent angles as shown in the figure (34-E18). Show that these paraxial electrons are refocused on the X-axis at a distance
√{(8π²mV)/(eB²)}.
ANSWER: Since the direction of the magnetic field is along the X-axis, the velocity of electrons moving along the X-axis will be unaffected. We resolve the velocity of other electrons emerging with a divergent angle say δ from the X-axis into two perpendicular components - one along the X-axis v.cosδ and another perpendicular to the X-axis v.sinδ. The electrons with a velocity component v.sinδ perpendicular to the magnetic field will move along a circular path with a speed v.sinδ. So at any instant, the electron will have two speeds, one along the X-axis =v.cosδ another v.sinδ along a circular path. Ignoring the v.cosδ, let us first find the time taken by the electron to touch the X-axis after emerging from the gun and following a path of a circular arc. This arc will be a semicircle.
The length of this semicircle = 2πr. Time taken by the electron to travel this distance,
t =2πr/v.sinδ,
But r =mv/eB,
and K.E. of an electron, eV =½mv²
→v² =2eV/m,
so r =(m/eB)√(2eV/m)
→r =√(2mV/eB²)
Now time t =2π√(2mV/eB²)/v.sinδ
=2π√(2mV/eB²)/{sinδ*√(2eV/m)}
=(2π/sinδ)*√{2mV*m/(2eV*eB²)}
=2πm/eB.sinδ
In this time the distance traveled by the electron along the X-axis
d =v.cosδ*t
=v.cosδ*2πm/eB.sinδ
=√(2eV/m)*2πm/eBtanδ
=√(8π²mV/eB²)*δ/tan
Since δ is a slightly divergent angle that means it is very very small.
Hence tanδ ≈δ, →δ/tanδ =1. Now,
d =√{(8π²mV)/(eB²)}.
We note that the expression for d is free from δ, hence all the paraxial electrons will be refocussed at distance d from the hole.
45. Two particles each having a mass m are placed at a separation d in a uniform magnetic field B as shown in figure (34-E19). They have opposite charges of equal magnitude q. At time t =0, the particles are projected towards each other each with a speed v. Suppose the Coulomb force between the charges is switched off. (a) Find the maximum value vₘ of the projection speed so that the two particles do not collide. (b) What would be the minimum and maximum separation between the particles if v =vₘ/2? (c) At what instant will a collision occur between the particles if v =2vₘ? (d) Suppose v = 2vₘ and the collision between the particles is completely inelastic. Describe the motion after the collision. 
The figure for Q-45
ANSWER: (a) Both the particles will deflect upward in the figure and move in circular paths. Just before the collision the radius of each circular path will be =d/2. See diagram below:-
Diagram for Q-45(a)
We know r =mv/qB. Hence,
d/2 =mvₘ/qB
→vₘ =qBd/2m.
(b) When v =vₘ/2 =qBd/4m, The radius of the circular paths will be,
r =mv/qB =(m/qB)*(qBd/4m)
→r =d/4.
The minimum separation will be when both the particles have moved a quarter of the circular path and it will be equal to,
d -d/4 -d/4
=d -d/2
=d/2.
The maximum separation will be when both the particles have moved three quarters of the circular path. The maximum sparation will be equal to,
d +d/4 +d/4
=d +d/2
=3d/2.
(c) When v =2vₘ =2*qBd/2m
→v =qBd/m,
radius of the circular path, r =mv/qB
→r =(m/qB)*(qBd/m)
→r =d. 
The diagram is shown for
one particle only with charge q,
other will have a similar motion
one particle only with charge q,
other will have a similar motion
The collision will occur when both the particles have moved a horizontal distance of d/2. In the figure, GP =d/2. Suppose at this instant the arc traveled by the particles make an angle δ at the center of the circular path. From geometry in the figure,
sin δ =GP/PE =(d/2)/d =1/2 =sin π/6
So δ =π/6.
Length of this arc =rδ =d*π/6 =πd/6.
Time spent in covering this arc length,
t =distance/speed
=(πd/6)/(qBd/m)
=πm/6qB.
