Physics-5. UNIT - 16 MAGNETIC EFFECTS OF CURRENT

UNIT - 16 MAGNETIC EFFECTS OF CURRENT 1. MAGNETIC FIELD DUE TO ELECTRIC CURRENT The magnetic field due to an electric current flowing through a wire can be calculated using Biot-Savart's law and Ampere's circuital law. Ampere's circuital law is used only in case of symmetry. 1.1 BIOT-SAVART'S LAW To get the magnetic field at a point using Biot-Savart's law, we need to understand the term current-element. Current element is the product of current and length of infinitesimal segment of current carrying wire. The current element is a vector quantity. Its direction is same as the direction of current. In the figure shown, there is a segment of a current carrying wire and P is a point where magnetic field is to be calculated. i represents the current element and , the position vector of the point ‘P’ with respect to the current element I . According to Biot-Savart's law, magnetic field at a point ‘P’ due to the current element i is given by the expression, Since entire segment in made-up of infinite such current elements, the magnetic field due to the entire wire segment can be found by integrating the magnetic field due to the current elements. Hence, Here μo is a proportionality constant, known as the permeability of free space. In SI units, the value of μo/4π is equal to 10-7 Tesla-meter/ampere. The SI units of magnetic fields are Tesla, weber/m2. In the above expression limits of the integral depend on the shape and size of the current carrying wire. 1.2 MAGNETIC FIELD DUE TO CURRENT IN A STRAIGHT WIRE The magnetic field due to a wire segment carrying current i at P, when the wire segment subtends angles α and β as shown, can be determined as follows: In vector form If the wire is infinitely long then α = 0 and β = 0, hence, B = Illustration 1: A current of 1A is flowing in the sides of an equilateral triangle of side 4.5 × 10 –2m. Find the magnetic field at the centroid of the triangle. Solution: Let be the current flowing in the sides of triangle. The magnetic field due to one side (say BC) of triangle at centroid O Here i = 1A α = β = 60°, R = OD = m ∴ The magnetic field due to whole triangle at centroid O is 1.3 MAGNETIC FIELD (B) AT THE CENTRE OF THE CURRENT CARRYING LOOP Magnetic field induction B = 1.4 MAGNETIC FIELD (B) AT THE CENTRE OF THE CURRENT CARRYING ARC We have to find magnetic field at point P due to arc which subtends an angle  at the centre (P). At the centre magnetic field induction B = Note: If the current in a loop is clockwise, then the magnetic field is directed inward perpendicular to the plane of the loop. If the current in a loop is anticlockwise, then the magnetic field is directed outward perpendicular to the plane of the loop. 1.5 MAGNETIC FIELD ON THE AXIS OF A CURRENT CARRYING LOOP Let the radius of the loop be a and axial distance be z. Note that from symmetry, the resultant field at P must be along z, therefore : wb/m2 (i) B varies non linearly as shown in the plot between z and B and is maximum when z = 0, i.e. the point at the centre of the coil and then B = , Which is same as magnetic field at the centre of a circular coil (ii) If z>>R, then B = = = but πR2 = A = Area cross section of the coil Then B = Or, Where = magnetic dipole moment of the current loop. The direction of is same as the direction of normal of the area of the loop. If the current is in clockwise sense then direction of magnetic dipole moment is along inward normal, otherwise it is along outward normal. If the coil consists of N turns then magnetic dipole moment of the loop will be given by M = NiA Illustration 2: In bhor model of hydrogen atom, the electron circulates around the nucleus in a path of radius 5.1 × 10 –11m at a frequency ‘v’ of 6.8 × 1015 rev/s. (a) What is the value of B at the centre of the orbit? (b) What is the equivalent dipole moment? Solution: (a) We know that Here e = 1.6 × 10 –19C and v = 6.8 × 1015 ∴ i = (1.6 × 10 –19) × (6.8 × 1015) = 1.1 × 10 – 3 amp. The field at the centre of a circular loop of current i is given by (b) Equivalent dipole moment = iA = i × πr2 = (1.1 × 10 – 3)π (5.1 × 10 – 11)2 = 90 × 10 – 24 Am2 2. MOTION OF A CHARGED PARTICLE IN A MAGNETIC FIELD The magnetic force on a charge q moving with velocity in a magnetic field is given by the expression. Here = velocity of the particle and = magnetic field On the basis of above expression, we can draw the following conclusion. (a) Stationary charge (i.e. = 0) experiences no magnetic force. (b) If is parallel or anti parallel to then the charged particle experiences no magnetic force. (c) Magnetic force is always perpendicular to both the and . (d) As magnetic force is always perpendicular to the , so magnetic force does not deliver power to the charged particle. (e) As magnetic force is always perpendicular to the , hence this force provides the necessary centripetal force to the charged particle to move on circular path. On the basis of expression the maximum value of magnetic force is equal to F=qvB, which occurs when the charge is projected perpendicular to the magnetic field. In this case path of charged particle is purely circular (in uniform ) and magnetic force provides necessary centripetal force. (f) If radius of the circular path is R then, where, m = mass of the particle (g) Time taken to complete one revolution is ⇒ T = Note: The time period is independent of the speed of the particle. If the charged particle is projected neither parallel nor perpendicular to the field then its velocity can be resolved into two components, one along the say v|| and the other perpendicular to say v⊥. It experiences a magnetic force and hence has a tendency to move on a circular path. Due to V|| it experience no force, and hence has a tendency to move on a straight path along the field. So in this case it moves along a helical path. (h) Radius of the helix is R = (i) (j) The distances moved by the charged particle along the magnetic field during one revolution is called pitch. Illustration 3: An α–particle is describing a circle of radius 0.45 m in a field of magnetic induction 1.2 weber/m2. Find its speed, frequency of rotation and kinetic energy. What potential difference will be required which will accelerate the particle, so as to give this much energy to it? The mass of α–particle is 6.8 × 10–27 kg and its charge is 3.2 × 10–27kg and its charge is 3.2 × 10–19 coulomb. Solution: We have or = 2.6 × 107 m/s The frequency of rotation = 9.2 × 106 sec–1. Kinetic energy of α–particle, = 2.3 × 10–12 joule. If V is accelerating potential of α–particle, then Kinetic energy = qV 14 × 106 eV = 2eV (since charge on α–particle = 2e) ∴ Illustration 4: What is the ratio of radii of paths when an electron and a proton enter at right angles to a uniform field with same (a) velocity (b) momentum and (c) kinetic energy? Given that (mp/me) = 1840. Solution: In case of circular motion of a charged particle in a uniform magnetic field, (a) As in this situation v, B and q are same for the two particles , so i.e., rp = 1840re (b) As in this situation, p, B and q are same for the two particles, so i.e., rp = re (c) As in this situation kinetic energy K, B and q are same for the two particle, so, i.e., rp ≈ 43re 2.1 FORCE ON A CURRENT CARRYING WIRE We know that We can say Actually, gives the force on the charge carriers within the length . However, this force is converted, by collisions, into a force on the wire as a whole, a force which, moreover, is capable of doing work on the wire. The net force on a wire is found by integrating along length. A corollary of this is there is no net force on a current carrying loop in a uniform magnetic field. 2.2 FLEMING’S LEFT-HAND RULE The direction of the force is given by the Fleming’s left hand rule. Close your left fist and then, “shoot your index finger in the direction of the magnetic field. Relax your middle finger in the direction of the current. The force on the conductor is shown by the direction of the erect thumb. Illustration 5: A wire bearing a current of 10A lies perpendicular to a uniform magnetic field. A force of 0.2 N is found to exist on a section of the wire 80cm long. Determine the magnetic induction B. Solution: For a straight segment of wire of length L, F = i B sin 90° or 0.2 = 10(0.80)B B = 0.025 T The direction of B will be normal to the plane of the force and the wire. Illustration 6: At the equator, the earth’s magnetic field is nearly horizontal, direction from the southern to northern hemisphere. Its magnitude is about 0.50 G. Find the force (direction and magnitude) on a 20–m wire carrying a current of 30A parallel to