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

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

The document Irodov Solutions: Electromagnetic Induction Maxwell’s Equations- 4 Notes | EduRev is a part of the NEET Course 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.

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

But in this case,

(b) In this

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 ,

then

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

To check this we calculate

This vanishes for x = 0 and for   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.

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

when the conduction current vanishes at the site.

We know that,

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

Thus,

On the other hand

differentiating and using

Thus,

This can be written as,

and   (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.

So,  cannot vanish.

(b) Here also, curl    cannot be uniform.

(c) Suppose for instance,

where  spatially and temporally fixed vector. Then  Generally speaking this contradicts the other equation curl   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   be time dependent.

Q. 356. Demonstrate that the law of electric charge conservation, i.e.  follows from Maxwell's equations.

Solution. 356. From the equation Curl

We get on taking divergence of both sides

But div

Q. 357. Demonstrate that Maxwell's equations  and  are compatible, i.e. the first one does not contradict the second one.

Solution. 357. From

we get on taking diveigence

This is compatible with div

Q. 358. In a certain region of the inertial reference frame there is magnetic field with induction B rotating with angular velocity ω. Find  in this region as a function of vectors ω and B.

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

Then curl,

So,

Hence,

where,

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   frame K. It experiences a force

In the frame K', moving with velocity   to K, the particle is at rest. This means that there must be an electric field   so that the particle experinces a force,

Thus,

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.

Solution. 360. Within the plate, there will appear a   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

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 . Neglecting the magnetic field of appearing charges, find their space and surface densities.

Solution. 361. Choose   along the z-axis, and choose   the cylindrical polar radius vector of a reference point (perpendicular distance from the axis). This point has the velocity,

and experiences a  force, which m ust b e counterbalanced by an electric field,

There must appear a space charge density,

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

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,

Here,   velocity of K', relative to the K frame, in which the particle has velocity

Clearly,   the second equation,

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   Then in K', considering nonrelativistic velocity

So,

In the relativistic case,

Now,

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.

In

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.

In

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,

These two equations can be combined to give,

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   he direction of the z-axis,  The frame K' is moving with velocity  in the x - z plane. Then in the frame K',

The vector along   and the perpendicular vector in the x - z plane is

(a) Thus using

So

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   the z direction, and the velocity   in the x - z plane, then in the K! frame,

We find similarly,

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,

But,
so,

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,   as the fields are mutually perpendicular. Also,

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

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.

and this has the following components,

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,

Also,

From these equations,

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## Physics Class 12

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