Irodov Solutions: Electric Current- 4 Notes | EduRev

Physics Class 12

JEE : Irodov Solutions: Electric Current- 4 Notes | EduRev

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Q. 201. A glass plate totally fills up the gap between the electrodes of a parallel-plate capacitor whose capacitance in the absence of that glass plate is equal to C = 20 nF. The capacitor is connected to a do voltage source V = 100 V. The plate is slowly, and without friction, extracted from the gap. Find the capacitor energy increment and the mechanical work performed in the process of plate extraction.

Solution. 201. Initially, capacitance of the system = C ε

So, initial energy of the system :  Irodov Solutions: Electric Current- 4 Notes | EduRev

and finally, energy of the capacitor : Irodov Solutions: Electric Current- 4 Notes | EduRev

Hence capacitance energy increment,

Irodov Solutions: Electric Current- 4 Notes | EduRev

From energy conservation

Irodov Solutions: Electric Current- 4 Notes | EduRev

(as there is no heat liberation)

Irodov Solutions: Electric Current- 4 Notes | EduRev

Irodov Solutions: Electric Current- 4 Notes | EduRev

Irodov Solutions: Electric Current- 4 Notes | EduRev


Q. 202. A cylindrical capacitor connected to a de voltage source V touches the surface of water with its end (Fig. 3.56). The separation d between the capacitor electrodes is substantially less than their mean radius. Find a height h to which the water level in the gap will rise. The capillary effects are to be neglected.

Irodov Solutions: Electric Current- 4 Notes | EduRev

Solution. 202. If C0 is the initial capacitance of the condenser before water rises in it then

Irodov Solutions: Electric Current- 4 Notes | EduRev

(R is the mean radius and / is the length of the capacitor plates.)

Suppose the liquid rises to a height h in it. Then the capacitance of the condenser is

Irodov Solutions: Electric Current- 4 Notes | EduRev

and energy of the capacitor and the liquid (including both gravitational and electrosatic contributions) is

Irodov Solutions: Electric Current- 4 Notes | EduRev

If the capacitor were not connected to a battery this energy would have to be minimized.
But the capacitor is connected to the battery and, in effect, the potential energy of the whole system has to be minimized. Suppose we increase h by bh. Then the energy of the capacitor and the liquid increases by

Irodov Solutions: Electric Current- 4 Notes | EduRev

and that of the cell diminishes by the quantity Acell which is the p rod uct o f charge flown and V

Irodov Solutions: Electric Current- 4 Notes | EduRev

In equilibrium, the two must balance; so

Irodov Solutions: Electric Current- 4 Notes | EduRev

Hence  Irodov Solutions: Electric Current- 4 Notes | EduRev


Q. 203. The radii of spherical capacitor electrodes are equal to a and b, with a < b. The interelectrode space is filled with homogeneous substance of permittivity ε and resistivity p. Initially the capacitor is not charged. At the moment t = 0 the internal electrode gets a charge q0. Find:
 (a) the time variation of the charge on the internal electrode;
 (b) the amount of heat generated during the spreading of the charge. 

Solution. 203. (a) Let us mentally islolate a thin spherical layer with inner and outer radii r and r + dr respectively. Lines of current at all the points of this layer are perpendicular to it and therefore such a layer can be treated as a spherical conductor of thickness dr and cross sectional area 4 πr2 . Now, we know that resistance,

Irodov Solutions: Electric Current- 4 Notes | EduRev    (1)

Integrating expression (1) between the limits,

Irodov Solutions: Electric Current- 4 Notes | EduRev     (2)

Capacitance of the network, Irodov Solutions: Electric Current- 4 Notes | EduRev    (3)

and   Irodov Solutions: Electric Current- 4 Notes | EduRev   (4)

also,   Irodov Solutions: Electric Current- 4 Notes | EduRev   (5)

From Eqs. (2), (3), (4) and (5) we get, 

Irodov Solutions: Electric Current- 4 Notes | EduRev

Integrating Irodov Solutions: Electric Current- 4 Notes | EduRev

Hence   Irodov Solutions: Electric Current- 4 Notes | EduRev

(b) From energy conservation heat generated, during the spreading of the charge, 

Irodov Solutions: Electric Current- 4 Notes | EduRev


Q. 204. The electrodes of a capacitor of capacitance C = 2.00 μF carry opposite charges q0 = 1.00 mC. Then the electrodes are interconnected through a resistance R = 5.0 MΩ. Find:
 (a) the charge flowing through that resistance during a time interval τ = 2.00 s;
 (b) the amount of heat generated in the resistance during the same interval. 

