AP Physics B– Multiple Choice Questions (for practice) on Atomic Physics and Quantum Effects

As promised in the post dated 23-2-08, I give below some typical multiple choice questions (MCQ) for practice:

(1) Five photons have the following energy values. Which one represents the visible light photon?

(a) 24.8 eV

(b)12.4 eV

(c) 6.2 eV

(d) 2.48 eV

(e) 1.24 eV

Photons in the visible region have wave length range from 4000 Ǻ to 7000 Ǻ (approximately). The product of the energy in eV and the wave length in Angstrom in the case of any photon is 12400 (very nearly).

The energy of the 4000 Ǻ photon in electron volt is 12400/4000 = 3.1 eV.

The energy of the 7000 Ǻ photon in electron volt is 12400/7000 = 1.77 eV.

Therefore, the photon in the visible region is given in option (d).

If you were asked to calculate the wave length of the photon of energy 2.48 eV, you will have

λ = 12400/2.48 = 5000 Ǻ

[You can calculate the wave length using the relation E = hc/λ where E is the energy in joule, h is Planck’s constant, c is the speed of light and λ is the wave length (in metre). In the above problem E = 2.48×1.6×10^–19 joule (on converting the energy in eV into joule), c = 3×10⁸ ms^–1and h = 6.6×10^–34 Js. But this is time consuming].

(2) A laser source gives light output of power P. If the wave length of the laser is λ, the number of photons emitted in a time t in terms of the given parameters and fundamental constants is

(a) Pλt/hc

(b) Phct/λ

(c) Pht/λc

(d) Pλ/hc

(e) Pct/hλ

The energy of a photon is hν = hc/λ where h is Planck’s constant, ν is the frequency of light, c is the speed of light and λ is the wave length.

Number of photons emitted per second = P/(hc/λ) = Pλ/hc.

Therefore, total number of photons emitted in time t = Pλt/hc.

(3) When electromagnetic radiations of wave length λ is incident on a photosensitive surface, the kinetic energy of the photoelectrons emitted from the surface is 2 eV. When the wave length of the incident radiations is 2λ, the kinetic energy of the photoelectrons emitted from the surface is 0.5 eV. The threshold wave length (maximum wave length) for photoelectric emission from the surface is

(a) λ/2

(b) λ

(c) 3λ/2

(d) 2λ

(e) 3λ

From Einstein’s equation, we have for the two cases

hc/λ = hc/λ₀ + 2 eV and

hc/2λ = hc/λ₀ + 0.5 eV

where h is Planck’s constant, c is the speed of light and λ₀ is the threshold wave length. We have expressed the kinetic energy in electron volt itself for convenience, with the understanding that all terms are in electron volt.

Multiplying the second equation by 4 and subtracting the first equation from it, we obtain

hc/λ = 3hc/λ₀ from which λ₀ = 3λ.

(4) According to the Bohr model, electrons of quantum number n = 4 in excited hydrogen atoms can undergo transitions to lower energy states in different ways and give rise to photons of discrete frequencies. How many discrete frequencies are possible in this case?

(a) 3

(b) 4

(c) 6

(d) 8

(e) 12

The possible transitions are shown in the adjoining figure. You will see 6 different transitions. Therefore, there will be six discrete frequencies.

You need not draw an energy level diagram to get the answer for similar questions. {If n is large, the method will be difficult). Remember that the possible number of transitions is n(n – 1)/2.

(5) Singly ionized helium (He⁺) atom is hydrogen like in the sense that a solitary electron revolves around a positively charged nucleus. If the energy of this electron in its first orbit (n = 1) is –54.4 eV, what will be its energy in the first excited state?

(a) –108.8 eV

(b) –27.2 eV

(c) –13.6 eV

(d) –6.8 eV

(e) –3.4 eV

First excited state means the electron is in the second orbit (n = 2).

For a hydrogen like atom, the energy of the electron in the orbit of quantum number ‘n’ is inversely proportional to n². Since the energy in the first orbit is –54.4 eV, the energy in the second orbit will be –54.4/2² eV = –13.6 eV

Few more questions will be posted from this section in the next post.

