Lecture 16 :

Key words :

Ampere’s law, Field due to solenoid, current sheet,.

Problems :

  1. Two long parallel wires carry 5 A current each in the opposite directions. The distance between the wires is 0.6 m. Find the magnetic field at a distance of 0.4 m from the wire to the right.
  1. Consider three long parallel wires whose intersection with the plane of the paper forms an equilateral triangle of side m. The wires carry 5 A current in the directions shown. Calculate the field at the location of the upper wire by the other two. What is the force per unit length on the upper wire?
  1. A very large number of long parallel wires carry current I each. The separation between adjacent wires is d and their current directions alternate. Find an expression for the field strength at the location of the leftmost wire. What would be the field strength if the number of wires were infinite?
  1. A coaxial cable has an inner conductor of radius Rwhich is surrounded by an outer conductor of inner radius 1.5R and outer radius 2R. a current I enters through the inner conductor and comes out through the outer conductor. The current is uniformly distributed across the cross sections. Find the magnetic field at a distance r from the axis.
  1. A cylindrical conductor of radius a has an off axis hole of radius b running through the cylinder. The distance between the axis of the cylinder and the hole is d. The cylinder carries a current I uniformly distributed over the cross section. Determine the magnetic field at a point inside the hole which is at a distance r from the axis of the cylinder.
  2. A cylindrical conductor of radius a has a non uniformly distributed current. The current density at a distance r from the axis of the cylinder is . Calculate the magnetic field strength both inside and outside the cylinder.
  1. Consider an semi-infinite current sheet of finite width w and negligible thickness which carries a current I along its length. Calculate the magnetic field at a distance d from one of the edges.
  1. A solenoid of radius 10 cm and length 1.5 m, carries a current 20 A in its windings. If the field at the centre of the solenoid is 100 Gauss, how many turns are there in the solenoid?
  1. A toroid of inner radius 18 cm and outer radius 20 cm, carries a current of 10 A. What are the maximum and minimum values of the field strength in the toroid, if it has 1000 turns?
  1. A long cylinder of radius 2R has a coaxial hole of radius R. The cylinder carries a current I distributed uniformly across its cross section. Calculate the field at a point on the axis.

Hints for solutions to Problem

  1. Fields subtract as P is situated to the right side of oppositely directed currents. Since the outward field is stronger, the net field is outward. T.
  1. Field due to each of the wires is perpendicular to the line joining the vertex to the wires. The net field is along the base from left corner to right (1 to 2) as the perpendicular components cancel. (The direction of the field due to the wires numbered 1 and 2 is indicated at the position of the top wire.

The field is T. The force on the top wire due to both wires is repulsive. Resolving parallel and perpendicular to the base, the components parallel to the base cancel, leaving only the perpendicular components. Force per unit length is N.

  1. Direct application of Ampere’s law gives . As the number of terms increase, the quantity within the bracket approaches .
  1. For , the problem is same as for a single wire with uniform distribution of current. The field is .For For , the cross sectional area of the outer cable is The cross sectional area of outer cable enclosed by the circle of radius r is The total current enclosed by a circle of radius is then . Using this in Ampere’s law, For the field is zero.
  1. Use the principle of superposition. Fill up the hole with equal and opposite current densities We then have a current in the original direction and . Use Ampere’s law and show that the field inside the hole is independent of r.
  1. Total current through the conductor is Thus for using Ampere’s law, For Thus for
  1. Take a current sheet of width dx. The current in the strip is . The field due to this strip is . Integrating, .
  1. Since the ratio of length to diameter is large, the solenoid could essentially be considered as an ideal one and edge effects may be neglected. Substituting the given values in

we get, , which gives the number of turns

  1. Ans. 0.011 T and 0.01 T
  1. Ans. Zero (see problem 5)

Lecture 17 :

Key words :

Vector Potential, Gauge Invariance, Coulomb Gauge, Magnetic Flux, Emf, Faraday’s Law.