Solution. 204. (a) Let, at any moment of time, charge on the plates be (q0 - q) then current through the resistor,  Irodov Solutions: Electric Current- 4 Notes | EduRev because the capacitor is discharging. 

or, Irodov Solutions: Electric Current- 4 Notes | EduRev

Irodov Solutions: Electric Current- 4 Notes | EduRev

Now, applying loop rule in the circuit,

Irodov Solutions: Electric Current- 4 Notes | EduRev

or,  Irodov Solutions: Electric Current- 4 Notes | EduRev

or,  Irodov Solutions: Electric Current- 4 Notes | EduRev

At t = 0, q = 0 and at t = τ, q = q

So,  Irodov Solutions: Electric Current- 4 Notes | EduRev

Thus   Irodov Solutions: Electric Current- 4 Notes | EduRev

(b) Amount of heat generated = decrement in capacitance energy

Irodov Solutions: Electric Current- 4 Notes | EduRev


Q. 205. In a circuit shown in Fig. 3.57 the capacitance of each capacitor is equal to C and the resistance, to R. One of the capacitors was connected to a voltage V0  and then at the moment t = 0 was shorted by means of the switch Sw. Find:
 (a) a current I in the circuit as a function of time t; 
 (b) the amount of generated heat provided a dependence I (t) is known.

Irodov Solutions: Electric Current- 4 Notes | EduRev

Solution. 205. Let, at any moment of time, charge flown be q then current  Irodov Solutions: Electric Current- 4 Notes | EduRev

Applying loop rule in the circuit, Irodov Solutions: Electric Current- 4 Notes | EduRev

Irodov Solutions: Electric Current- 4 Notes | EduRev
Irodov Solutions: Electric Current- 4 Notes | EduRev

Irodov Solutions: Electric Current- 4 Notes | EduRev

Hence,  Irodov Solutions: Electric Current- 4 Notes | EduRev

Now, heat liberaled,

Irodov Solutions: Electric Current- 4 Notes | EduRev


Q. 206. A coil of radius r = 25 cm wound of a thin copper wire of length l = 500 m rotates with an angular velocity ω = 300 rad/s about its axis. The coil is connected to a ballistic galvanometer by means of sliding contacts. The total resistance of the circuit is equal to R = 21 Ω. Find the specific charge of current carriers in copper if a sudden stoppage of the coil makes a charge q = 10 nC flow through the galvanometer. 

Solution. 206. in a rotating frame, to first order in ω, the main effect is a coriolis force  Irodov Solutions: Electric Current- 4 Notes | EduRev This unbalanced force will cause electrons to react by setting up a magnetic field Irodov Solutions: Electric Current- 4 Notes | EduRev that the magnetic force  Irodov Solutions: Electric Current- 4 Notes | EduRev balances the coriolis force. 

Thus  Irodov Solutions: Electric Current- 4 Notes | EduRev

The flux associated with this is

Irodov Solutions: Electric Current- 4 Notes | EduRev

where  Irodov Solutions: Electric Current- 4 Notes | EduRev is the number of turns of the ring. If ω changes (and there is time for the electron to rearrange) then B also changes and so Irodov Solutions: Electric Current- 4 Notes | EduRev An emf will be induced and a current will flow. This is

Irodov Solutions: Electric Current- 4 Notes | EduRev

The total charge flowing through the ballastic galvanometer, as the ring is stopped, is

Irodov Solutions: Electric Current- 4 Notes | EduRev

So, Irodov Solutions: Electric Current- 4 Notes | EduRev


Q. 207. Find the total momentum of electrons in a straight wire of length l = 1000 m carrying a current I = 70 A. 

Solution. 207. Let, n0 be the total number of electoms then, total momentum of electoms,

Irodov Solutions: Electric Current- 4 Notes | EduRev     (1)

Now,   Irodov Solutions: Electric Current- 4 Notes | EduRev

Here Sx = Cross sectional area, p = electron charge density, V = volume of sample From (1) and (2)

Irodov Solutions: Electric Current- 4 Notes | EduRev


Q. 208. A copper wire carries a current of density j = 1.0 A/mm2. Assuming that one free electron corresponds to each copper atom, evaluate the distance which will be covered by an electron during its displacement l = 10 mm along the wire. 

Solution. 208. By definition

ne vd = j (where vd is drift velocity, n is number density of electrons.)

