Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

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Q. 219. A current I = 1.00 A circulates in a round thin-wire loop of radius R = 100 mm. Find the magnetic induction
 (a) at the centre of the loop; 
 (b) at the point lying on the axis of the loop at a distance x = 100 mm from its centre.

Solution. 219. 

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev
Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

(b) From Biot-Savart’s law : 

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev
Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev
and  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Here  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev is a unit vector perpendicular to the plane containing the current loop (Fig.) and in the direction of  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Thus we get  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 220. A current I flows along a thin wire shaped as a regular polygon with n sides which can be inscribed into a circle of radius R. Find the magnetic induction at the centre of the polygon. Analyse the obtained expression n → ∞

Solution. 220.  As  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev or perpendicular distance of any segment from centre equals Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev Now magnetic induction at O, due to die right current carrying element AB

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

(From Biot- Savart’s law, the magnetic field at O due to any section such as AB is perpendicular to the plane of the figure and has the magnitude.)

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

As there are n number of sides and magnetic induction vectors, due to each side at O, are equal in magnitude and direction. So,

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRevConstant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 221. Find the magnetic induction at the centre of a rectangular wire frame whose diagonal is equal to d = 16 cm and the angle between the diagonals is equal to φ = 30°; the current flowing in the frame equals I = 5.0 A.

Solution. 221. We know that magnetic induction due to a straight current carrying wire at any point, at a perpendicular distance from it is given by :

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

where r is the perpendicular distance of the wire from the point, considered, and θ1 is the angle between the line, joining the upper point of straight wire to the considered point and the perpendicular drawn to the wire and θ2 that from the lower point of the straight wire.

Here,  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

and   Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Hence, the magnitude of total magnetic induction at O,

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 222. A current l = 5.0 A flows along a thin wire shaped as shown in Fig. 3.59. The radius of a curved part of the wire is equal to R = 120 mm, the angle 2φ = 90°. Find the magnetic induction of the field at the point O. 

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Solution. 222. Magnetic induction due to the arc segment at O,

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

and magnetic induction due to the line segment at O,

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

So, total magnetic induction at O,

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 223. Find the magnetic induction of the field at the point O of a loop with current I, whose shape is illustrated
 (a) in Fig. 3.60a, the radii a and b, as well as the angle φ are known;
 (b) in Fig. 3.60b, the radius a and the side b are known. 

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Solution. 223. (a) From the Biot-Savart law,

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

So, magnetic field induction due to the segment 1 at O,

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

also  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

and  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

and B5 = 0

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 224. A current I flows along a lengthy thin-walled tube of radius R with longitudinal slit of width h. Find the induction of the magnetic field inside the tube under the condition h ≪ R. 

Solution. 224. The thin walled tube with a longitudinal slit can be considered equivalent to a full tube and a strip carrying the same current density in the opposite direction. Inside the tube, the former does not contribute so the total magnetic field is simply that due to the strip. It is

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

where r is the distance of the field point from the strip.


Q. 225. A current I flows in a long straight wire with cross-section having the form of a thin half-ring of radius R (Fig. 3.61). Find the induction of the magnetic field at the point O. 

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Solution. 225. First of all let us find out the direction of vector  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev point O. For this purpose, we divide the entire conductor into elementary fragments with current di. It is obvious that the sum of any two symmetric fragments gives a resultant along  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev shown in the figure and consequently, vector Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev  will also be directed as shown

So,  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev   (1)

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Hence   Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 226. Find the magnetic induction of the field at the point O if a current-carrying wire has the shape shown in Fig. 3.62 a, b, c. The radius of the curved part of the wire is R, the linear parts are assumed to be very long.

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev 

Solution. 226. (a) From symmetry

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

(b) From symmetry

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 227. A very long wire carrying a current I = 5.0 A is bent at right angles. Find the magnetic induction at a point lying on a perpendicular to the wire, drawn through the point of bending, at a distance l = 35 cm from it.

Solution. 227.

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 228. Find the magnetic induction at the point 0 if the wire carrying a current I = 8.0 A has the shape shown in Fig. 3.63 a, b, c. The radius of the curved part of the wire is R = 100 mm, the linear parts of the wire are very long. 

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Solution. 228. 

(a)  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

So,  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

So,

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


(c) Here using tbe law of parallel resistances 

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

So,  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Hence  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Thus  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRevConstant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Thus,  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 229. Find the magnitude and direction of the magnetic induction vector B
 (a) of an infinite plane carrying a current of linear density i; the vector i is the same at all points of the plane;
 (b) of two parallel infinite planes carrying currents of linear densities i and —i; the vectors i and —i are constant at all points of the corresponding planes. 

Solution. 229. (a) We apply circulation theorem as shown. The current is vertically upwards in the plane and the magnetic field is horizontal and parallel to file plane.

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

(b) Each plane contributes  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev between the planes and outside the plane that cancel.

Thus Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 230. A uniform current of density j flows inside an infinite plate of thickness 2d parallel to its surface. Find the magnetic induction induced by this current as a function of the distance x from the median plane of the plate. The magnetic permeability is assumed to be equal to unity both inside and outside the plate.

Solution. 230. We assume that the current flows perpendicular to the plane of the paper, by circulation theorem,

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

or,  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 231. A direct current I flows along a lengthy straight wire. From the point O (Fig. 3.64) the current spreads radially all over an infinite conducting plane perpendicu- lar to the wire. Find the magnetic induction at all points of space.