CORRECTIONS

  1. In several places (indicated below) has been printed as vecA. These have to be replaced.
  2. Page 2 : line 2 : Equation : , should read . (a minus sign was missing in the equation)
  3. Page 3 : three places line 1, line 2 and line 12 : vecA
  4. Inside Exercise 1 : lines 3 and 4 : vecA. Also remove a big horizontal line on this page.
  5. Correct all example numbers.
  6. Example 16 : Inside page 2 : vecA, also on page 4 of the example line 4 from bottom : vecA
  7. Example17 : inside : page 1, line 5 : Equation : insidesolenoid inside solenoid also next line a space between 0 and outside.
  8. Inside example 17 : Page 2 line 3 : vecA in two places.
  9. Page 5 : Remove a big horizontal line. Line 5 from bottom vecA

Problems :

  1. The magnetic field B in a region is given by , where c is a constant. Calculate the value of c.
  2. Magnetic field in a region is given by the expression . Obtain an expression for the vector potential corresponding to this field.
  3. Obtain an expression for the vector potential corresponding to the magnetic field produced by a current sheet in the xy plane carrying a linear current density K in the x direction.
  4. Two long, parallel wires carrying current I in opposite directions are at a distance d from each other. The currents run parallel to z axis, with one of the wires at x=0 while the other at x=d. Calculate the vector potential at a point P in the xy plane at a position as shown.
  1. A rectangular loop of sides lies in the xy plane. A uniform magnetic field B exists everywhere in the region directed perpendicular to the plane of the loop. Calculate the emf generated when
  2. The strength of the magnetic field is decreased from its maximum value B0 to zero steadily over a time T.
  3. The loop is moved with a speed v in the xy plane.
  4. The loop is rotated about one of the sides with a uniform angular speed .
  5. A circular coil of radius 40 cm lies in the xy plane in a region where the magnetic field is given by

Find the emf induced in the loop. If the loop has a resistance of 10 , what would be the rms value of the induced current?

Hints for solutions to Problem

  1. B is solenoidal, i.e. gives .
  2. Take . Use

This gives with . (and could also have an additive constant which is a function of x alone but this will not be feasible because such constants will not be consistent with the y and z components of the curl.) Taking y and z components give us two pairs of relations, withand with . These equations are consistent if . Thus . Thus .

  1. Use superposition principle. The vector potential is along z direction and is sum of two vector potentials, one due to each wire. Using the expression for the vector potential worked out in Example 1, we get . For .
  2. The normal to the loop is along the z direction. Thus the flux through the loop is .
  3. If the magnetic field varies with time
  4. Zero, as the flux does not change.
  5. As the plane rotates, the normal to the plane makes an angle with the magnetic field, so that The emf is .
  6. Emf is given by . Maximum current is 4.73 A, rms current is 3.35 A.

Lecture 18 :

Key words :

Motional emf, Induced current, Lenz’s Law.

Problems :