Then  Irodov Solutions: Electric Current- 4 Notes | EduRev

So distance actually travelled

Irodov Solutions: Electric Current- 4 Notes | EduRev

(<v> = mean velocity of thermal motion of an electron)


Q. 209. A straight copper wire of length l = 1000 m and crosssectional area S = 1.0 mm2 carries a current I = 4.5 A. Assuming that one free electron corresponds to each copper atom, find:
 (a) the time it takes an electron to displace from one end of the wire to the other;
 (b) the sum of electric forces acting on all free electrons in the given wire. 

Solution. 209. Let, n be the volume density of electrons, then from  Irodov Solutions: Electric Current- 4 Notes | EduRev

Irodov Solutions: Electric Current- 4 Notes | EduRev

So,  Irodov Solutions: Electric Current- 4 Notes | EduRev

(b) Sum of electric forces

Irodov Solutions: Electric Current- 4 Notes | EduRevwhere p is resistivity of die material.

Irodov Solutions: Electric Current- 4 Notes | EduRev


Q. 210. A homogeneous proton beam accelerated by a potential difference V = 600 kV has a round cross-section of radius r = 5.0 mm. Find the electric field strength on the surface of the beam and the potential difference between the surface and the axis of the beam if the beam current is equal to I = 50 mA. 

Solution. 210. From Gauss theorem field strength at a surface of a cylindrical shape equals, Irodov Solutions: Electric Current- 4 Notes | EduRev where X is the linear chaige density.

Now, Irodov Solutions: Electric Current- 4 Notes | EduRev

Also,  Irodov Solutions: Electric Current- 4 Notes | EduRev

or,   Irodov Solutions: Electric Current- 4 Notes | EduRev

Hence  Irodov Solutions: Electric Current- 4 Notes | EduRev

(b) For the point, inside the solid charged cylinder, applying Gauss’ theorem,

Irodov Solutions: Electric Current- 4 Notes | EduRev

or,   Irodov Solutions: Electric Current- 4 Notes | EduRev

So,from  Irodov Solutions: Electric Current- 4 Notes | EduRev

Irodov Solutions: Electric Current- 4 Notes | EduRev

or,  Irodov Solutions: Electric Current- 4 Notes | EduRev

Hence,  Irodov Solutions: Electric Current- 4 Notes | EduRev


Q. 211. Two large parallel plates are located in vacuum. One of them serves as a cathode, a source of electrons whose initial velocity is negligible. An electron flow directed toward the opposite plate produces a space charge causing the potential in the gap between the plates to vary as φ = ax4/3, where a is a positive constant, and x is the distance from the cathode. Find: 
 (a) the volume density of the space charge as a function of x;
 (b) the current density. 

Solution. 211. Between the plates  Irodov Solutions: Electric Current- 4 Notes | EduRev

or,   Irodov Solutions: Electric Current- 4 Notes | EduRev

Irodov Solutions: Electric Current- 4 Notes | EduRev

or,   Irodov Solutions: Electric Current- 4 Notes | EduRev

Let the charge on the electron be - e,

then  Irodov Solutions: Electric Current- 4 Notes | EduRev

as the electron is initially emitted with neligible energy.

Irodov Solutions: Electric Current- 4 Notes | EduRev

So,  Irodov Solutions: Electric Current- 4 Notes | EduRev

(j is measued from the anode to cathode, so the - ve sign.)


Q. 212. The air between two parallel plates separated by a distance d = 20 mm is ionized by X-ray radiation. Each plate has an area S = 500 cm2. Find the concentration of positive ions if at a voltage V = 100 V a current I = 3.0 μ,A flows between the plates, which is well below the saturation current. The air ion mobilities are Irodov Solutions: Electric Current- 4 Notes | EduRev = 1.37 cm2/(V•s) andIrodov Solutions: Electric Current- 4 Notes | EduRev = 1.91 cm2/(V•s). 

 

Solution. 212. 

Irodov Solutions: Electric Current- 4 Notes | EduRev

So by the definition of the mobility

Irodov Solutions: Electric Current- 4 Notes | EduRev

and   Irodov Solutions: Electric Current- 4 Notes | EduRev

(The negative ions move towards the anode and the positive ion towards the cathode and the total current is the sum of the currents due to them.) On the other hand, in equilibrium n+ = n_

So, Irodov Solutions: Electric Current- 4 Notes | EduRev

Irodov Solutions: Electric Current- 4 Notes | EduRev


Q. 213. A gas is ionized in the immediate vicinity of the surface of plane electrode l (Fig. 3.58) separated from electrode 2 by a distance l. An alternating voltage varying with time t as V = V0 sin cot is applied to the electrodes. On decreasing the frequency ω it was observed that the galvano- meter G indicates a current only at ω < ω0 where ω0 is a certain cut-off frequency. Find the mobility of ions reaching electrode 2 under these conditions. 