You will find a useful post on hydrogen like atoms at physicsplus

AP Physics B– Atomic Physics and Quantum Effects – Equations to be Remembered

The topics included under atomic physics and quantum effects are the following:

(i) Photons, the photoelectric effect, Compton scattering, x-rays

(ii) Atomic energy levels

(iii) Wave-particle duality

Remember the following:

(1) Energy of a photon, E = hν = hc/λ where ‘h’ is Planck’s constant, ν is the frequency of light, ‘c’ is the speed of light and λ is the wave length. [Note that the speed and the wave length of light depend on the medium where as the frequency is independent of the medium].

Since the energy of a photon of wave length 1000 Angstrom is very nearly 12.4 electron volt, in the case of any photon, the product of energy in eV and the wave length in Ǻ is 12400. This can be used to calculate the energy or wave length of a photon.

(2) Momentum of photon, p = hν/c = h/λ = E/c

(3) Einstein’s photoelectric equation relates the energy of incident photon to the maximum kinetic energy (K_max) of the photo electron and the work function (Ф) of the photosensitive surface and can be written as

hν = K_max + Ф

Or, hν = K_max + hν₀ where ν₀ is the threshold frequency (minimum frequency of light to initiate photo electron emission from the surface).

The above equation can be written as K_max = h(ν – ν₀) = hc[(1/λ) – (1/λ₀)] where λ₀ is the threshold wave length (maximum wave length of light to initiate photo electron emission from the surface).

The cut-off or stopping potential is the minimum negative potential to be applied on the anode (plate) so that the photoelectric current becomes zero (stops).

In terms of the stopping potential V_s, Einstein’s equation can be written as

eV_s = h(ν – ν₀), since K_max = eV_s.

(4) de Broglie wave length is l = h/p = h/mv where λ is the wave length associated with a particle of momentum ‘p’.

[In the case of an electron accelerated by a small voltageV (so that relativistic mass increase is negligible), the wave length in Angstrom is very nearly √(150/V)]

(5) The minimum wave length (λ) of X-rays produced by an X-ray tube is inversely proportional to the anode voltage V.

Since the entire energy of the impinging electron is converted in to the energy of the X-ray photon of minimum wave length, the energy of the photon must be V electron volt. Therefore, the minimum wave length (λ_min ) in Angstrom of X-rays produced by an X-ray tube operating with anode voltage V volt must be given by

λ_min×V = 12400 [See (1) above].

(6) In Compton effect, the change (dλ) in the wave length of the scattered x-ray photon is given by

dλ = (h/mc) (1– cosφ) where ‘h’ is Planck’s constant, ‘m’ is the mass of the electron, ‘c’ is the speed of light in free space and ‘φ’ is the angle of scattering.

The above equation shows that the maximum change in wave length of the scattered X-ray photon is 2h/mc which happens when the photon is turned back (φ = 180º).

(7) In a hydrogen like atom of atomic number Z, the energy (E_n) of the electron in the n^th orbit is given by

E_n = – 13.6 Z²/n² electron volt. Here n = 1,2,3,4 etc.

Note that the energy is inversely proportional to the square of the quantum number n.

The above energy is the total energy of the electron and is made of negative potential energy and positive kinetic energy. The potential energy value is twice the kinetic energy value (as in the case of planetary motion) and hence the total energy is negative.

(8) Momentum of the electron in the n^th orbit is nh/2π.

(9) The Bohr radius (r₀) is the radius of the innermost orbit in a hydrogen atom and is given by

r₀ = h²ε₀/πme² = 0.53 Ǻ (nearly)

(10) The radius (r) of the n^th orbit in a a hydrogen like atom is given by

r = n²r₀/Z

Note that the orbital radius is directly proportional to the square of the quantum number n.

(11) The orbital period T is related to the orbital radius r as

T² α r³, as in planetary motion.

(12) When an electron undergoes a transition from an orbit of energy E₂ to an orbit of energy E₁, the frequency of the radiation emitted is given by Bohr’s frequency condition,

ν = (E₂ – E₁)/h

Since the emitted photon has linear momentum p = hν/c = h/λ = E/c, the atom receives an equal and opposite recoil momentum.

[When an atom is excited by absorbing a photon of energy hν, the electron in the atom undergoes transition from lower energy E₁ to higher energy E₂ and the same frequency condition (Bohr’s) holds].

