# Irodov Solutions: Electromagnetic Induction: Maxwell’s Equations - 1 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

## JEE: Irodov Solutions: Electromagnetic Induction: Maxwell’s Equations - 1 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

The document Irodov Solutions: Electromagnetic Induction: Maxwell’s Equations - 1 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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Q. 288. A wire bent as a parabola y = ax2 is located in a uniform magnetic field of induction B, the vector B being perpendicular to the plane x, y. At the moment t = 0 a connector starts sliding translationwise from the parabola apex with a constant acceleration w (Fig. 3.78). Find the emf of electromagnetic induction in the loop thus formed as a function of y.

Solution. 288. Obviously, from Lenz's law, the induced current and hence the induced e.m.f. in the loop is anticlockwise.

From Faraday's law of electromagnetic indcution,

Q. 289. A rectangular loop with a sliding connector of length l is located in a uniform magnetic field perpendicular to the loop plane (Fig. 3.79). The magnetic induction is equal to B. The connector has an electric resistance R, the sides AB and CD have resistances R1 and R2 respectively. Neglecting the self-inductance of the loop, find the current flowing in the connector during its motion with a constant velocity v.

Solution. 289. Let us assume,    directed into the plane of the loop. Then the motional e.m.f.

and directed in the same of

As R1 and R2 are in parallel connections.

Q. 290. A metal disc of radius a = 25 cm rotates with a constant angular velocity ω = 130 rad/s about its axis. Find the potential difference between the centre and the rim of the disc if
(a) the external magnetic field is absent;
(b) the external uniform magnetic field of induction B = 5.0 mT is directed perpendicular to the disc.

Solution. 290. (a) As the metal disc rotates, any free electron also rotates with it with same angular velocity ω, and that's why an electron must have an acceleration ω2r directed towards the disc's centre, where r is separation of the electron from the centre of the disc. We know from Newton's second law that if a particle has some acceleration then there must be a net effecetive force on it in the direction of acceleration. We also know that a charged particle can be influenced by two fields electric and magnetic. In our problem magnetic field is absent hence we reach at the conslusion that there is an electric field near any electron and is directed opposite to the acceleration of the electron.
If E be the electric field strength at a distance r from the centre of the disc, we have from Newton's second law.

and the notential difference

Thus

(b) When field   present, by definition, of motional e.m.f.

Hence the sought potential difference,

Thus

(In general   so we can neglect the effect discussed in (1) here).

Q. 291. A thin wire AC shaped as a semi-circle of diameter d = 20 cm rotates with a constant angular velocity ω = 100 rad/s in a uniform magnetic field of induction B = 5.0 mT, with  The rotation axis passes through the end A of the wire and is perpendicular to the diameter AC. Find the value of a line integral  along the wire from point A to point C. Generalize the obtained result.

Solution. 291. By definition,

So,

But, v = ωr, where r is the perpendicular distance of the point from A.

Hence,

This result can be generalized to a wire AC of arbitary planar shape. We have

d being AC and  measured from A.

Q. 292. A wire loop enclosing a semi-circle of radius a is located on the boundary of a uniform magnetic field of induction B (Fig. 3.80). At the moment t = 0 the loop is set into rotation with a constant angular acceleration p about an axis 0 coinciding with a line of vector B on the boundary. Find the emf induced in the loop as a function of time t. Draw the approximate plot of this function. The arrow in the figure shows the emf direction taken to be positive.

Solution. 292. Flux at any moment of time,

where φ is the sector angle, enclosed by the field.

Now, magnitude of induced e.m.f. is given by,

where ω is the angular v elocity o f the disc. But as it starts rotating from rest at t = 0 with an angular acceleration p its angular velocity ω (f) = βt. So,

According to Lenz law the first half cycle current in the loop is in anticlockwise sense, and in subsequent half cycle it is in clockwise sense.

Thus in general,   where n in number of half revolutions.

The plot     is shown in the answer sheet.

