Questions involving Magnetic Fields

The following M.C.Q. which appeared in the All India Pre –Medical/Dental Entrance-2005 (C.B.S.E) test paper will be interesting to you:
An electron moves in a circular orbit with a uniform speed ‘v’. It produces a magnetic field ‘B’ at the centre of the circle. The radius of the circle is proportional to
(a) √(B/v) (b) B/v (c) √(v/B) (d) v/B
The electron, moving along the circular path is equivalent to a current I = e/T = ev/(2πr) where ‘e’ is the electronic charge, T is the period, ‘v’ is the speed and ‘r’ is the radius of the circular motion. The magnetic field at the centre of the circle, B = μ₀I/2r = μ₀ev/4πr² so that r α √(v/B). The correct option is (c).
Now, consider the following simple question. (Be careful however, so that you won’t pick out a wrong answer):
A long straight conductor P carries a current ‘I’ while another parallel long straight conductor Q carries a current 2I in the same direction. The magnetic field midway between them is B. If the current 2I in the conductor Q is switched off, the magnetic field midway between the conductors will be
(a) B/3 (b) B (c) –B (d) –B/2 (e) 2B
The magnetic fields produced by the conductors are in opposition in the region between them. The field produced by Q is twice the field produced by P. If the field produced by Q is 2B, the field produced by P is –B. This is indeed the case, since the net field is B when both currents are switched on. When the current 2I in Q is switched off, the field due to Q alone is present. The answer therefore is –B [Option (c)].
The following question also is simple. It is designed to check your understanding of certain basic things:
A straight conductor of length ‘L’ carrying a current ‘I’ is bent in the form of a semicircle. The magnetic flux density (in tesla) at the centre of the semicircle is approximately
(a) 10^-6I/L (b) 10-7I/L (c) 10^-6π²I/L (d) 10⁷π²I/L (e) π²I/L
The magnetic field at the centre of a single turn circular current carrying coil is μ₀I/2r and hence the magnetic field at the centre of a semicircle is μ₀I/4r. The radius of the semicircle obtained by bending the straight conductor of length L is, r = L/π. On substituting this value, the magnetic field at the centre of the semicircle is μ₀πI/ 4L. Since μ₀ = 4π ×10^-7, the field is 10^-7π² I/L. But π² is approximately equal to 10 so that the answer is 10^-6I/L [Option (a)].
Let us now consider the following M.C.Q. which appeared in the Karnataka Common Entrance test paper of 2005:
The electrons in the beam of a television tube move horizontally from south to north. The vertical component of the earth’s magnetic field points down. The electron is deflected towards
(a) west (b) no deflection (c) east (d) north to south.
It will be convenient to use Fleming’s left hand rule here. You have to hold the fore-finger, middle finger and thumb of your left hand along mutually perpendicular directions so that the fore-finger points along the magnetic field and the middle finger along the direction of the conventional current (and hence opposite to the direction of the electron). The thumb will then give you the direction of deflection, which you can easily obtain as along the east [Option (c)]. You will find more multiple choice questions (with solution) at physicsplus: Magnetic Force on Moving Charges

Kerala Medical Entrance Test 2006- Two Questions on Heat

The following two questions appeared in the Kerala Medical Entrance 2006 test paper:
(1) A thermos flask made of stainless steel contains several tiny lead shots. If the flask is quickly shaken up and down several times, the temperature of lead shots
(a) increases by adiabatic process (b) increases by isothermal process (c) decreases by adiabatic process (d) remains the same (e) first decreases and then increases.
This is a simple question aimed at testing your understanding of a basic concept. In an adiabatic process, there is no heat transfer to or from the system. The system in the question is the lead shots contained in the thermos flask which is thermally insulated from outside. The temperature of the lead shots is therefore increased by adiabatic process [Option (a)].
(2) A body cools from 50˚C to 49.9˚C in 5 s. How long will it take to cool from 40˚C to 39.9˚C? Assume the temperature of the surroundings to be 30˚C and Newton’s law of cooling to be valid.
(a) 2.5 s (b) 5 s (c) 20 s (d) 10 s (e) 15 s
As per Newton’s law of cooling, the rate of cooling of a body is directly proportional to the mean excess of temperature of the body over the surroundings (if the absolute temperature of the body is nearly equal to the absolute temperature of the surroundings).
Therfore, we have (50 – 49.9)/5 α (49.95 – 30) and
(40 – 39.9)/t α (39.95 – 30)
Dividing the first one by the second, we obtain t/5 = 2, very nearly, so that the required time for cooling is 10 s.