(13) Rydberg’s relation for the wave number`ν (number of waver per metre) of the spectral line emitted by a hydrogen atom is

`ν = 1/ λ = R[(1/n₁²) – (1/n₂²)] where n₁ and n₂ are the quantum numbers of the inner and outer orbits.

(14) The important spectral series in the case of hydrogen atom are Lyman, Balmer, Paschen, Brackett and Pfund series. Of these, the Balmer series contains spectral lines in the visible region.

Wave numbers of the spectral lines in the Lyman series are obtained by putting n₁ =1 and n₂ = 2,3,4,5…etc. in the Rydberg relation (since the spectral lines in this series arise due to transitions from outer orbits to the innermost orbit).

For Balmer series, n₁ = 2 and n₂ = 3,4,5,6.…etc. (since the spectral lines in this series arise due to transitions from outer orbits to the second orbit).

For Paschen series, n₁ = 3 and n₂ = 4,5,6.…etc. (since the spectral lines in this series arise due to transitions from outer orbits to the third orbit).

For Brackett series, n₁ = 4 and n₂ = 5,6,7..…etc. (since the spectral lines in this series arise due to transitions from outer orbits to the fourth orbit).

For Pfund series, n₁ = 5 and n₂ = 6,7,8..…etc. (since the spectral lines in this series arise due to transitions from outer orbits to the fifth orbit).

In the next post, I’ll discuss questions from this section. Meanwhile, find some useful multiple choice questions here.

AP Physics B & C- Answer to Free-Response Question on Magnetic Fields

"Only two things are infinite, the universe and human stupidity, and I'm not sure about the former."

- Albert Einstein

 

A free response question (for AP Physics B & C aspirants) involving magnetic fields was posted on 11^th February 2008 for your practice. This was the question:

(a) An electron enters an environment of a non-uniform static magnetic field varying in magnitude and direction from point to point, and comes out of it following an irregular path. Would its final speed be equal to the initial speed if it suffered no collisions with anything during its irregular motion in the magnetic field? Justify your answer.

(b) In a cloud chamber photograph of the electron-positron pair production by a high energy gamma ray photon, two circular tracks emerging from a common point can be seen. The tracks curve in opposite directions in a plane normal to the magnetic field applied in the chamber. But if this event is photographed in a liquid hydrogen bubble chamber, two spiral tracks (instead of circular tracks) are seen. Explain why.

(c) Explain why a charged particle projected into a uniform magnetic field at an arbitrary angle θ with respect to the field follows in general, a helical path. Derive an expression for the ‘pitch’ of the helix.

(d) A positively charged particle of mass ‘m’ and charge ‘q’ traveling with constant velocity ‘v’ in the positive X-direction enters a uniform magnetic field B directed along the negative Z-direction extending from x = x₁ to x = x₂ (fig.). Derive an expression (in terms of the given parameters) for the minimum velocity required for the particle to cross the magnetic field.

As promised, I give the answer below:

(a) The magnetic force on a moving charge is always perpendicular to the magnetic field and hence no work is done by the field. The kinetic energy and the speed of the charged particle therefore remain unchanged.

[You might have noted the adjective ‘static’ with the magnetic field. If the magnetic field has a time variation, there will be induced electromotive force, which will change the speed of the charged particle (as in a betatron)].

(b) In a liquid, the charged particles lose energy by collision to a much greater extent than in a gas. (They cause much greater ionization in a liquid medium). The velocity of the charged particle goes on decreasing and therefore, the radius (r) of the circular path goes on decreasing in accordance with the expression for the radius,

r = mv/qB where ‘m’ is the mass, ‘q’ is the charge and ‘v’ is the speed of the particle and B is the magnetic flux density.Instead of a circular path, a spiral path is therefore followed by the particle.

(c) When a charged particle is projected with velocity ‘v’ in to a magnetic field at some arbitrary angle θ with respect to the field, the velocity component v cosθ, which is along the direction of the magnetic field makes the particle move forward where as the velocity component v sinθ, which is perpendicular to the direction of the magnetic field makes the particle move along a circle of radius ‘r’ given by the equation,

(mv² sin² θ)/r = q(v sinθ)×B

[We have equated the centripetal force to the magnetic force].

Therefore, r = (mv sinθ)/qB

The charged particle therefore moves along a helical path within the magnetic field.