Q. 293. A long straight wire carrying a current I and a II-shaped conductor with sliding connector are located in the same plane as shown in Fig. 3.81. The connector of length l and resistance R slides to the right with a constant velocity v. Find the current induced in the loop as a function of separation r between the connector and the straight wire. The resistance of the H-shaped conductor and the selfinductance of the loop are assumed to be negligible.

Solution. 293. Field, due to the current carrying wire in the region, right to it, is directed into the plane of the paper and its magnitude is given by,

where r is the perpendicular distance from the wire.

As B is same along the length of the rod thus motional e.m.f.

and it is directed in the sense of

So, current (induced) in the loop,

Q. 294. A square frame with side a and a long straight wire carrying a current I are located in the same plane as shown in Fig. 3.82. The frame translates to the right with a constant velocity v. Find the emf induced in the frame as a function of distance x.

Solution. 294. Field, due to the current carrying wire, at a perpendicular distance x from it is given by,

Motional e.m.f is given by

There will be no induced e.m.f. in the segments (2) and (4)

as,   magnitude of e.m.f. induced in 1 and 3, will be

respectively, and their sense will be in the direction of

So, e.m .f., induced in the network

Q. 295. A metal rod of mass m can rotate about a horizontal axis O, sliding along a circular conductor of radius a (Fig. 3.83). The arrangement is located in a uniform magnetic field of induction B directed perpendicular to the ring plane. The axis and the ring are connected to an emf source to form a circuit of resistance R. Neglecting the friction, circuit inductance, and ring resistance, find the law according to which the source emf must vary to make the rod rotate with a constant angular velocity ω.

Solution. 295. As the rod rotates, an emf.

is induced in it The net current in the conductor is then

A magnetic force will then act on the conductor of magnitude BI per unit length. Its direction will be normal to B and the rod and its torque will be

Obviously both magnetic and mechanical torque acting on the C.M. of the rod must be equal but opposite in sense. Then

for equilibrium at constant ω

(The answer given in the book is incorrect dimensionally.)

Q. 296. A copper connector of mass m slides down two smooth copper bars, set at an angle a to the horizontal, due to gravity (Fig. 3.84). At the top the bars are interconnected through a resistance R. The separation between the bars is equal to l. The system is located in a uniform magnetic field of induction B, perpendicular to the plane in which the connector slides. The resistances of the bars, the connector and the sliding contacts, as well as the self-inductance of the loop, are assumed to be negligible. Find the steady-state velocity of the connector.

Solution. 296. From Lenz’s law, the current through the connector is directed form A to B. Here   between A and B

where v is the velocity of the rod at any moment.

For the rod, from Fx = mwx

or,

For steady state, acceleration of the rod must be equal to zero.

Hence,    (1)

But,

From (1) and (2)

Q. 297. The system differs from the one examined in the foregoing problem (Fig. 3.84) by a capacitor of capacitance C replacing the resistance R. Find the acceleration of the connector.

Solution. 297. From Lenz’s 1 aw, the current through the copper bar is directed from 1 to 2 or in other words, the induced crrrent in the circuit is in clockwise sense.

Potential difference across the capacitor plates,

Hence, the induced current in the loop,

But the variation of magnetic flux through the loop is caused by the movement of the bar. So, the induced e.m.f.

and,

Hence,

Now, the forces acting on the bars are the weight and the Ampere’s force, where

From Newton’s second law, for the rod, Fx = mwx

or,

Hence

Q. 298. A wire shaped as a semi-circle of radius a rotates about an axis OO' with an angular velocity ω in a uniform magnetic field of induction B (Fig. 3.85). The rotation axis is perpendicular to the field direction. The total resistance of the circuit is equal to R. Neglecting the magnetic field of the induced current, find the mean amount of thermal power being generated in the loop during a rotation period.

Solution. 298. Flux of   at an arbitrary moment of time t :

and induced current,

N ow thermal power, generated in the circuit, at the moment t = t :

and mean thermal power generated,

Note : The claculation of  which can also be checked by using motional emf is correct even though the conductor is not a closed semicircle , for the flux linked to the rectangular part containing the resistance R is not changing. The answer given in the book is off by a factor 1/4.

