Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

: Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

The document Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev is a part of the Course I. E. Irodov Solutions for Physics Class 11 & Class 12.

Q. 331. Two thin concentric wires shaped as circles with radii a and b lie in the same plane. Allowing for a ≪ b, find:
 (a) their mutual inductance;
 (b) the magnetic flux through the surface enclosed by the outside wire, when the inside wire carries a current I. 

Solution. 331. The direct calculation of the Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev is a rather complicated problem, since the configuration of the field itself is complicated. However, the application of the reciprocity theorem simplifies the solution of the problem. Indeed, let the same current i flow through loop 2. Then the magnetic flux created by this current through loop 1 can be easily found.

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Magnetic induction at the centre of the loop,   Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

So, flux throug loop  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

and from reciprocity theorem, 

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev


Q. 332. A small cylindrical magnet M (Fig. 3.95) is placed in the centre of a thin coil of radius a consisting of N turns. The coil is connected to a ballistic galvanometer. The active resistance of the whole circuit is equal to R. Find the magnetic moment of the magnet if its removal from the coil results in a charge q flowing through the galvanometer. 

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Solution. 332. Let  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev be the magnetic moment of the magnet Af. Then the magnetic field due to this magnet is,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

The flux associated with this, when the magnet is along the axis at a distance x from the centre, is

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev
Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

When the flux changes, an e.m.f.   Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev  induced and a current  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev  flows. The total charge q, flowing, as the magnet is removed to infinity from x = 0 is,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

or,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev


Q. 333. Find the approximate formula expressing the mutual inductance of two thin coaxial loops of the same radius a if their centres are separated by a distance l, with l ≫ a.

Solution. 333. If a current l flows in one of the coils, the magnetic field at the centre of the other coil is,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

The flux associated with the second coil is then approximately  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Hence,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev


Q. 334. There are two stationary loops with mutual inductance L12. The current in one of the loops starts to be varied as I1  = αt, where α is a constant, t is time. Find the time dependence I2 (t) of the current in the other loop whose inductance is L2 and resistance R.

Solution. 334. When the current in one of the loop is  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev is induced in the other loop. Then if the current in the other loop is I2 we must have,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

This familiar equation has the solution,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev  which is the required current


Q. 335. A coil of inductance L = 2.0 µH and resistance R = 1.0 Ω is connected to a source of constant Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev = 3.0 V (Fig. 3.96). A resistance R0  = 2.0 Ω is connected in parallel with the coil. Find the amount of heat generated in the coil after the switch Sw is disconnected. The internal resistance of the source is negligible. 

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Solution. 335. Initially, after a steady current is set up, the current is flowing as shown.

In steady condition  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

When the switch is disconnected, the current through R0 changes from i10 to the right, to i20 to the left. (The current in the inductance cannot change suddenly.). We then have the equation,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

This equation has the solution  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

The heat dissipated in the coil is, 

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev
Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev


Q. 336. An iron tore supports N = 500 turns. Find the magnetic field energy if a current I = 2.0 A produces a magnetic flux across the tore's cross-section equal to Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Solution. 336. To find the magnetic field energy we recall that the flux varies linearly with current Thus, when the flux Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev for current i, we can write Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev The total energy inclosed in the field, when the current is /, is

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

The characteristic factor 1/2  appears in this way.


Q. 337. An iron core shaped as a doughnut with round cross-section of radius a = 3.0 cm carries a winding of N = 1000 turns through which a current I = 1.0 A flows. The mean radius of the doughnut is b = 32 cm. Using the plot in Fig. 3.76, find the magnetic energy stored up in the core. A field strength H is supposed to be the same throughout the cross-section and equal to its magnitude in the centre of the cross-section. 

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Solution. 337. We apply circulation theorem

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Thus the total energy,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Given N , l , b we know H , and can find out B from the B - H curve. Then W can be calculated.


Q. 338. A thin ring made of a magnetic has a mean diameter d = 30 cm and supports a winding of N = 800 turns. The crosssectional area of the ring is equal to S = 5.0 cm2. The ring has a cross-cut of width b = 2.0 mm. When the winding carries a certain current, the permeability of the magnetic equals μ = 1400. Neglecting the dissipation of magnetic flux at the gap edges, find:
 (a) the ratio of magnetic energies in the gap and in the magnetic;
 (b) the inductance of the system; do it in two ways: using the flux and using the energy of the field. 

Solution. 338. From  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Also,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Since B is continuous across the gap, B is given by,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev both in the magnetic and the gap. 

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev
Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRevIrodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

So, Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Energy wise; total energy

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

The L, found in the one way, agrees with that, found in the other way. Note that, in calculating the flux, we do not consider the field in the gap, since it is not linked to the winding. But the total energy includes that of the gap.


Q. 339. A long cylinder of radius a carrying a uniform surface charge rotates about its axis with an angular velocity ω. Find the magnetic field energy per unit length of the cylinder if the linear charge density equals λ, and μ, = 1.

Solution. 339. When the cylinder with a linear charge density λ rotates with a circular frequency co, a surface current density (charge / length x time) of  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

The direction of the surface current is normal to the plane of paper at Q and the contribution of this current to the magnetic field at P is  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev is the direction of the current. In magnitude,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev and the  direction of  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev shown.

It’s component,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev cancels out by cylindrical symmetry. The component that survives is,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

where we have used  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev the total solid angle around any point.