The ‘pitch’ of the helix is obtained by multiplying the forward velocity v cosθ with the period (T) of the circular component of motion. The period of the circular component of motion is given by

T = (2πr/v sinθ) = [2π(mv sinθ/qB)] /v sinθ) = 2πm/qB

Therefore, pitch of the helix = T×v cosθ = (2πm/qB)×v cosθ = (2πmv cosθ)/qB

(d) The charged particle will move along a semicircular path just before crossing the magnetic field (fig.). The direction of the curve is obtained by applying Fleming’s left hand rule for the magnetic force. If the velocity of the particle is slightly greater than the velocity for traversing the semicircle, it will cross the magnetic field. Therefore, the required limiting velocity is the velocity for the motion along the semicircle within the magnetic field region.

The radius of the semicircle is x₂ – x₁. Therefore, the limiting velocity ‘v’ is given by

mv²/r = qvB where r = x₂ – x₁

Therefore, mv²/(x₂ – x₁) = qvB, from which v = qB(x₂ – x₁)/m

AP Physics B & C- Free-Response Practice Question on Magnetic Fields

Here is a free-response question meant for those appearing for AP Physics B as well as AP Physics C examination:

(a) An electron enters an environment of a non-uniform static magnetic field varying in magnitude and direction from point to point, and comes out of it following an irregular path. Would its final speed be equal to the initial speed if it suffered no collisions with anything during its irregular motion in the magnetic field? Justify your answer.

(b) In a cloud chamber photograph of the electron-positron pair production by a high energy gamma ray photon, two circular tracks emerging from a common point can be seen. The tracks curve in opposite directions in a plane normal to the magnetic field applied in the chamber. But if this event is photographed in a liquid hydrogen bubble chamber, two spiral tracks (instead of circular tracks) are seen. Explain why.

(c) Explain why a charged particle projected into a uniform magnetic field at an arbitrary angle θ with respect to the field follows in general, a helical path. Derive an expression for the ‘pitch’ of the helix.

(d) A positively charged particle of mass ‘m’ and charge ‘q’ traveling with constant velocity ‘v’ in the positive X-direction enters a uniform magnetic field B directed along the negative Z-direction extending from x = x₁ to x = x₂ (fig.). Derive an expression (in terms of the given parameters) for the minimum velocity required for the particle to cross the magnetic field.

The above question carries 15 points (3+2+5+5). Try to answer it within 15 minutes. I’ll be back soon with a model answer for you.

AP Physics C- Answer to Free-Response Question Involving Magnetic Field

A free response question involving magnetic field was posted on 7^th January 2008 for your practice. This was the question:

A rectangular, plane, single turn coil ABCD of length 2ℓ and breadth ℓ is arranged as shown, with the plane of the coil in the XY plane. O is the origin of coordinates and the side AB of the coil is along the X-axis OX such that the X-coordinate of the side DA is 2ℓ. A Magnetic field acing in the negative Z-direction and of magnitude B = K/x where K is a constant and x is the x-coordinate, acts in the entire region of the coil.

Now, answer the following questions:

(a) Calculate the magnetic flux linked with the coil in terms of K, x and ℓ.

(b) Calculate the change of magnetic flux through the coil when the coil is rotated through 180º about the side AB.

(c) The coil is moved in the positive X-direction, maintaining its plane in the XY plane. What is the direction of the current induced in the coil? (Put a tick mark against the correct direction)

Clockwise------

Anticlockwise------

Give reason for your answer.

(d) If the coil is moved in the positive Y-direction, maintaining its orientation in the XY plane, what is the emf induced in the coil? Justify your answer.

(e) How will you produce a magnetic field (in the region of the coil) of the type given in this question? (You must consider the magnitude as well as the direction of the magnetic field).

As promised, I give the answer below:

(a) Consider a strip of the area (of the coil) parallel to the side DA at distance x from the Y-axis (fig.).If the width dx of this strip is small, the magnetic field over the entire strip can be assumed constant (equal to K/x). The magnetic flux through the strip is (K/x) ℓdx, since the area of the strip is ℓdx and the magnetic field is directed perpendicular to the plane of the strip.

The total magnetic flux (Ф) through the entire coil is obtained by integrating the above flux between the limits 2ℓ and 4ℓ.