Questions on Bohr Model of Hydrogen Atom

Questions based on the Bohr model of hydrogen atom are inevitable in any Medical and Engineering test paper. Consider the following two questions which appeared in A.I.I.M.S.- 2005 test paper:
(1) Solid targets of different elements are bombarded by highly energetic electron beams. The frequency (f) of the characteristic X-rays emitted from different targets varies with atomic number Z as
(a) f α √Z (b) f α Z² (c) f α Z (d) f α Z^3/2
As you might have noted, X-rays are produced by electron transitions from outer orbits to inner orbits. But atoms of high atomic number are required for the production of X-rays since the X-ray photon has much greater energy compared to light photon. The energy of the electron in an orbit is directly proportional to Z². [Note that in a hydrogen like atom, the energy is -13.6 Z²/n² electron volt]. The energy difference between levels also is directly proportional to Z². Since the energy difference is equal to hν where ‘ν’ is the frequency of the radiation emitted, the correct option is (b).
[More rigorous treatment shows that ν α (Z-b)² where b is a constant for a given spectral series].
(2) The ground state energy of hydrogen atom is -13.6 eV. What is the potential energy of the atom in this state?
(a) 0 eV (b) -27.2 eV (c) 1eV (d) 2 eV
The correct option is (b) since the potential energy is twice the kinetic energy. You should note that in all cases of central field motion under inverse square law attractive force, the total energy and potential energy are negative and the potential energy is twice the total energy.
Suppose we modify this question as follows:
The ground state energy of hydrogen atom is -13.6 eV. What is the kinetic energy of the electron in this state?
(a) -13.6 eV (b) -27.2 eV (c) 0 eV (d) 13.6 eV

The correct option is (d) since the kinetic energy is always positive and its magnitude is the same as that of the total energy. This is true in all cases of central field motion under an attractive inverse square law force, as in the case of the motion of a satellite around a planet.

You will find more multiple choice questions (with solution) at physicsplus: Questions on Bohr Atom Model

Optics - Questions on Reflection and Refraction

The average student is usually scared of questions in ray optics because of the wrong notion that the sign convention is some thing that adds to one’s confusion! The best thing to do is to work out as many questions as possible. You will definitely realize that there is nothing to worry about. Let us discuss the following questions:
(1) A glass slab of refractive index 1.5 and thickness 4.5cm is kept close to a concave mirror of focal length 20cm so that the face of the slab is perpendicular to the principal axis of the mirror. A point object is placed on the principal axis so that its image coincides with itself. Then, the distance of the object from the mirror is
(a) 40cm (b) 41.5cm (c) 43cm (d) 37cm (e) 47.5cm
If the glass slab is not interposed, the object is to be placed at the center of curvature of the mirror (at 40cm). When the glass slab is present, the distance of the object is to be increased by the apparent shift t - (t/μ) produced by the slab, so that the ray emerging from the slab appears to come from the center of curvature of the mirror.
Therefore, distance of the object = 40+ 4.5 – (4. 5/1. 5) = 41.5cm [option (b)].

(2) A glass sphere of radius 2cm and refractive index 1.5 has a small air bubble at a distance of 1.2cm from the surface through which it is viewed normally. Its apparent distance from the surface is
(a) 0.8cm (b) 1cm (c) 1.33cm (d) 1.6cm (e) 2.4cm
Don’t be tempted to pick out option (a), under the impression that this is similar to the normal refraction at a glass slab. You have to use the equation, μ₂/v - μ₁/u = (μ₂- μ₁)/R where μ₂= 1, μ₁= 1.5, R = -2cm and u = -1.2cm.
Note that we have imagined the ray from the object (air bubble) to be proceeding in the positive X-direction so that the object distance measured from the pole (which is imagined to be at the origin) to the object is negative. The radius of curvature of the spherical surface also is negative since we have to measure it from this pole to the centre of curvature. We leave the unknown quantity ‘v’ in the equation as it is, with out worrying about its sign.
Substituting these values, v = -1cm. The negative sign shows that the image, which is virtual, is on the same side of the refracting surface as the object is.
(3)An astronomical telescope has an eye piece of focal length 10cm. It gives a magnification of 20 in normal adjustment. The distance between the objective and the eye piece is
(a) 30cm (b) 60cm (c) 110cm (d) 200cm
In normal adjustment, magnification, m = fo/fe where fo and fe are the focal lengths of the objective and eye piece respectively. Therefore, fo = m.fe = 20´10 = 200cm.
Distance between objective and eye piece = fo + fe = 200+10 = 210cm.
(4) The plane faces of two identical plano-convex lenses each having a focal length of 50cm are placed against each other to form a usual biconvex lens. The distance from this lens combination at which an object must be placed to obtain a real, inverted image which has the same size as the object is
(a) 50cm (b) 25cm (c) 100cm (d) 40cm (e) 125cm
This simple question appeared in Kerala Engineering Entrance test paper of 2006. When two identical lenses are kept in contact, the focal length of the combination is reduced to half the value of each lens. (This is true even if the plane surface of one plano-convex lens is placed in contact with the curved surface of the other). The focal length of the combination therefore is 25cm.
A converging lens will produce a real, inverted image of the same size as the object when the object distance is 2f. So, the answer is 50cm.

You will find more multiple choice questions (with solution) from Optics here as well as at physicsplus: Questions (MCQ) on Refraction at Spherical Surfaces