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.
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,
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.
Solution. 244. Take an element of length rd θ containing turns. Its magnetic moment is
normal to the plane of cross section. We resolve it along OA and OB. The moment along OA integrates to
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.
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,
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,
where is the contribution of one turn of radius r and dN is the number of turns in the interval (r, r + dr)
Substituting in equation (1) and integrating the result over r between a and b, we obtain,
(b) The m agnetic moment of a turn of radius and of all turns,
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,
So, magnetic induction at the centre, due to this element :
and hence, from symmetry
(b) Magnetic moment of the element, considered,
Hence, the sought magnetic moment,
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.
The equivalent current due to the ring element,
and magnetic induction due to this loop element at the centre of the sphere, O,
[Using Q.219 (b)]
Hence, the total magnetic induction due to the sphere at the centre, O,
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 The rotation of the sphere causes this charge to revolve around the axis and constitute a current.
Its magnetic moment will be
So the total magnetic moment is
The mechanical moment is
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
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)
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)
Its contribution to the magnetic field at the centre is
and the total magnetic field is
As for the volume charge density consider a circle of radius r, radial thickness dr and length dx.
The current is
The total magnetic field due to the volume charge distribution is
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,
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.
Solution. 251. (a) The magnetic field at O is only due to the curved path, as for the line element,
(b) In this part, magnetic induction will be effective only due to the two semi infinite segments of wire. Hence
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.
where da is the angle subtended by the element at the centre.
The wire experiences a stress
This must equals the breaking stress σm for rupture. Thus,
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°.
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.
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,
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.
Solution. 254. We know that the torque acting on a magnetic dipole.
But, 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.
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.
So, 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).
Now, the Ampere force on (1),
and that on (3),
So, the resultant force on the frame = F1 - F3, (as they are opposite in nature.)
(b) Work done in turning the frame through some angle, where is the flux through the frame in final position, and that in the the initial position.
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 we must take into account the electric force 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.
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
where i is the current in the wire.
Now, from the relation,
λ, = C φ, where C is the capacitance of the wires per unit lengths and is given in problem Q.108 and φ = iR
Dividing (2) by (1) and then substuting the value of
The resultant force of interaction vanishes when this ratio equals unity. This is possible when R = R0, where
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.
Solution. 257. Use Q.225
The magnetic field due to the conductor with semicircular cross section is
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.
Solution. 258. We know that Ampere’s force per unit length on a wire element in a magnetic field is given by.
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,
From Eqs. (1) and (2)
(because the current through the element equals
Hence the magnetic force on the conductor :
Then according to the Newton’s third law the magnitude of sought magnetic interaction force
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 = μ0 i,
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 The force on the plate is,
(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.
Solution. 260. (a) The external field must be which when superposed with the internal field (of opposite sign on the two sides of the plate) must give actual field. Now
(b) Here, the external field must be upward with an internal field, upward on the left and downward on the right Thus,
(c) Our boundary condition following from
The external field parallel to the plate must be
(The perpendicular component B1 cos θ1, does not matter since the corresponding force is tangential)
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.
The Current density is is the length of the section. The difference in pressure produced must be,
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
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,
If B (I') α I' , then this becomes,
In the present case, and this acts on n l ampere turns per unit length, so,
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,
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,
Note that Flim, here, is actually a force and not a stress.