Ф = _2ℓ ∫^4ℓ (K/x) ℓdx = Kℓ [_2ℓ∫^4ℓ (dx/x)] = Kℓ[ln(4ℓ) – ln(2ℓ)]

Or, Ф = Kℓ[ln(4ℓ/2ℓ)] = Kℓ ln2

(b) When the coil is rotated through 180º, the magnetic flux linked with the coil changes from Ф to – Ф so that the change of flux is 2Kℓln2.

(c) When the coil is moved in the positive X-direction, maintaining its plane in the XY plane, the flux through the coil decreases. Lenz’s law demands that this decrease of flux is to be opposed. Therefore, a current is induced in the coil so as to produce a magnetic field in the negative Z-direction, thereby trying to increase the flux. Applying the right hand palm rule, we find that the induced current must be clockwise.

(d) If the coil is moved in the positive Y-direction, maintaining its orientation in the XY plane, the flux through the coil is the same in all positions. The change of flux and hence the induced emf is therefore zero.

(e) The magnetic field given in this question is of the form B = μ₀I/2πr, which is the field produced at distance ‘r’ by a long straight conductor carrying a current I. The straight conductor must be placed along the Y-axis and the current in it must flow in the positive Y- direction.

You will find some useful posts (MCQ) in this section here

AP Physics C- Free Response Question Involving Magnetic Field

The following free response practice question is meant for those preparing for AP Physics C exam:

A rectangular, plane, single turn coil ABCD of length 2ℓ and breadth ℓ is arranged as shown, with the plane of the coil in the XY-plane. O is the origin of coordinates and the side AB of the coil is along the X-axis OX such that the X-coordinate of the side DA is 2ℓ. A Magnetic field acing in the negative Z-direction and of magnitude B = K/x where K is a constant and x is the x-coordinate, acts in the entire region of the coil.

Now, answer the following questions:

(a) Calculate the magnetic flux linked with the coil in terms of K, x and ℓ.

(b) Calculate the change of magnetic flux through the coil when the coil is rotated through 180º about the side AB.

(c) The coil is moved in the positive X-direction, maintaining its plane in the XY plane. What is the direction of the current induced in the coil? (Put a tick mark against the correct direction)

Clockwise------

Anticlockwise------

Give reason for your answer.

(d) If the coil is moved in the positive Y-direction, maintaining its orientation in the XY plane, what is the emf induced in the coil? Justify your answer.

(e) How will you produce a magnetic field (in the region of the coil) of the type given in this question? (You must consider the magnitude as well as the direction of the magnetic field).

The above question carries 15 points and you will get 15 minutes to answer it. The distribution of points may be as 5+2+3+2+3 for parts a,b,c,d and e respectively.

Take it as an exercise and try to answer it within the stipulated time. I’ll be back with the answer shortly.

AP Physics B & C – Multiple Choice Questions involving Magnetic fields

The following questions will be of common interest to students preparing for AP Physics B as well as AP Physics C examinations:

(1) The magnetic field produced at a point P by a long, thick, straight cylindrical copper wire of radius R carrying a steady current I is plotted against the distance of the point P from the centre of the wire. Which graph among the five options shown in the figure represents the correct plot?

Normally, you will consider the magnetic field due to thin straight conductors and you know that the field outside at distance r is given by

B =µ₀I/2πr

If you remember that the field is directly proportional to the current I and inversely proportional to the distance r, you can solve the above problem.

Since the field is inversely proportional to the distance, if the current is constant, the graph must be a hyperbola, as is the case for points outside the wire. In the case of points within the wire ( where r is less than the radius R)

i = I(πr²/ πR²) = I r²/R²

[We have assumed that the current is uniformly distributed across the section of the wire. In the case of any point, we need consider the contribution due to the current passing through a circular section of radius r only, remembering that the magnetic field within a current carrying pipe is zero everywhere. Try and prove it!]

In the case of points within the wire, the magnetic field is obtained by replacing the constant current I with the variable current i in the equation, B =µ₀I/2πr.

Thus, B = µ₀ (I r²/R²) /2πr = µ₀I r/ 2πR²

The field within the wire is thus directly proportional to the distance r.

The correct plot is shown in figure (e).