Q. 299. A small coil is introduced between the poles of an electromagnet so that its axis coincides with the magnetic field direction. The cross-sectional area of the coil is equal to S = 3.0 mm2, the number of turns is N = 60. When the coil turns through 180° about its diameter, a ballistic galvanometer connected to the coil indicates a charge q = 4.5 μC flowing through it. Find the magnetic induction magnitude between the poles provided the total resistance of the electric circuit equals R = 40 Ω.

Solution. 299. The flux through the coil changes sign. Initially it is BS per turn.

Finally it is - BS per turn. Now if flux is  an intermediate state then the current at that moment will be

So charge that flows during a sudden turning of the coil is

Hence,

Q. 300. A square wire frame with side a and a straight conductor carrying a constant current I are located in the same plane (Fig. 3.86). The inductance and the resistance of the frame are equal to L and R respectively. The frame was turned through 180° about the axis OO' separated from the current-carrying conductor by a distance b. Find the electric charge having flown through the frame.

Solution. 300. According to Ohm’s law and Faraday’s law of induction, the current i0 appearing in the frame, during its rotation, is determined by the formula,

Hence, the required amount of electricity (charge) is,

Since the frame has been stopped after rotation, the current in it vanishes, and hence Δ i0 = 0. It remains for us to find the increment of the flux   through the frame

Let us choose the normal   the plane of the frame, for instance, so that in the final position,   directed behind the plane of the figure  (along

Then it can be easily seen that in the final position,   while in the initial position,  (th e normal is op posite to   and  turns out to be simply equal to the fulx  through the surface bounded b y the final and initial positions of the frame :

where B is a function of r, whose form can be easily found with the help of the theorem of circulation. Finally omitting the minus sign, we obtain,

Q. 301. A long straight wire carries a current I0. At distances a and b from it there are two other wires, parallel to the former one, which are interconnected by a resistance R (Fig. 3.87). A connector slides without friction along the wires with a constant velocity v. Assuming the resistances of the wires, the connector, the sliding contacts, and the self-inductance of the freme to be negligible, find:
(a) the magnitude and the direction of the current induced in the connector;
(b) the force required to maintain the connector's velocity constant.

Solution. 301. As   due to the straight current carrying wire, varies along the rod (connector) and enters linerarly so, to make the calculations simple   made constant by taking its average value in the range [a, b].

(a) The flux of    through the loop due to the movement of the connector. According to Lenz’s law, the current in the loop will be anticlockwise. The magnitude of motional e.m.f.

So, induced current

(b) The force required to maintain the constant velocity of the connector must be the magnitude equal to that of Ampere’s acting on the connector, but in opposite direction.

So,

and will be directed as shown in the (Fig.)

Q. 302. A conducting rod AB of mass m slides without friction over two long conducting rails separated by a distance l (Fig. 3.88). At the left end the rails are interconnected by a resistance R. The system is located in a uniform magnetic field perpendicular to the plane of the loop. At the moment t = 0 the rod AB starts moving to the right with an initial velocity vo. Neglecting the resistances of the rails and the rod AB, as well as the self-inductance, find:
(a) the distance covered by the rod until it comes to a standstill;
(b) the amount of heat generated in the resistance R during this process.

Solution. 302. (a) The flux through the loop changes due to the movement of the rod AB. Recording to Lenz’s law current should be anticlockwise in sense as we have assumed   directed into the plane o f the loop. The motion e.m .f

and,  induced current

From Newton’s law in projection form fx =  mwx

or,

(b) From equation of energy conservation; Ef - Ei + Heat liberated = A cell + Aext

Q. 303. A connector AB can slide without friction along a Hshaped conductor located in a horizontal plane (Fig. 3.89). The connector has a length l, mass m, and resistance R. The whole system is located in a uniform magnetic field of induction B directed vertically. At the moment t = 0 a constant horizontal force F starts acting on the connector shifting it translationwise to the right. Find how the velocity of the connector varies with time t. The inductance of the loop and the resistance of the II-shaped conductor are assumed to be negligible.

