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

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Q. 242. Fig. 3.65 shows a toroidal solenoid whose cross-section is rectangular. Find the magnetic flux through this cross-section if the current through the winding equals I = 1.7 A, the total number of turns is N = 1000, the ratio of the outside diameter to the inside one is η = 1.6, and the height is equal to h = 5.0 cm. 

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

Solution. 242. Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev

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

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


Q. 243. Find the magnetic moment of a thin round loop with current if the radius of the loop is equal to R = 100 mm and the magnetic induction at its centre is equal to B = 6.0 μT. 

Solution. 243. Magnetic moment of a current loop is given by pm - n i S ( where n is the number of turns and 5, the cross sectional area.) In our problem,

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

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


Q. 244. Calculate the magnetic moment of a thin wire with a current I = 0.8 A, wound tightly on half a tore (Fig. 3.66). The diameter of the cross-section of the tore is equal to d = 5.0 cm, the number of turns is N = 500. 

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

Solution. 244. Take an element of length rd θ containing  Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev turns. Its magnetic moment is

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

normal to the plane of cross section. We resolve it along OA and OB. The moment along OA integrates to

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

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

Q. 245. A thin insulated wire forms a plane spiral of N = 100 tight turns carrying a current I = 8 mA. The radii of inside and outside turns (Fig. 3.67) are equal to a = 50 mm and b = 100 mm. Find:
 (a) the magnetic induction at the centre of the spiral;
 (b) the magnetic moment of the spiral with a given current. 

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

Solution. 245. (a) From Biot-Savart’s law, the magnetic induction due to a circular current carrying wire loop at its centre is given by,

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

The plane spiral is made up of concentric circular loops, having different radii, varying from a to b. Therefore, the total magnetic induction at the centre,

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

where Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRevis the contribution of one turn of radius r and dN is the number of turns in the interval (r, r + dr)

i.e.    Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev

Substituting in equation (1) and integrating the result over r between a and b, we obtain, 

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

(b) The m agnetic moment of a turn of radius   Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev and of all turns,

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


Q. 246. A non-conducting thin disc of radius R charged uniformly over one side with surface density σ rotates about its axis with an angular velocity ω. Find:
 (a) the magnetic induction at the centre of the disc;
 (b) the magnetic moment of the disc. 

Solution. 246. (a) Let us take a ring element of radius r and thickness dr, then charge on the ring element., d q = σ 2 π r dr

and current, due to this element,  Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev

So, magnetic induction at the centre, due to this element :  Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev

and hence, from symmetry    Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev

(b) Magnetic moment of the element, considered,

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

Hence, the sought magnetic moment,

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


Q. 247. A non-conducting sphere of radius R = 50 mm charged uniformly with surface density σ = 10.0 µC/m2  rotates with an angular velocity ω = 70 rad/s about the axis passing through its centre. Find the magnetic induction at the centre of the sphere. 

Solution. 247. As only the outer surface of the sphere is charged, consider the element as a ring, as shown in the figure.

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

The equivalent current due to the ring element,

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

and magnetic induction due to this loop element at the centre of the sphere, O,

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

[Using Q.219 (b)]

Hence, the total magnetic induction due to the sphere at the centre, O,

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

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


Q. 248. A charge q is uniformly distributed over the volume of a uniform ball of mass m and radius R which rotates with an angular velocity e.) about the axis passing through its centre. Find the respective magnetic moment and its ratio to the mechanical moment.

Solution. 248. The magnetic moment must clearly be along the axis of rotation. Consider a volume elem ent dV. It contains a charge  Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev The rotation of the sphere causes this charge to revolve around the axis and constitute a current. 

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

Its magnetic moment will be

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

So the total magnetic moment is

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

The mechanical moment is

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


Q. 249. A long dielectric cylinder of radius R is statically polarized so that at all its points the polarization is equal to P = αr, where α is a positive constant, and r is the distance from the axis. The cylinder is set into rotation about its axis with an angular velocity ω. Find the magnetic induction B at the centre of the cylinder. 

Solution. 249. Because of polarization a space charge is present within the cylinder. It’s density is

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

Since the cylinder as a whole is neutral a surface charge density σp must be present on the surface of the cylinder also. This has the magnitude (algebraically)

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

When the cylinder rotates, currents are set up which give rise to magnetic fields. The contribution of Pp and σp can be calculated separately and then added.
For the surface charge the current is (for a particular element)

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

Its contribution to the magnetic field at the centre is

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

and the total magnetic field is

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

As for the volume charge density consider a circle of radius r, radial thickness dr and length dx.

The current is Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev

The total magnetic field due to the volume charge distribution is

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

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


Q. 250. Two protons move parallel to each other with an equal velocity v = 300 km/s. Find the ratio of forces of magnetic and electrical interaction of the protons.