The magnetic field vanishes outside the cylinder by similar argument.

The total energy per unit length of the cylinder is,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev


Q. 340. At what magnitude of the electric field strength in vacuum the volume energy density of this field is the same as that of the magnetic field with induction B = 1.0 T (also in vacuum).

Solution. 340. Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev for the electric field,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev for the magnetic field.

Thus,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

when  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev  


Q. 341. A thin uniformly charged ring of radius a = 10 cm rotates about its axis with an angular velocity ω = 100 rad/s. Find the ratio of volume energy densities of magnetic and electric fields on the axis of the ring at a point removed from its centre by a distance l = a. 

Solution. 341. The electric field at P is, 

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

To get the magnetic field, note that the rotating ring constitutes a current i - q ω/2 π, and the corresponding magnetic field at P is,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev


Q. 342. Using the expression for volume density of magnetic energy, demonstrate that the amount of work contributed to magnetization of a unit volume of para- or diamagnetic, is equal to A = — JB/2. 

Solution. 342. The total eneigy of the magnetic field is,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

The second term can be interpreted as the energy of magnetization, and has the density

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev


Q. 343. Two identical coils, each of inductance L, are interconnected (a) in series, (b) in parallel. Assuming the mutual inductance of the coils to be negligible, find the inductance of the system in both cases. 

Solution. 343. (a) In series, the current I flows through both coils, and the total e.m.f. induced, when the current changes is,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

or,  L' = 2L

(b) In parallel, the current flowing through either coil isIrodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev the e.m.f. induced is

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Equating this to  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev


Q. 344. Two solenoids of equal length and almost equal crosssectional area are fully inserted into one another. Find their mutual inductance if their inductances are equal to L1 and L2

Solution. 344. We use  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

So,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev


Q. 345. Demonstrate that the magnetic energy of interaction of two current-carrying loops located in vacuum can be represented as Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRevare the magnetic inductions within a volume element dV, produced individually by the currents of the first and the second loop respectively.

Solution. 345. The interaction energy is 

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Here, if  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev the magnetic field produced by the first of the current carrying loops, and  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev that of the second one, then the magnetic field due to both the loops will be Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev


Q. 346. Find the interaction energy of two loops carrying currents I1 and I2 if both loops are shaped as circles of radii a and b, with a ≪ b. The loops' centres are located at the same point and their planes form an angle θ between them. 

Solution. 346. We can think of the smaller coil as constituting a magnet of dipole moment,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Its direction is normal to the loop and makes an angle θ with the direction of the magnetic field, due to the bigger loop. This magnetic Geld is,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

The interaction energy has the magnitude,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Its sign depends on the sense of the currents.


Q. 347. The space between two concentric metallic spheres is filled up with a uniform poorly conducting medium of resistivity p and permittivity ε. At the moment t = 0 the inside sphere obtains a certain charge. Find:
 (a) the relation between the vectors of displacement current density and conduction current density at an arbitrary point of the medium at the same moment of time;
 (b) the displacement current across an arbitrary closed surface wholly located in the medium and enclosing the internal sphere, if at the given moment of time the charge of that sphere is equal to q. 

Solution. 347. (a) There is a radial outward conduction current Let Q be the instantaneous charge on the inner sphere, then,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Then,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

and  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

The surface integral must be - ve because  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev


Q. 348. A parallel-plate capacitor is formed by two discs with a uniform poorly conducting medium between them. The capacitor was initially charged and then disconnected from a voltage source. Neglecting the edge effects, show that there is no magnetic field between capacitor plates.

Solution. 348. Here also we see that neglecting edge effects,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev Thus Maxwell’s equations reduce  to ,div  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

A general solution of this equation is Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev can be thought of as an  extraneous extraneous magnetic field. If it is zero, Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev


Q. 349. A parallel-plate air condenser whose each plate has an area S = 100 cm2  is connected in series to an ac circuit. Find the electric field strength amplitude in the capacitor if the sinusoidal current amplitude in lead wires is equal to Im. = 1.0 mA and the current frequency equals ω = 1.6-107  s-1.

Solution. 349. Given I = Im sin ωt. We see that
Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev is the amplitude of the electric field and is 7V / cm


Q. 350. The space between the electrodes of a parallel-plate capacitor is filled with a uniform poorly conducting medium of conductivity σ and permittivity ε. The capacitor plates shaped as round discs are separated by a distance d. Neglecting the edge effects, find the magnetic field strength between the plates .at a distance r from their axis if an ac voltage V = Vm, cos cot is applied to the capacitor.

Solution. 350. The electric field between the plates can be written as,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

This gives rise to a conduction current,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

and a displacement current,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

The total current is,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

where,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev on taking the real part of the resultant.

The corresponding magnetic field is obtained by using circulation theorem,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

or, H = Hm cos (ωt + α), where,Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev


Q. 351. A long straight solenoid has n turns per unit length. An alternating current I = Im sin ωt flows through it. Find the displacement current density as a function of the distance r from the solenoid axis. The cross-sectional radius of the solenoid equals R. 

Solution. 351. Inside the solenoid, there is a magnetic field,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

Since this varies in time there is an associated electric field. This is obtained by using,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

For  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

For  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

The associated displacement current density is,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 3 Notes | EduRev

The answer, given in the book, is dimensionally incorrect without the factor ε0.

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