the earth (a) from east to west, (b) from north to south. Solution: 1 Tesle = 104 Gauss (a) F = iLB sin θ = 30(20)(5 × 10–5)(1) = 0.030 N, down (b) F = 30(20)(5 × 10–5)(0) = zero Illustration 7: A charge of 20 μc moves with a speed of 2.0 × 106 m/s along the positive x-axis. A magnetic force of strength (0.2 + 0.4 ) T exists in space. Find the force acting on the charge. Solution: Velocity = 2.0 ×106 F = qv × B = = 2.3 FORCE BETWEEN TWO PARALLEL CURRENT CARRYING WIRES To calculate the force between two infinite parallel current carrying wires separated by a distance r, we take an arbitrary pt. ‘P’ on the second wire; then the magnetic field at this point due to other wire is Force on an elementary length of the second wire is We note that wires carrying current in the same direction attract each other. (Verify using Fleming’s left hand rule). Illustration 8: A long horizontal wire P carries a current of 50 A. It is rigidly fixed. Another fine wire Q is placed directly above and parallel to P. The weight of wire Q is 0.075 N m–1 and carries a current of 25 A. Find the position of wire Q from P so that the wire Q remains suspended due to the magnetic repulsion. Also indicate the direction of current in Q with respect to P. Solution: As force per unit length between two parallel current–carrying wires separated by a distance d is given by: and is repulsive if the current in the wires is in opposite direction (otherwise attractive). So, in order that wire Q may remain suspended, the force F on it must be repulsive and equal to its weight, i.e,. the current in the two wires must be in opposite directions and F = mg, i.e., or, or, 3. TORQUE ON A CURRENT CARRYING PLANAR LOOP IN A UNIFORM MAGNETIC FIELD Case I. When place of the loop is perpendicular to magnetic field Length of AB = DC = And that of BC = AD = b Forces experienced by all the sides are shown in the figure ∴ Force on AB and DC are equal and opposite to the each other and that on BC and AD too. ⇒ ΣF = 0 Since the line of action of the forces on AB and DC is same and also the line of action of the forces BC and AD is same, therefore torque is zero. Case II. When the plane of the loop is inclined to the magnetic field. In this case again ΣF = 0 Lines of action of the forces on AB and DC are different, therefore this forms a couple and produces a torque. Side view of the loop is shown in the figure. Torque = Bil(bsinθ) = Bi(lb)sinθ ⇒ BiAsinθ. If loop has N turns then τ = BNiA sin θ In vector form , where, Energy needed to rotate the loop through an angle dθ is dU = τdθ ⇒ ΔU = ΔU = MB(cosθ1 − cosθ2), if we choose at θ1 such that at θ = θ1, U1 = 0 This is the energy stored in the loop. U = − If a current carrying loop is placed in a uniform magnetic field it experiences a torque which is given by the expression, Illustration 9: A circular coil of wire 8 cm in diameter has 12 turns and carries a current of 5A. The coil is in a field where the magnetic induction is 0.6 T. (i) What is the maximum torque on the coil? (ii) In what position would the torque be one half as great as in (i)? Solution: (i) The dipole moment of the coil = NiA Here N = 12 turns, i = 5 amp, and A = πr2 = π(4 × 10 – 2)2 ∴ Dipole moment = 12 × 5 × π × (4 × 10 – 2)2 = 0.30 Am2 Maximum torque = Mc B = 0.302 × 0.60 = 0. 181 n–m (ii) If θ be the angle between the axis of the coil and the field ∴ torque = Mc B sinθ According to the problem, torque Mc B /2 ∴ ∴ Thus the normal to the coil is at 30° to the field. Illustration 10: A rectangular loop of sides 20 cm and 10 cm carries a current of 5.0A. A uniform magnetic field of magnitude 0.2 T exists parallel to the longer side of the loop. What is the force acting on the loop? Solution: i = 5 A,  = 20 cm b = 10 cm = 0.2 T Forces on sides AB & CD are zero since are either parallel or anti-parallel to each other respectively along the two sides. Forces on side BC. F1 = i (10-1 ) = = Force on side AD F2 = = Hence the net force on loop is zero. 4. TORQUE ON A BAR MAGNET Due to equal and opposite forces acting on the poles of a bar magnet placed in a uniform magnetic field B, a torque acts on it about an axis passing through the centre and perpendicular to the length given by =MB sinθ where θ is the angle between the magnetic field and the magnetic moment . Since the net force on the magnet is zero, the torque on the magnet causes oscillations about the equilibrium position θ = 0°, when the magnet slightly displaced and set free. If is the moment of inertia of the magnet and M its magnetic moment, for small angular displacement the time period of oscillations will be given by T = 2π where moment of inertia = for a thin magnet. 