Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Physics Class 12

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Q. 352. A point charge q moves with a non-relativistic velocity v = const. Find the displacement current density jd at a point located at a distance r from the charge on a straight line
 (a) coinciding with the charge path;
 (b) perpendicular to the path and passing through the charge. 

Solution. 352. In the non-relativistic limit.

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

(a) On a straight line coinciding with the chaige path,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

But in this case,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

(b) In this Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev


Q. 353. A thin wire ring of radius a carrying a charge q approaches the observation point P so that its centre moves rectilinearly with a constant velocity v. The plane of the ring remains perpendicular to the motion direction. At what distance xm, from the point P will the ring be located at the moment when the displacement current density at the point P becomes maximum? What is the magnitude of this maximum density? 

Solution. 353. We have ,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

then  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

This is maximum, when  x = xm = 0,  and minimum at some other value. The maximum displacement current density is

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

To check this we calculate  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

This vanishes for x = 0 and for  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev The latter is easily shown to be a smaller local minimum (negative maximum).


Q. 354. A point charge q moves with a non-relativistic velocity v = const. Applying the theorem for the circulation of the vector H around the dotted circle shown in Fig. 3.97, find H at the point A as a function of a radius vector r and velocity v of the charge.

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Solution. 354. We use Maxwell's equations in the form ,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

when the conduction current vanishes at the site.

We know that,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

where, 2k (1 - cos θ) is the solid angle, formed by the disc like surface, at the charge.

Thus,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

On the other hand Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

differentiating and using  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Thus,   Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

This can be written as,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

and  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev (The sense has to be checked independently.)


Q. 355. Using Maxwell's equations, show that
 (a) a time-dependent magnetic field cannot exist without an electric field;
 (b) a uniform electric field cannot exist in the presence of a timedependent magnetic field;
 (c) inside an empty cavity a uniform electric (or magnetic) field can be time-dependent.

Solution. 355. 

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

So, Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev cannot vanish.

(b) Here also, curl   Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev cannot be uniform.

(c) Suppose for instance,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

where Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev spatially and temporally fixed vector. Then Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev Generally speaking this contradicts the other equation curl  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev for in this case the left hand side is time independent but RHS. depends on time. The only exception is when f (r) is linear function. Then a uniform field Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev  be time dependent.


Q. 356. Demonstrate that the law of electric charge conservation, i.e. Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev follows from Maxwell's equations. 

Solution. 356. From the equation Curl  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

We get on taking divergence of both sides  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

But div  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev


Q. 357. Demonstrate that Maxwell's equations Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev and Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev are compatible, i.e. the first one does not contradict the second one.

Solution. 357. From  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev 

we get on taking diveigence

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

This is compatible with div  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev


Q. 358. In a certain region of the inertial reference frame there is magnetic field with induction B rotating with angular velocity ω. Find Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev in this region as a function of vectors ω and B. 

Solution. 358. A rotating magnetic field can be represented by,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Then curl,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

So,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Hence,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

where,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev


Q. 359. n the inertial reference frame K there is a uniform magnetic field with induction B. Find the electric field strength in the frame K' which moves relative to the frame K with a non-relativistic velocity v, with v⊥B. To solve this problem, consider the forces acting on an imaginary charge in both reference frames at the moment when the velocity of the charge in the frame K' is equal to zero. 

Solution. 359. Consider a particle with charge ey moving with velocity  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev frame K. It experiences a force  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

In the frame K', moving with velocity  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev to K, the particle is at rest. This means that there must be an electric field Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev  so that the particle experinces a force,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Thus,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev


Q. 360. A large plate of non-ferromagnetic material moves with a constant velocity v = 90 cm/s in a uniform magnetic field with induction B = 50 mT as shown in Fig. 3.98. Find the surface density of electric charges appearing on the plate as a result of its motion. 

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Solution. 360. Within the plate, there will appear a  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev force, which will cause charges inside the plate to drift, until a countervailing electric field is set up. This electric field is related to By by E = eB, since v & B are mutually perpendicular, and E is perpendicular to both. The charge density ± σ, on the force of the plate, producing this electric field, is given by

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev


Q. 361. A long solid aluminum cylinder of radius a = 5.0 cm rotates about its axis in a uniform magnetic field with induction B = 10 mT. The angular velocity of rotation equals ω = 45 rad/s, with Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev. Neglecting the magnetic field of appearing charges, find their space and surface densities. 