Irodov Solutions: Electric Current- 4 Notes | EduRev

Solution. 213. Velocity = mobility x field

Irodov Solutions: Electric Current- 4 Notes | EduRev which is positive for 0 < ω t < π 

So, maximum displacement in one direction is

Irodov Solutions: Electric Current- 4 Notes | EduRev

Irodov Solutions: Electric Current- 4 Notes | EduRev

Thus  Irodov Solutions: Electric Current- 4 Notes | EduRev


Q. 214. The air between two closely located plates is uniformly ionized by ultraviolet radiation. The air volume between the plates is equal to V = 500 cm3, the observed saturation current is equal to Isat = 0.48 μA. Find : 
 (a) the number of ion pairs produced in a unit volume per unit time;
 (b) the equilibrium concentration of ion pairs if the recombination coefficient for air ions is equal to r = 1.67.10-6  cm3/s. 

Solution. 214. When the current is saturated, all the ions, produced, reach the plate. 

Then,  Irodov Solutions: Electric Current- 4 Notes | EduRev

(Both positive ions and negative ions are counted here) 

The equation of balance is Irodov Solutions: Electric Current- 4 Notes | EduRev

The first term on the right is the production rate and the second term is the recombination rate which by the usual statistical arguments is proportional to n2 (= no of positive ions x no. of -ve ion). In eauilibrium.

Irodov Solutions: Electric Current- 4 Notes | EduRev

so,  Irodov Solutions: Electric Current- 4 Notes | EduRev


Q. 215. Having been operated long enough, the ionizer producing ni = 3.5.109  cm-3 •s -1  of ion pairs per unit volume of air per unit time was switched off. Assuming that the only process tending to reduce the number of ions in air is their recombination with coefficient r = 1.67.10-6  cm3/s, find how soon after the ionizer's switching off the ion concentration decreases η= 2.0 times. 

Solution. 215. Initially  Irodov Solutions: Electric Current- 4 Notes | EduRev

Since we can assume that the long exposure to the ionizer has caused equilibrium to be set up. Afer the ionizer is switched off,

Irodov Solutions: Electric Current- 4 Notes | EduRev

 or  Irodov Solutions: Electric Current- 4 Notes | EduRev

But  Irodov Solutions: Electric Current- 4 Notes | EduRev

The concentration will decrease by a factor η when

Irodov Solutions: Electric Current- 4 Notes | EduRev

or,   Irodov Solutions: Electric Current- 4 Notes | EduRev


Q. 216. A parallel-plate air capacitor whose plates are separated by a distance d = 5.0 mm is first -charged to a potential difference V = 90 V and then disconnected from a de voltage source. Find the time interval during which the voltage across the capacitor decreases by η = 1.0%, taking into account that the average number of ion pairs formed in air under standard conditions per unit volume per unit time is equal to ni = 5.0 cm-3 •s-1 and that the given voltage corresponds to the saturation current.

Solution. 216. Ions produced will cause charge to decay. Clearly,

Irodov Solutions: Electric Current- 4 Notes | EduRev decrease of charge  Irodov Solutions: Electric Current- 4 Notes | EduRev

or,   Irodov Solutions: Electric Current- 4 Notes | EduRev

Note , that ni, here, is the number of ion pairs produced.


Q. 217. The gap between two plane plates of a capacitor equal to d is filled with a gas. One of the plates emits v0 electrons per second which, moving in an electric field, ionize gas molecules; this way each electron produces α new electrons (and ions) along a unit length of its path. Find the electronic current at the opposite plate, neglecting the ionization of gas molecules by formed ions. 

Solution. 217. If v = number of electrons moving to the anode at distance x, then

Irodov Solutions: Electric Current- 4 Notes | EduRev

Assuming saturation,   Irodov Solutions: Electric Current- 4 Notes | EduRev


Q. 218. The gas between the capacitor plates separated by a dist- ance d is uniformly ionized by ultraviolet radiation so that ni electrons per unit volume per second are formed. These electrons moving in the electric field of the capacitor ionize gas molecules, each electron producing α new electrons (and ions) per unit length of its path. Neglecting the ionization by ions, find the electronic current density at the plate possessing a higher potential.

Solution. 218. Since the electrons are produced uniformly through the volume, the total current attaining saturation is clearly,

Irodov Solutions: Electric Current- 4 Notes | EduRev

Thus,  Irodov Solutions: Electric Current- 4 Notes | EduRev

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