Solution. 250. Force of magnetic interaction,  Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev

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

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

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

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

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


Q. 251. Find the magnitude and direction of a force vector acting on a unit length of a thin wire, carrying a current I = 8.0 A, at a point O, if the wire is bent as shown in
 (a) Fig. 3.68a, with curvature radius R = 10 cm;
 (b) Fig. 3.68b, the distance between the long parallel segments of the wire being equal to l = 20 cm.

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

Solution. 251. (a) The magnetic field at O is only due to the curved path, as for the line element,  Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev

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

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

(b) In this part, magnetic induction  Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev will be effective only due to the two semi infinite segments of wire. Hence

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

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


Q. 252. A coil carrying a current I = 10 mA is placed in a uniform magnetic field so that its axis coincides with the field direction. The single-layer winding of the coil is made of copper wire with diameter d = 0.10 mm, radius of turns is equal to R = 30 mm. At what value of the induction of the external magnetic field can the coil winding be ruptured? 

Solution. 252. Each element of length dl experiences a force Bl dl. This causes a tension T in the wire.

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

For equilibrium,

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

where da is the angle subtended by the element at the centre.

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

The wire experiences a stress

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

This must equals the breaking stress σm for rupture. Thus,

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


Q. 253. A copper wire with cross-sectional area S = 2.5 mm2 bent to make three sides of a square can turn about a horizontal axis OO' (Fig. 3.69). The wire is located in uniform vertical magnetic field. Find the magnetic induction if on passing a current I = 16 A through the wire the latter deflects by an angle θ = 20°. 

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

Solution. 253. The Ampere forces on the sides OP and O'P' are directed along the same line, in opposite directions and have equal values, hence the net force as well as the net torque of these forces about the a x is OO ' is zero. The Am pere-force on the segm ent PP ’ and the corresponding moment of this force about the axis OO' is effective and is deflecting in nature.

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

In equilibrium (in the dotted position) the deflecting torque must be equal to the restoring torque, developed due to the weight of the shape.
Let, the length of each side be l and p be the density of the material then,

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


Q. 254. A small coil C with N = 200 turns is mounted on one end of a balance beam and introduced between the poles of an electromagnet as shown in Fig. 3.70. The cross-sectional area of the coil is S = 1.0 cm2, the length of the arm OA of the balance beam is l = 30 cm. When there is no current in the coil the balance is hi equilibrium. On passing a current I = 22 mA through the coil the equilibrium is restored by putting the additional counterweight of mass Δm = 60 mg on the balance pan. Find the magnetic induction at the spot where the coil is located.

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

Solution. 254. We know that the torque acting on a magnetic dipole.

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

But,   Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev is the normal on the plane of the loop and is directed in the direction of advancement of a right handed screw, if we rotate the screw in the sense of current in the loop.

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

On passing a current through the coil, this torque acting on the magnetic dipol, is counterbalanced by the moment of additional weight, about O. Hence, the direction of current in the loop must be in the direction, shown in the figure.

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

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

So,  Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev  on putting the values.


Q. 255. A square frame carrying a current I = 0.90 A is located in the same plane as a long straight wire carrying a current I0 = 5.0 A. The frame side has a length α = 8.0 cm. The axis of the frame passing through the midpoints of opposite sides is parallel to the wire and is separated from it by the distance which is η = 1.5 times greater than the side of the frame. Find:
 (a) Ampere force acting on the frame;
 (b) the mechanical work to be performed in order to turn the frame through 180° about its axis, with the currents maintained constant.

Solution. 255. (a) As is clear from the condition, Ampere's forces on the sides (2) and (4) are equal in magnitude but opposite in direction. Hence the net effective force on the frame is the resultant of the forces, experienced by the sides (1) and (3).

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

Now, the Ampere force on (1),

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

and that on (3),

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

So, the resultant force on the frame = F1 - F3, (as they are opposite in nature.)

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

(b) Work done in turning the frame through some angle,  Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev where Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev is the flux through the frame in final position, and Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev that in the the initial position.

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

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

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

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


Q. 256. Two long parallel wires of negligible resistance are connected at one end to a resistance R and at the other end to a de voltage source. The distance between the axes of the wires is η = 20 times greater than the cross-sectional radius of each wire. At what value of resistance R does the resultant force of interaction between the wires turn into zero? 

Solution. 256. There are excess surface charges on each wire (irrespective of whether the current is flowing through them or not). Hence in addition to the magnetic force  Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev we must take into account the electric force  Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev Suppose that an exc ess charge X corresponds to a unit length of the wire, then electric force exerted per unit length of the wire by other wire can be found with the help of Gauss’s theorem.

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

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

where l is the distance between the axes of the wires. The magnetic force acting per unit length of the wire can be found with the help of the theorem on circulation of vector  Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev

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

where i is the current in the wire. 