Solution. 361. Choose  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev along the z-axis, and choose  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev the cylindrical polar radius vector of a reference point (perpendicular distance from the axis). This point has the velocity,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

and experiences a Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev force, which m ust b e counterbalanced by an electric field,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

There must appear a space charge density,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Since the cylinder, as a whole is electrically neutral, the surface of the cylinder must acquire a positive charge of surface density,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev


Q. 362. A non-relativistic point charge q moves with a constant velocity v. Using the field transformation formulas, find the magnetic induction B produced by this charge at the point whose position relative to the charge is determined by the radius vector r. 

Solution. 362. In the reference frame K', moving with the particle,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Here,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev velocity of K', relative to the K frame, in which the particle has velocity Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Clearly,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev the second equation,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev


Q. 363. Using Eqs. (3.6h), demonstrate that if in the inertial reference frame K there is only electric or only magnetic field, in any other inertial frame K' both electric and magnetic fields will coexist simultaneously, with E' ⊥ B'

Solution. 363. Suppose, there is only electric field  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev Then in K', considering nonrelativistic velocity  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

So,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

In the relativistic case,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Now,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRevIrodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev


Q. 364. In an inertial reference frame K there is only magnetic field with induction B = b (yi — xj)/ (x2 + y2), where b is a constant, i and j are the unit vectors of the x and y axes. Find the electric field strength E' in the frame K' moving relative to the frame K with a constant non-relativistic velocity v = vk; k is the unit vector of the z axis. The z' axis is assumed to coincide with the z axis. What is the shape of the field E'

Solution. 364. Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

In  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

The electric field is radial  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev


Q. 365. In an inertial reference frame K there is only electric field of strength E = a (xi + yj)/(x2 + y2), where a is a constant, i and j are the unit vectors of the x and y axes. Find the magnetic induction B' in the frame K' moving relative to the frame K with a constant non-relativistic velocity v = vk; k is the unit vector of the z axis. The z' axis is assumed to coincide with the z axis. What is the shape of the magnetic induction B'?

Solution. 365. Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

In  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev  

The magnetic lines are circular.


Q. 366. Demonstrate that the transformation formulas (3.6h) follow from the formulas (3.6i) at v0 ≪ c. 

Solution. 366. In the non relativistic limit, we neglect v2/ c2 and write,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

These two equations can be combined to give,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev


Q. 367. In an inertial reference frame K there is only a uniform electric field E = 8 kV/m in strength. Find the modulus and direction
 (a) of the vector E', (b) of the vector B' in the inertial reference frame K' moving with a constant velocity v relative to the frame K at an angle α = 45° to the vector E. The velocity of the frame K' is equal to a β = 0.60 fraction of the velocity of light.

Solution. 367. Choose  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev he direction of the z-axis, Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev The frame K' is moving with velocity Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev in the x - z plane. Then in the frame K',

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

The vector along  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev and the perpendicular vector in the x - z plane is 

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

(a) Thus using  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

So  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev


Q. 368. Solve a problem differing from the foregoing one by a magnetic field with induction B = 0.8 T replacing the electric field.

Solution. 368. Choose  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev the z direction, and the velocity  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev in the x - z plane, then in the K! frame,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

We find similarly,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev


Q. 369. Electromagnetic field has two invariant quantities. Using the transformation formulas (3.6i), demonstrate that these quantities are
 (a) EB; (b) E2  — c2B2

Solution. 369. (a) We see that,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

But,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev
so,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev
Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev
Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev


Q. 370. In an inertial reference frame K there are two uniform mutually perpendicular fields: an electric field of strength E = 40 kV/m and a magnetic field induction B = 0.20 mT. Find the electric strength E' (or the magnetic induction B') in the reference frame K' where only one field, electric or magnetic, is observed.
 Instruction. Make use of the field invariants cited in the foregoing problem.

Solution. 370. In this case,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev as the fields are mutually perpendicular. Also,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Thus, we can find a frame, in which E' = 0, and

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRevIrodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev


Q. 371. A point charge q moves uniformly and rectilinearly with a relativistic velocity equal to a β  fraction of the velocity of light (β = v/c). Find the electric field strength E produced by the charge at the point whose radius vector relative to the charge is equal to r and forms an angle θ with its velocity vector.

Solution. 371. Suppose the charge qr moves in the positive direction of the x-axis of the frame K. Let us go over to the moving frame K', at whose origin the charge is at rest. We take the x and x' axes of the two frames to be coincident, and the y & y' axes, to be parallel.

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

and this has the following components,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Now let us go back to the frame K. At the moment, when the origins of the two frames coincide, we take t * 0. Then,

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Also,  Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

From these equations,   Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev

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