(2) A free rectangular current carrying loop ABCD is placed as shown, near a long straight conductor PQ carrying a current I. The plane of the loop is the same as the plane containing the straight conductor and two sides of the loop are parallel to the straight conductor. The loop will

(a) move towards the straight conductor

(b) move away from the straight conductor

(c) rotate clockwise

(d) rotate anticlockwise

(e) have translational as well as rotational motion

Since PQ and DA carry like currents (currents in the same direction), they attract each other. You can easily show this by Fleming’s left hand rule (motor rule).. Since PQ and CB carry unlike currents (currents in opposite directions), they repel each other. The attractive force is greater than the repulsive force in this case since the separation between PQ and DA is greater than that between PQ and CB. [Remember that the force is directly proportional to the product of currents and inversely proportional to the separation]. The loop therefore will move towards the straight conductor.

At the loop, the magnetic field produced by PQ is directed perpendicularly into the plane of the figure. Therefore, the magnetic force on side AB is upwards and that on side CD is downwards. Since the magnitude is the same, the resultant of these two forces is zero The lines of action of these two forces are not separated . So, there is no torque. The lines of action of the forces on BC and AD also are not separated and again there is no torque.

The correct option therefore is (a)

(3) Two negative ions of the same charge but different masses are projected vertically upwards with the same velocity from point O (fig.) in a region of space where a uniform magnetic field directed perpendicularly in to the plane of the figure exists. If the gravitational pull is ignored, the trajectory of the heavier ion is

(a) OA

(b) OB

(c) OC

(d) OD

(e) OE

This is a question in which Flemings left hand rule (motor rule) will help you. It is better that you treat the ions to be positive (for the time being) so that you can hold the middle finger of your left hand along the direction of projection (upwards). The forefinger should indicate the direction of the magnetic field and hence must point into the plane of the figure. Then the direction of the thumb which is held perpendicular to the fore finger as well as the middle finger gives the direction of the magnetic force. The direction is leftwards for the positive ion. Since the ions in the question are negative, The deflection is towards right. Since the ions are of the same charge and the same velocity, the heavier ion is deflected less (because of greater inertia). Therefore its path is OD.

[If you remember the expression for radius r, which is

r = mv/qB, you will immediately note that r is greater for the larger mass m.

(4) In a mass spectrograph, positive ions moving with different velocities in the positive X-direction are admitted into a velocity filter in which an electric field of magnitude E is applied in the negative Y-direction. What should be the direction of the magnetic field so that ions of the same velocity pass through the velocity filter without deflection?

(a) Positive Z-direction

(b) Positive Y-direction

(c) Positive X-direction

(d) Negative Y-direction

(e) Negative Z-direction

As you might be knowing, the velocity filter is an arrangement in which electric and magnetic fields are applied at right angles so that the deflection produced by the electric field is cancelled by the deflection produced by the magnetic field. In other words, the electric force and the magnetic force on the ion are equal and opposite.

Since the electric field is in the negative Y-direction, the electric force (on the positive ion) is in the negative Y-direction. The magnetic force must therefore be in the positive Y-direction.

Now, to apply Fleming’s left hand rule, hold the thumb, fore finger and middle finger of your left hand in mutually perpendicular directions such that the middle finger is in the positive X-direction (to indicate the velocity of positive charge) and the thumb is in the positive Y-direction (to indicate the required direction of the magnetic force). The fore finger will then point along the required direction of the magnetic field. You will get it along the negative Z-direction

(5) In the above question, if the electric field is E, what should be the magnitude of the magnetic field so that ions of velocity v and charge q proceed undeflected through the velocity filter?

(a) q/E

(b) Eq/v

(c) E/v

(d) v/E

(e) vq/E

The electric force and the magnetic force on the ion must be equal and opposite. Equating the magnitudes, we have

qE = qvB

Therefore, B = E/v

You can expect more questions in this section in due course.

Meanwhile, find some useful posts in this section at http://www.physicsplus.in

AP Physics B & C - Magnetic Field - Equations to be Remembered

The section ‘Magnetic Fields’ in the AP Physics syllabus contains the following sub sections:

(1) Forces on moving charges in magnetic fields

(2) Forces on current-carrying wires in magnetic fields

(3) Fields of long current-carrying wires

(4) Biot-Savart’s law and Ampere's law (For AP Physics C only)

AP Physics B carries 4% of the total points in this section while AP Physics C carries 10%.