  1. A circular loop of radius 10 cm, having a resistance of 100 , lies on the plane of the paper. A magnetic field B acts perpendicular to the plane of the loop and points inward. If the magnetic field decreases at a uniform rate from its initial value 500 Gauss to a value of 50 Gauss in 25 ms, find the current in the loop. Which direction does the current point as seen from above the plane of the paper?
  2. A rectangular conducting loop of dimensions and resistance R is taken out of the magnetic field B with a constant speed v, as shown. Determine the emf generated and the direction of the current flow.
  1. A circular loop of radius 20 cm and resistance 10  symmetrically surrounds a long solenoid having a radius of 2.5 cm and 1600 turns per meter. The plane of the coil is perpendicular to the axis of the solenoid. The current in the solenoid changes from a value of 10 A to 30 A in 2 seconds. Calculate the maximum current induced in the circular coil.
  1. A rectangular conducting loop of sides 50 cm x 30 cm lies in a magnetic field directed perpendicular to its plane. A 1.5 V light bulb is in the loop, the resistance of rest of the loop can be neglected. At what uniform rate should the magnetic field strength be increased so that the light bulb shines with maximum brightness. What is the direction of the induced current in the loop?
  1. A square loop of side a rotates such that the angle between the normal to the loop and a fixed magnetic field B varies with time as Find the maximum emf in the loop.
  2. A conducting rod AB is moving with a speed of 0.1 m/s at right angles to its length, in contact with two parallel rails of a circuit which are separated by 0.5 m. A uniform magnetic field of strength 1 T is directed perpendicular to the plane of the circuit. The rod has a resistance of 0.2  while the resistance of the other parts of the circuit may be neglected. How much work is needed to be done to keep the rod moving with a constant speed? Show that this amount of work is dissipated as Joule heat.
  1. A conducting rod I of length a and resistance R s pivoted at a point in the region of a uniform magnetic field B. The rod is free to rotate in a plane perpendicular to the direction of the magnetic field B. If the rod rotates with a constant angular velocity w, find the emf between the two ends of the rod. If the circuit is completed by connecting the pivot by resistanceless wire so that it is in the shape of a sector of a circle a, what is the force and torque that acts on the rod?
  2. A long straight wire carries a current I. A square loop of side a having a resistance R and lying in the plane of the paper moves with a constant velocity v parallel to the wire. Calculate the emf induced in the loop.
  1. Consider the same arrangement as in Problem 8 but the square loop is moving perpendicular to the wire. What is the emf in this case when the distance from the wire is b?
  2. A conductingrodof resistance R slides frictionlessly along a pair of parallel conducting rails on a table. The rails are separated by a length a and are connected by a fixed conducting section to complete a circuit. The rails and the connecting section have negligible resistance. The rod is connected to a mass by a massless string which passes over a pulley. A uniform magnetic field of strength B is directed perpendicular to the surface of the table. Obtain an expression for the instantaneous velocity v of the mass. What is the terminal speed?

Hints for solutions to Problem

  1. The time dependence of the field can be expressed by the equation , where B is expressed in Tesla and t in seconds. The area of the loop is . Thus the emf V. The induced current is 0.57 mA. As the magnetic flux is decreasing downwards, the induced current must be so as to increase it in the downward direction. The current should thus be clockwise as viewed from above.
  2. The instantaneous area of the loop which is inside the magnetic field is . The emf is . Since the magnetic field is out of the page, theemf is positive, implying the current is positive, i.e. it is in the counterclockwise direction. (convince yourself that this is consistent with Lenz’s law.)
  3. The field due to solenoid is confined within the solenoid and is given by The flux enclosed by the circular loop is , where r is the radius of the solenoid (not of the loop). emf V. The current is 0.395 A.
  4. For full brightness the emf in the circuit should be 1.5 V. dB/dt required for this is 10 T/s.
  5. The flux through the loop is . Emf is . Thus the maximum emf is .
  6. The emf in the circuit is V, the current flowing in the circuit is 0.05/0.2=0.25 A. The force acting on the rod is N. The direction of the force is to the left, opposite to the direction of the velocity of the rod. To keep the rod moving at constant speed, the power required is W. To see that this is the rate of Joule heat, note that Joule heat is W.
  7. Considering an imaginary sector of which the rod is a radial section, the flux through the sector having an instantaneous angle  is . Thus the emf developed is . The current in the circuit is . The force acting on the rod is . The torque acting is
  8. The flux enclosed does not change as the rectangle moves. Emf is zero.
  9. Consider a strip of width dx which is located at a distance x from the current. The strip has an area The flux through the strip is . Suppose at a particular instant the nearer edge of the square is at a distance bfrom the current, the flux through the entire loop is . The emf is obtained from Faraday’s law.
  1. As the rod slides, a motional emf develops, resulting in a current in the circuit. This provides a leftward force on the sliding road, which, in turn, provides an upward tensile force on the mass. This acts in addition to the gravitational force mg acting downward. The equation of motion is . The differential equation can be solved by direct integration. Assuming that the mass is released at the velocity is given by . The terminal velocity can be found by taking limit of this expression, or more directly, by equating the net force to zero.