(d) Since the motion is completely inelastic, both the particles will stick togather after collision. Just bfore this instant, each particle will have a horizonat component of velocity =v.cosδ but opposite in direction. Suppose after collision, the joint horizontal velocity =v'. Applying the conservation of momentum principle in horizontal direction,
mv.cosδ +m(-v.cosδ) =2m*v'
→2mv' =0
→v' =0.
So after sticking togethe the joint particle has no velocity in the horizontal direction. Now suppose that the joint velocity in upward direcion (in the plane of figure) =v". Just before the collision, each particle has an upward velocity,
= v.sin δ
From the conservation of momentum in upward direction,
2mv" =mv.sin δ +mv.sin δ
→v" =v.sin δ
=2vₘ*sin π/6
=2vₘ*(1/2)
=vₘ
So after collision the joint mass will move upward in the plane of figure from the point of collision P in a straight line with a velocity vₘ.
46. A uniform magnetic field of magnitude 0.20 T exists in space from east to west. With what speed should a particle of mass 0.010 g and having a charge 1.0x10⁻⁵ C be projected from south to north so that it moves with a uniform velocity?
ANSWER: From the right hand rule, the magnetic force on the particle will be in the vertically upward direction. The particle will move with a uniform velocity only if the weight of the particle balances the magnetic force on it so that the resultant force on the particle is zero.
Let the required speed of the particle be v. Then,
qvB =mg
→v =mg/qB
=(1.0x10⁻⁵)*9.8/(1.0x10⁻⁵)*0.2 m/s
=49 m/s.
47. A particle moves in a circle of diameter 1.0 cm under the action of a magnetic field of 0.40 T. An electric field of 200 V/m makes the path straight. Find the charge/mass ratio of the particle.
ANSWER: The magnetic force on the particle F =qvB.
Since the radius of the circular path under the magnetic field is,
r =mv/qB →v =qBr/m =qBd/2m,
{r =d/2}
So F =qB*qBd/2m
=q²B²d/2m
Since the given magnetic field makes the path straight, so the coulumb force is equal in magnitude but opposite in direction on the particle. The coulumb force,
F' =qE
Here F =F'
→q²B²d/2m =qE
→q/m =2E/B²d
=2*200/(0.40²*0.01)
=2.5x10⁵ C/kg.
It is the charge/mass ratio of the particle.
48. A proton goes undeflected in a crossed electric and magnetic field (the fields are perpendicular to each other) at a speed of 2.0x10⁵ m/s. The velocity is perpendicular to both fields. When the electric field is switched off, the proton moves along a circle of radius 4.0 cm. Find the magnitude of the electric and the magnetic fields. Take the mass of the proton =1.6x10⁻²⁷ kg.
ANSWER: Given, v =2.0x10⁵ m/s,
q =e =1.6x10⁻¹⁹ C, under magnetic field only, r =4.0 cm =0.04 m. Then,
E =? and B =?
r =mv/qB =mv/eB -------- (i)
→B =mv/er
=1.6x10⁻²⁷*2x10⁵/(1.6x10⁻¹⁹*0.04) T
=0·05 T.
Magnetic force F =evB
Under the both fields in action, the path of the proton is a straight line. So both forces are equal and opposite.
Coulumb force, F' =eE
So, F' =F
→eE =evB
→E =vB
=2·0x10⁵*0.05 V/m
=1·0x10⁴ V/m.
49. A particle having a charge of 5.0 µC and mass of 5.0x10⁻¹² kg is projected with a speed of 1.0 km/s in a magnetic field of magnitude 5.0 mT. The angle between the magnetic field and the velocity is sin⁻¹(0.90). Show that the path of the particle will be a helix. Find the diameter of the helix and its pitch.
ANSWER: The angle between the magnetic field and the velocity,
δ =sin⁻¹(0·90)
→sin δ =0·90.