5. WORK DONE IN ROTATING A BAR MAGNET The work done in rotating a bar magnet against the torque acting on it is W = = MB (cosθ1 − cosθ2) The potential energy U = − = − MB cosθ As work done is equal to change in potential energy W = U2 − U1 = MB(cosθ1 − cosθ2) Illustration 11: A Magnet is suspended in the magnetic meridian with a untwisted wire. The upper end of the wire is rotated through 180° to deflect the magnet by 30° from magnetic meridian. Now this magnet is replaced by another magnet. Now the upper end of the wire is rotated through 270° to deflect the magnet by 30° from magnetic meridian. Compare the magnetic moments of magnets. Solution: If φ be the twist of the wire, then , where C being restoring couple per unit twist of wire. Here φ1 = 180° – 30° = 150° = (150 × π/150) radian φ2 = (270° – 30° )= 240° = (240 × π/180) radian If M be the magnetic moment and H, the horizontal component of earth’s field, then = MH sinθ = MH sinθ If M1 and M2 be the magnetic moments of the two magnets respectively, then 1 = M1H sinθ for first magnet 2 = M2H sinθ for second magnet ∴ M1 : M2 = 5 : 8 6. AMPERE'S CIRCUITAL LAW The line integral of the magnetic field around a closed loop is proportional to the current linked by the loop. The proportionality constant is equal to μo (=4π × 10−7 H/m). Mathematical form of Ampere’s law is as follows This law is useful for us only in the case of symmetry. For unsymmetrical cases the Biot-Savart law is more useful. 6.1 MAGNETIC FIELD (B) DUE TO AN INFINITELY LONG STRAIGHT CONDUCTOR In the figure there is a long straight wire perpendicular to the plane of the paper. The inward direction of the current is shown by a cross. There is a point P at a distance r from the wire where the magnetic field is to be calculated. Draw a circle of radius r and symmetrically around the wire. This loop also represents magnetic lines of force due to the wire. The sense of lines of force can be found by using the right hand grip rule. At any point on this loop, is tangential. d is an elementary segment of this loop. This is also directed along the tangent. Hence = Bd Here = line integral of over closed path. i is the net current linked by the Amperes loop. From symmetry, the magnitude of the magnetic field at any point on the Amperes loop is always the same and so ⇒ B.2πr =μoi ⇒ B = Illustration 12: A current i flows along a long thin walled tube of radius r with a longitudinal slit of width h. Find the induction of magnetic field inside the tube (h < a is (D) The field at a distance r < a is 10. A straight thin walled tube of radius a has a current i flowing through it. If B (r) is the magnitude of the magnetic field at a distance r from the axis of the tude then, (A) B(r) = 0 for 0 ≤ r < a (B) B(r) ∝ for 0 ≤ r < a (C) B(r) ∝ for r > a (D) B(r) = 0 for r > a 11. A charged particle of specific charge α moves with a velocity in a magnetic field . Then (specific charge = charge per unit mass) (A) path of the particle is a helix (B) path of the particle is a circle (C) distance moved by particle in time (D) velocity of particle after time 12. A particle of mass m and charge q moves with a constant velocity v along the positive x–direction. It enters a region containing a uniform magnetic field B direction along the negative z–direction, extending form x = a to x = b. The minimum value of v required so that the particle can just enter the region x > b is: (A) (B) (C) (D) 13. A particle of charge q and mass m moves in a circular orbit of radius of r with angular velocity ω. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on (A) ω and q (B) ω, q and m (C) q and m (D) ω and m 14. An ionized gas contains both positive and negative ions. If it is subjected simultaneously to an electric field along the +x direction and a magnetic field along the +z direction, then (A) Positive ions deflect towards +y direction and negative ions towards –y direction (B) All ions deflect towards +y direction (C) All ions deflect towards –y direction (D) Positive ions deflect towards –y direction and negative ion towards +y direction. 