Here are the equations to be remembered in this section:

(1) The magnetic force ‘F’ acting on a particle of charge ‘q’ moving with velocity ‘v’ making an angle ‘θ’ with a magnetic field ‘B’ (fig.) is given by

F = qvB sinθ

The force F is perpendicular to both v and B. In vector form, the above equation is

F = qv×B

Note that bold face characters are used to represent vectors.

When electric and magnetic fields act simultaneously on a charge, the total force on the charge is given by Lorentz force equation,

F = q(v×B + E) where E is the electric field.

(2) The path of a charged particle of mass ‘m’ projected with a velocity ‘v’ perpendicular to a magnetic field B is a circle of radius ‘r’ given by

qvB = mv²/r, where ‘q’ is the charge

[Note that we have equated the magnetic force to the centripetal force]

Therefore, r = mv/qB

If the particle is projected into the magnetic field at an angle other than zero or 90º, the path is a helix of radius ‘r’ given by

r = mv sinθ/qB

(3) The period of circular motion as well as helical motion of a charged particle in a magnetic field is

T = 2πm/qB

(4) The frequency of revolution along the circular path (or helical path) is

f = qB/2πm

This is called the cyclotron frequency

(5) The magnetic force ‘dF’ acting on an elemental length dℓ of a conductor carrying a current ‘I’ placed in a magnetic field B, making an angle ‘θ’ with the magnetic field is given

dF = IdℓB sinθ

The force dF is perpendicular to both dℓ and B. In vector form, the above equation is

dF = I dℓ×B, treating the elemental length dℓ as a vector.

In the case of a straight conductor of length ℓ, the magnetic force is

F= IℓB sinθ which can be written in vector form as

F = I ℓ×B

(6) Force per unit length between two infinitely long parallel current carrying conductors is given by

F = µ₀I₁I₂/2πd, where µ₀ is the permeability of free space, ‘d’ is the separation between the conductors and I₁ and I₂ are the currents in the conductors.

(7) Torque on a plane current carrying coil (current loop) placed in a magnetic field B is

τ = nIAB sinθ, where ‘n’ is the number of turns in the coil, A is the area of the coil, I is the current in the coil and θ is the angle between the area vector and the magnetic field vector. Remember that the area is a vector which has magnitude equal to the area and direction perpendicular to the area.

(8) Magnetic field (dB) due to a current element of length dℓ at a point P distant ‘r’ from the current element is given by Biot-Savart law:

dB = (µ₀/4π) Idℓ sinθ/r²

This magnetic field is perpendicular to the plane containing the current element and the point P.

In vector notation the above equation is

dB = (µ₀/4π) Idℓ ×r/r³

Here r is a vector of length r directed from the current element to the

point P.

The length of the current element also is treated as a vector dℓ of length dℓ with its direction same as that of the current.

(9) The magnetic field due to a straight infinitely long current carrying conductor
at a point P at a perpendicular distance ‘r’ from the conductor is

B =µ₀I/2πr

(10) The magnetic field due to a plane circular current carrying

coil of ‘n’ turns and radius R at a point P on the axis at a distance

‘x’ from the centre of the coil is

B = µ₀nR²I /2(R² + x²)^3/2

The magnetic field at the centre of the coil is

B= µ₀nI/2R

(11) The magnetic field on the axis of an infinitely long straight solenoid at a point P well within the solenoid is

B= µ₀nI, where ‘n’ is the number of turns per metre of the solenoid.

(12) The magnetic field inside a toroid of ‘n’ turns per metre is

B= µ₀nI

(13) Ampere’s circuital law states that the line integral of magnetic flux density over any closed curve is equal to µ₀ times the total current passing through the surface enclosed by the closed curve. This is stated mathematically as

∫B. dℓ = µ₀I (The integration is over the closed path)

Amperes circuital law as modified by Maxwell to accommodate the displacement current flowing through even free space is

∫B. dℓ = µ₀ [I+ ε₀ (dφ_E/dt)], where ε₀ (dφ_E/dt) is the displacement current resulting from the rate of change of electric flux φ_E. ε₀ is the permittivity of free space.

In the next post we will discuss questions in this section.