Since the velocity is not perpendicular to the magnetic field, it will have uniform velocity along the magnetic field
=v.cos δ
Due to the perpendicular component of the velocity v.sin δ, it will have an circular motion. But oweing to the two types of motion the circular path will not be closed and the actual path will be a helix.
For the circular motion under magnetic field,
r =mv.sinδ/qB
→d =2mv/qB
=2*5·0x10⁻¹²*10³*0·90/(5·0x10⁻⁶*5x10⁻³) m
=0·36 m
=36 cm =Diameter of the helix.
To find out the pitch of the helix we first need to know the time period of one revolution. Distance traveled in one revolution =πd =0·36π m.
Speed =v.sinδ =1000*0·9 m/s
=900 m/s.
Time needed to cover one revolution
t =Distance/speed
=0·36π/900 s
=1·25x10⁻³ s
The pitch of the helix will be the distanced moved along the magnetic field in this time t. Uniform component of the velocity along the magnetic field =v.cos δ =v√(1-sin²δ)
=1000*√{1 -(0·90)²} m/s
=436 m/s
Hence pitch =436*1·25x10⁻³ m
=0·545 m
≈55 cm.
50. A proton projected in a magnetic field of 0.020 T travels along a helical path of radius 5.0 cm and pitch 20 cm. Find the components of the velocity of the proton along and perpendicular to the magnetic field. Take the mass of the proton =1.6x10⁻²⁷ kg.
ANSWER: Let the component of velocity along the magnetic field =v. Given that pitch of the helical path =20 cm =0·20 m.
Hence time taken to complete one revolution, t =0·20/v.
Radius of the helical path =5 cm
=0·05 m
The distance covered in one revolution =2πr
=2π*0·05 m =0·10π m.
So the speed of the particle in the circular motion,
v' =distance/time
→v' =0·10π/(0·20/v)
=πv/2 m/s.
Given B =0·020 T
For the circular motion of the proton under magnetic field,
r =mv'/eB =πmv/2eB
→v =2eBr/πm
=2*1.6x10⁻¹⁹*0·02*0·05/(π*1.6x10⁻²⁷)
=6·4x10⁴ m/s.
As assumed in the beginning it is the velocity component along the magnetic field.
The velocity component perpendicular to the magnetic field, v' =πv/2
→v' =π*6·4x10⁴/2
=1·0x10⁵ m/s.
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CHAPTER- 34- Magnetic Field
CHAPTER- 33- Thermal and Chemical Effects of Electric Current
CHAPTER- 34- Magnetic Field
CHAPTER- 33- Thermal and Chemical Effects of Electric Current
CHAPTER- 32- Electric Current in ConductorsCHAPTER- 31- CapacitorsCHAPTER- 30- Gauss's Law
CHAPTER- 29- Electric Field and Potential
CHAPTER- 28- Heat Transfer
OBJECTIVE -I
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Questions for Short Answer
OBJECTIVE-I
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EXERCISES - Q-11 to Q-18
CHAPTER- 24-Kinetic Theory of Gases
CHAPTER- 23 - Heat and Temperature
CHAPTER- 21 - Speed of Light
CHAPTER- 20 - Dispersion and Spectra
CHAPTER- 19 - Optical Instruments
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 32- Electric Current in Conductors
CHAPTER- 31- Capacitors
CHAPTER- 30- Gauss's Law
CHAPTER- 29- Electric Field and Potential
CHAPTER- 28- Heat Transfer
CHAPTER- 26-Laws of Thermodynamics
CHAPTER- 25-CALORIMETRY
Questions for Short Answer
OBJECTIVE-I
OBJECTIVE-II
CHAPTER- 24-Kinetic Theory of Gases
CHAPTER- 23 - Heat and Temperature
CHAPTER- 21 - Speed of Light
CHAPTER- 20 - Dispersion and Spectra
CHAPTER- 19 - Optical Instruments
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
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CHAPTER- 5 - Newton's Laws of Motion
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CHAPTER- 4 - The Forces
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CHAPTER- 2 - "Physics and Mathematics"
CHAPTER- 2 - "Physics and Mathematics"