Lecture 19 :

Key words :

Induced Electric Field, Mutual Inductance, Self Inductance, Magnetic Energy

Problems :

  1. A solenoid of length 1m and radius 10 cm has 5000 turns. The strength of the induced electric field at a distance of 20 cm from the axis of the solenoid is 40 mV/m. Calculate the rate at which the current through the coils of the solenoid is changing with time.
  2. Two solenoids have radii R and 2R respectively. The number of turns in the former is N while the number of turns in the latter is 2n per unit length. If the narrower solenoid is completely enclosed by the other one, determine the mutual inductance of the pair.
  1. A solenoid of length 1 m and radius 5 cm has 5000 turns of wire. A circular coil of radius 1 cm having 100 turns is placed coaxially within the solenoid. Calculate (a) the mutual inductance of the pair and (b) the induced electric field in the circular coil.
  2. Current in a coil is changing at a rate of 150 A/s. If the emf developed in the coil is 0.5 V, what is the self inductance of the coil?
  3. A rectangular loop of dimensions a x b is at a distance d from a long straight wire. Calculate the mutual inductance.
  1. A conductor consists of two long thin conducting shells of radii a and b. A current I goes in through the inner conductor and come out through the outer conductor. Calculate the magnetic energy per unit length stored in the conductor and hence calculate the self inductance of the conductor.
  2. A rod of length L is moving with a velocity v in a plane perpendicular to a uniform magnetic field. What is the induced electric field at any point of the rod?:
  3. A conducting rod OA of length L rotates with a constant angular speed in the xy plane, about one of its ends which is pivoted at the origin. A uniform magnetic field B points in the +z direction. Find the induced electric field at any point on the rod and by taking the line integral of the electric field, show that this leads to the value of emf developed, as calculated in Problem 7 of Lecture 18.
  4. A cylindrical volume of radius R has a uniform axial magnetic field B which is increasing at a constant rate . A conducting rod rests in the cylinder in a plane perpendicular to its axis, as shown in the cross sectional view. Determine the electric field at any point of the rod, measuring the distance from one end. By taking the line integral of the electric field, determine the potential difference between the two ends of this rod and show that this is consistent with the emf calculation using Faraday’s law.
  1. A long cylindrical wire of radius R carries a current I, distributed uniformly over its cross section. Find the magnetic energy density in the wire.

Hints for solutions to Problem

  1. The field inside the solenoid is . The flux enclosed by a circle of radius is . Substituting data, the current changes at 255 A/s.
  2. Consider a current flowing through the inner conductor. The entire flux due to this is captured by the outer conductor. The field due to the inner solenoid is confined to the inner solenoid only. The flux captured by 2n turns of the outer solenoid is . The mutual inductance is .
  3. The flux trapped by the circular coil is . The mutual inductance is H.
  4. Ans. 3.33 mH.
  5. If the current flowing through the wire is I, the flux trapped by the rectangle is . The mutual inductance is given by dividing this expression by current I.
  6. Using Ampere’s law one can see that non-zero magnetic field only exists in the region between the two conducting shells and is given by . In this region consider a cylindrical shell lying between r and r+dr. The energy per unit volume being the energy contained in a shell of length lis. Total energy is obtained by integrating this expression from a to b, giving . Equating this to , one gets the self inductance per unit length of the conductor.
  7. The motional emf developed in the rod is . Since the rod is symmetrical, the electric field must be constant over its length. As emf is the line integral of the electric field, the field is
  8. Consider a strip of length dr at a distance r from the pivot. The velocity of this strip is . Thus the electric field at this point is . The emf is .