Now, from the relation,Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev

λ, = C φ, where C is the capacitance of the wires per unit lengths and is given in problem Q.108 and φ = iR

Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev    (3)

Dividing (2) by (1) and then substuting the value of Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev

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

The resultant force of interaction vanishes when this ratio equals unity. This is possible when R = R0, where

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


Q. 257. A direct current I flows in a long straight conductor whose cross-section has the form of a thin half-ring of radius R. The same current flows in the opposite direction along a thin conductor located on the "axis" of the first conductor (point 0 in Fig. 3.61). Find the magnetic interaction force between the given con- ductors reduced to a unit of their length. 

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

Solution. 257. Use Q.225

The magnetic field due to the conductor with semicircular cross section is

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

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


Q. 258. Two long thin parallel conductors of the shape shown in Fig. 3.71 carry direct currents I1 and I2. The separation between the conductors is a, the width of the right-hand conductor is equal to b. With both conductors lying in one plane, find the magnetic interaction force between them reduced to a unit of their length. 

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

Solution. 258. We know that Ampere’s force per unit length on a wire element in a magnetic field is given by.

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

Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev is the unit vector along the direction of current.   (1)

Now, let us take an element of the conductor i2, as shown in the figure. This wire element is in the magnetic field, produced by the current i1, which is directed normally into the sheet of the paper and its magnitude is given by,

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

From Eqs. (1) and (2)

Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev (because the current through the element equals Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev

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

Hence the magnetic force on the conductor :

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

Then according to the Newton’s third law the magnitude of sought magnetic interaction force

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


Q. 259. A system consists of two parallel planes carrying currents producing a uniform magnetic field of induction B between the planes. Outside this space there is no magnetic field. Find the magnetic force acting per unit area of each plane.

Solution. 259. By the circulation theorem B = μi,

where i = current per unit length flowing along the plane perpendicular to the paper. Currents flow in the opposite sense in the two planes and produce the given field B by superposition.

The field due to one of the plates is just  Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev The force on the plate is,

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

(Recall the formula F = BIl on a straight wire)


Q. 260. A conducting current-carrying plane is placed in an external uniform magnetic field. As a result, the magnetic induction becomes equal to B1 on one side of the plane and to B2, on the other. Find the magnetic force acting per unit area of the plane in the cases illustrated in Fig. 3.72. Determine the direction of the current in the plane in each case. 

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

Solution. 260. (a) The external field must be  Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev which when superposed with the internal field Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev (of opposite sign on the two sides of the plate) must give actual field. Now

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

Now

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

(b) Here, the external field must be  Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev upward with an internal field, Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev upward on the left and downward on the right Thus,

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

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

(c) Our boundary condition following from 

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

The external field parallel to the plate must be  Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev

(The perpendicular component B1 cos θ1, does not matter since the corresponding force is tangential)

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

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

The direction of the current in the plane conductor is perpendicular to the paper and beyond the drawing.


Q. 261. In an electromagnetic pump designed for transferring molten metals a pipe section with metal is located in a uniform magnetic field of induction B (Fig. 3.73). A current I is made to flow across this pipe section in the direction perpendicular both to the vector B and to the axis of the pipe. Find the gauge pressure produced by the pump if B = 0.10 T, I = 100 A, and a = 2.0 cm.

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

Solution. 261. 

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

The Current density is  Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev is the length of the section. The difference in pressure produced must be, 

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


Q. 262. A current I flows in a long thinwalled cylinder of radius R. What pressure do the walls of the cylinder experience? 

Solution. 262. Let t - thickness of the wall of the cylinder. Then, J = I / 2 π R t along z axis. The magnetic field due to this at a distance r

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

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

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

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

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

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

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


Q. 263. What pressure does the lateral surface of a long straight solenoid with n turns per unit length experience when a current I flows through it? 

Solution. 263. When self-forces are involved, a typical factor of 1/2 comes into play. For example, the force on a current carrying straight wire in a magnetic induction B is BIl. If the magnetic induction B is due to the current itself then the force can be written as,

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

If B (I') α I' , then this becomes,   Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev

In the present case, Constant Magnetic Field Magnetics (Part - 2) - Electrodynamics, Irodov JEE Notes | EduRev and this acts on n l ampere turns per unit length, so,

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


Q. 264. A current I flows in a long single-layer solenoid with crosssectional radius R. The number of turns per unit length of the solenoid equals n. Find the limiting current at which the winding may rupture if the tensile strength of the wire is equal to Flim

Solution. 264. The magnetic induction B in the solenoid is given by B = μ0 nl. The force on an element dl of the current carrying conductor is,

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

This is radially outwards. The factor 1/2 is explained above.

To relate dF to the tensile strength Flim we proceed as follows. Consider the equilibrium of the element dl. The longitudinal forces F have a radial component equal to,

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

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

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

Note that Flim, here, is actually a force and not a stress.

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