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Solution. 231. It is easy to convince oneself that both in the regions. 1 and 2, there can only be a circuital magnetic field (i.e. the component Bφ). Any radial field in region 1 or any Bz away from the current plane will imply a violation of Gauss’ law of magnetostatics, Bφ must obviously be symmetrical about the straight wire. Then in 1,

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

or,   Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

In 2,  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 232. A current I flows along a round loop. Find the integral Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev along the axis of the loop within the range from  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev Explain the result obtained. 

Solution. 232. Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Thus,  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

The physical interpretation of this result is that  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev dx can be thought of as the circulation of B over a closed loop by imaging that the two ends of the axis are connected, by a line at infinity (e.g. a semicircle of infinite radius).


Q. 233. A direct current of density j flows along a round uniform straight wire with cross-section radius R. Find the magnetic induction vector of this current at the point whose position relative to the axis of the wire is defined by a radius vector r. The magnetic permeability is assumed to be equal to unity throughout all the space. 

Solution. 233. By circulation theorem inside the conductor

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

i.e.,     Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Similarly outside the conductor,

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

So, Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 234. Inside a long straight uniform wire of round cross-section there is a long round cylindrical cavity whose axis is parallel to the axis of the wire and displaced from the latter by a distance 1. A direct current of density j flows along the wire. Find the magnetic induction inside the cavity. Consider, in particular, the case I = 0. 

Solution. 234. We can think of the given current which will be assumed uniform, as arising due to a negative current, flowing in the cavity, superimposed on the true current, everywhere including the cavity. Then from the previous problem, by superposition.

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev so that the cavity is concentric with the conductor, there is no magnetic field in the cavity.


Q. 235. Find the current density as a function of distance r from the axis of a radially symmetrical parallel stream of electrons if the magnetic induction inside the stream varies as B = brα, where b and α are positive constants. 

Solution. 235. By Circulation theorem, 

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

or using Bφ = brα inside the stream,

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

So by differentiation,

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Hence,  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 236. A single-layer coil (solenoid) has length l and cross-section radius R, A number of turns per unit length is equal to n. Find the magnetic induction at the centre of the coil when a current I flows through it. 

Solution. 236. On the surface of the solenoid there is a surface current density

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Then,  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

where  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev is the vector from the current element to the field point, which is the centre of the solenoid, 

Now , 

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Thus.  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 237. A very long straight solenoid has a cross-section radius R and n turns per unit length. A direct current I flows through the solenoid. Suppose that x is the distance from the end of the solenoid, measured along its axis. Find:
 (a) the magnetic induction B on the axis as a function of x; draw an approximate plot of B vs the ratio x/R;
 (b) the distance xo  to the point on the axis at which the value of B differs by η = 1% from that in the middle section of the solenoid.

Solution. 237. We proceed exactly as in the previous problem. Then (a) the magnetic induction on the axis at a distance x from one end is clearly,

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

x > 0 menas that the field point is outside the solenoid. B then falls with x. x < 0 means that the field point gets more and more inside the solenoid. B then increases with (x) and eventually becomes constant, equal to μ0 nI. The B - x graph is as given in the answer script.

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

or,  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Since η is small Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev must be negative. Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

and Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

or,  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 238. A thin conducting strip of width h = 2.0 cm is tightly wound in the shape of a very long coil with cross-section radius R = 2.5 cm to make a single-layer straight solenoid. A direct current I = 5.0 A flows through the strip. Find the magnetic induction inside and outside the solenoid as a function of the distance r from its axis. 

Solution. 238. If the strip is tightly wound, it must have a pitch of A. This means that the current will flow obliquely, partly along Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev and partly along Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev Obviously, the surface current density is,

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

By comparision with the case of a solenoid and a hollow straight conductor, we see that field inside the coil

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Outside, only the other term contributes, so

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

or,  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Note - Surface current density is defined as current flowing normally acrpss a unit length over a surface.


Q. 239. N = 2.5.103  wire turns are uniformly wound on a wooden toroidal core of very small cross-section. A current I flows through the wire. Find the ratio η of the magnetic induction inside the core to that at the centre of the toroid.

Solution. 239. Suppose a is the radius of cross section of the core. The winding has a pitch 2πR/N, so the surface current density is

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

where  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev  a unit vector along the cross section of the core and Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev a unit vector along its length. The magnetic field inside the cross section of the core is due to first term above, and is given byConstant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

(NI is total current due to the above surface current (first term.))

Thus, Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

The magnetic field at the centre of the core can be obtained from the basic formula. 

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev and is due to the second term.

So, Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

or,  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

The ratio of the two magnetic field, is  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 240. A direct current I = 10 A flows in a long straight round conductor. Find the magnetic flux through a half of wire's crosssection per one metre of its length.

Solution. 240. We need the flux through the shaded area.

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

Now by Ampere's theorem; 

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

or,    Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

The flux through the shaded region is,

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev


Q. 241. A very long straight solenoid carries a current I. The cross-sectional area of the solenoid is equal to S, the number of turns per unit length is equal to n. Find the flux of the vector B through the end plane of the solenoid.

Solution. 241. Using Q.237, the magnetic field is given by,

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

At the end  Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

is the field deep inside the solenoid. Thus,

Constant Magnetic Field Magnetics (Part - 1) - Electrodynamics, Irodov JEE Notes | EduRev

is the flux of the vector B through the cross section deep inside the solenid. 

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