15. Two particles A and B of masses mA and mB respectively and having the same charge are moving in plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are vA and vB respectively and the trajectories are as shown in the figure. Then (A) mAvA < mB vB (B) mAvA > mBvB (C) mA < mB and vA < vB (D) mA = mB and vA = vB CMP: A wire of resistance R in the form of a semicircle lies on the top of a smooth table. A uniform magnetic find B is confined to the region as shown. The ends of the semicircle are attached to spring C & D whose other ends are fixed. If r the radius of the semicircle and K is the force constant for each spring. 16. Net. Force constant of ring is (A) K/2 (B) 2K (C) K (D) K/3 17. Force applied on the semicircle wire of radius r due to current (A) 4 iBr (B) iBr (C) 6 iBr (D) 2iBr 18. What will be the extension x of ring is. (A) (B) (C) (D) 19. An electric current passes through a long straight wire. At a distance 5 cm from the wire, the magnetic field is B. The field at 20 cm from the wire would be (A) 2B (B) B/4 (C) B/2 (D) B 20. Two parallel wires of length 9 m each, are separated by a distance of 0.15 m. If they carry equal currents in the same direction and exert a total force of 30 x 10–7 N on each other, the value of the current must be (A) 1.5 Amperes (B) 2.25 Amperes (C) 0.5 Ampere (D) 0.25 Ampere 21. An electron beam is allowed to pass normally through magnetic and electric field which are mutually perpendicular. When the magnetic field induction and electric field strength are 0.0004 Wb/m2 and 3000 V/m receptively, the beam suffers no deflection. Then the velocity of electron is, (A) 7.5 x 106 m/sec (B) 7.5 x 104 m/sec (C) 7.5 x 102 m/sec (D) 1.2 x 106 m/sec 22. A galvanometer has a resistance of and a current of 0.01 A will cause full scale deflection. To convert this into an ammeter with full deflection for 5A, we have to connect approximately. (A) 0.1 in series (B) 0.1 in parallel (C) 0.2 in series (D) 0.2 in parallel 23. When two infinitely long parallel wires separated by a distance of 1m, each carry a current of 3 Amperes, the force in newtons/meter length experienced by each will be, (given = 4 x10–7 SI units) (A) 2 x 10 –7 (B) 3 x 10 –7 (C) 6 x 10 –7 (D) 18 x 10 –7 24. A doubly ionised He+2 atom travels at right angles to a magnetic field of induction 0.4 T at a velocity of 105 m/s describing a circle of radius 'r'. A proton travelling with same eed in the same direction in the same field will describe a circle of radius (A) r/4 (B) r/2 (C) r (D) 2r 25. Two long parallel copper wires carrying currents in the opposite direction of 5A each. If the wires are separated by a distance of 0.5m then the force between the two wires is: (A) 10–5 N/m attractive force (B) 10–5 N/m repulsive force (C) 2 x 10–5 N/m attractive force (D) 2 x 10–5 N/m repulsive force 26. A Charge of 10–6 C is describing a circular path of radius 1 cm making 5 revolution per second. The magnetic induction field at the centre of the circle is (A) (B) (C) (D) 27. A galvanometer has coil resistance 50 and shows full deflection at 100 . The resistance to be added for the galvanometer to work as an ammeter of range 20 mA is (A) 0.5 in series (B) 0.5 in Parallel (C) 50 in series (D) 50 in Parallel 28. If an electron revolves in the path of a circle of radius of 0.5x10–10 m at a frequency of 5x1015 cycles/s, the electric current in the circle is (charge of an electron 1.6 x 10–19 C) (A) 0.4 mA (B) 0.8 mA (C) 1.2 mA (D) 1.6 mA 29. A galvanometer of 25 Resistance can read a maximum current of 6 mA. It can be used as voltmeter to measure maximum of 6 V by connecting a resistance to the galvanometer. Identify the correct choice on the given answer. (A) 1025 in series (B) 1025 in parallel (C) 975 in series (D) 975 in parallel 30. A beam particles with ecific charge 108 C/Kg is entering with velocity 3 x 105 m/s by making an angle of with the uniform magnetic field of 0.3 tesla. Radius of path of particle is: (A) 0.5 cm (B) 0.02 cm (C) 1.25 cm (D) 2 cm 10. ANSWER SHEET 1. C 11. B, C 21. A 2. C, D 12. B 22. B 3. C 13. A 23. D 4. D 14. B 24. B 5. A, B, D 15. B 25. B 6. A, B, D 16. B 26. A 7. A, B 17. D 27. B 8. A, B, C, D 18. C 28. B 9. A, C, D 19. B 29. C 10. A, C 20. C 30. A