Solution. 303. With the help of the calculation, done in the previous problem, Ampere’s force on the connector,

Now from Newton’s second law,

or,

Thus

Q. 304. Fig. 3.90 illustrates plane figures made of thin conductors which are located in a uniform magnetic field directed away from a reader beyond the plane of the drawing. The magnetic induction starts diminishing. Find how the currents induced in these loops are directed.

Solution. 304. According to Lenz, the sense of induced e.m.f. is such that it opposes the cause of change of flux. In our problem, magnetic field is directed away from the reader and is diminishing.

So, in figure (a), in the round conductor, it is clockwise and there is no current in the connector

In figure (b) in the outside conductor, clockwise.

In figure (c) in both the conductor, clockwise; and there is no current in the connector to obey the charge conservation.

In figure (d) in the left side of the figure, clockwise.

Q. 305. A plane loop shown in Fig. 3.91 is shaped as two squares with sides a = 20 cm and b = 10 cm and is introduced into a uniform magnetic field at right angles to the loop's plane. The magnetic induction varies with time as B = B0  sin ωt, where B0 = 10 mT and ω = 100 s-1 . Find the amplitude of the current induced in the loop if its resis- tance per unit length is equal to p = 50 mΩ/m. The inductance of the loop is to be neglected.

Solution. 305. The loops are connected in such a way that if the current is clockwise in one, it is anticlockwise in the other. Hence the e.m.f. in loop b opposes the e.m.f. in loop a.

Hence, net e.m.f. in the circuit as both the e.m.f’s are in opposite  sense, and resistance of the circuit = 4 (a + b) p

Therefore, the amplitude of the current

Q. 306. A plane spiral with a great num- ber N of turns wound tightly to one another is located in a uniform magnetic field perpendicular to the spiral's plane. The outside radius of the spiral's turns is equal to a. The magnetic induction varies with time as B = B0 sin ωt, where Band ω are constants. Find the amplitude of emf induced in the spiral

Solution. 306. The flat shape is made up of concentric loops, having different radii, varying from 0 to a.

Let us consider an elementary loop of radius r, then e.m.f. induced due to this loop

and the total induced e.m.f.,

(1)

where π r2 ω cos ωt is the contribution of one turn of radius r and dN is the number of turns in the interval [r, r + dr].

So,      (2)

From (1) and (2),

Maximum value of e.m.f. amplitude

Q. 307. A H-shaped conductor is located in a uniform magnetic field perpendicular to the plane of the conductor and varying with time at the rate  A conducting connector starts moving with an acceleration w = 10 cm/s2 along the parallel bars of the conductor. The length of the connector is equal to l = 20 cm. Find the emf induced in the loop t = 2.0 s after the beginning of the motion, if at the moment t = 0 the loop area and the magnetic induction are equal to zero. The inductance of the loop is to be neglected.

Solution. 307. The flux through the loop changes due to the variation in   time and also due to the movement of the connector.

So ,

But, B, after t sec. of beginning of motion - Bt, and S becomes   as connector starts moving from rest with a constant acceleration w.

So,

Q. 308. In a long straight solenoid with cross-sectional radius a and number of turns per unit length n a current varies with a constant velocity . Find the magnitude of the eddy current field strength as a function of the distance r from the solenoid axis. Draw the approximate plot of this function.

Solution. 308.

Then, from the law of electromagnetic induction

So, for r < a

For r>a

The meaning of minus sign can be deduced from Lenz's law.

Q. 309. A long straight solenoid of cross-sectional diameter d = 5 cm and with n = 20 turns per one cm of its length has a round turn of copper wire of cross-sectional area S = 1.0 mm2 tightly put on its winding. Find the current flowing in the turn if the current in the solenoid winding is increased with a constant velocity 100 A/s. The inductance of the turn is to be neglected.

Solution. 309.  The e.m.f. induced in the turn is

The resistance is

So, the current is   where p is the resistivity of copper.

The document Irodov Solutions: Electromagnetic Induction: Maxwell’s Equations - 1 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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