Moving Charges and Magnetism

Moving Charges and Magnetism OERSTED EXPERIMENT AND OBSERVATIONS (i) Oerseted performed an experiment in 1819 whose arrangement is shown in the following figure. Following observations were noted from this experiment. (ii) Oersted concluded from this experiment that on passing a current through the conducting wire, a magnetic field is produced around this wire. As a result the magnetic needle is deflected. This phenomenon is called magnetic effect of current. (iii) From another experiment, it is found that the magnetic lines of force due to the current flowing in the wire are in the form of concentric circles around the conducting wire.  (a) When no current is passed through the wire AB, the magnetic needle remains undefelcted. (b) When current is passed through the wire AB, the magnetic needle gets deflected in a particular direction and the deflection increases as the current increases. (c) When the current flowing in the wire is reversed, the magnitude needle gets deflected in the opposite direction and its deflection increases as the current increases. MAGNETIC FIELD (B) (Magnitude, Direction and unit) (i) When a charged particle q moving with a velocity   v passes through a magnetic field B (when electric field is not present), a force acts on it. It is called magnetic force. The force acting on a charged    paritcle due to its motion is given by F = q(v × B)  Its magnitude, | F | = qvB sin   where  is the angle between B and v . 1  (ii) It is found from this relation that the direction of F  is perpendicualr to the plane of B and v . If v = 0 or   B = 0 or v is parallel to B , then F = 0. (iii) If keeping the magnitude of v constant its direction is continuously changed, then the magnitude   of F changes but the direction of F always remains perpendicular to v . That direction of v for which  the magnitude F is zero or no force acts on the moving  charge, is the direction of vector B . (iv) If the test charge q moves with a velocity v perpendicular to the magnetic field, then the magnitude of F will be maximum, i.e., Fmax = qvB The magintude of B can be defined from B = Moving Charges and Magnetism Magnetizing Field (i) The magnetic induction B depends upon the external current which produces field as well as the magnetization of the medium. In absence of medium, that is in vacuum where effect of medium is zero, it the magnetic induction is B0, then B0 represents a vector which depends upon the external currents only. This vecot is called magnetizing field  H  (ii) For solenoid, the magnetising field is given by Fmax qv If q = 1C and v = 1 m/s, then B = Fmax Thus the magnetic force acting on a charge of 1C moving normal to the magnetic field with a velocity of 1 m/s is equal to the magnitude of the intensity of magnetic field. where n is the number of turns per unit length of solenoid and i is the current flowing in the solenoid. (v) If a charged particle passes through such region where both electric and magnetic fields are present, then the resultant force acting on the particle is the vector sum of electric and magnetic forces, 1 ampere/metre = 4 × 10–3 oersted. i.e.          (iii) Units : In C.G.S. system-oersted In M.K.S. system - ampere/metre BIOT AND SAVARTS LAW (i) The intensity of magnetic field due to current carrying conductor is determined with the help of this law. (ii) This law is obtained on the basis of experiments. This force is called Lorentz force. It is a microscopic force.  (vi) Unit of B in M.K.S. system is weber/meter2 or tesla or newton/ampere-meter. It is defined as If a charge of 1 coulomb is moving in a direction perpendicular to the direction of magnetic field with a velocity of 1 m/s and one newton force acts on it, then the magnitude of magnetic intensity of field is 1 weber/meter2.  In C.G.S. system the unit of B is maxwell/ cm2 or gauss. The relation of these units is 1 tesla = 104 maxwell/cm2 4 = 10 gauss (iii) The intensity of magnetic field (B) at a certain point due to current carrying element of length ( ) of the conductor depends upon the following (a) B is directly proportional to the current flowing through the element, i.e. B  i. (b)  B is directly proportional to the length of the element of the conductor, i.e.,  B  . 2 Moving Charges and Magnetism (c) B is inversely proportional to the square of the distance r of the observation point P from the element, i.e., B  1/r2 (d) dB is directly proportional to the sine of the angle   between the position vector r of the observation point P with respect to the element carrying element is given by (in rationalized M.K.S. system)  r2 where 0 is magnetic permeability of vacuum and its value is 4 × 10 –7 = 12.57 × 10 –7 weber/ amperemeter or henry/meter (H/m). (v) In vector representation  r3  (xi) Direction of magnetic field : The direction of magnetic field is determined with the help of the following simple laws : (a) Maxwell’s cork screw rule : According to this law if a right handed cork screw is rotated in such a way that it moves forward in the direction of current in the conductor, then the direction of the rotation of the screw will show the direction of lines of foce. (fig.-a)  2 The direction of the element  is taken in the direction of flow of current. (vi) The direction of magnetic field is always perpendiular to the plane formed by the vectors (fig.-a) (fig.-b)   and r (vii) The intenstiy of magnetic field due to entire length of the conductor will be  r2 (fig.-c) (b) Right hand palm rule : According to this rule if or a current carrying conductor is held in the right hand r2 such that the thumb of the hand represents the direction of current flow, then the direction of folding (viii) If  = 0, that is the observation point is on the fingers will represent the direction of magnetic lines current carrying element, of force. (fig.-b)    B = 0 or B = 0 (minimum) (c) Right hand palm rule for circular currents : (ix) If the observation point is on the line According to this rule if the direction of current in perpendicular to the current carrying element, then circular conducting coil is in the direction of folding  = 90º and sin  = 1 fingers of right hand, then the direction of magnetic field will be in the direction of stretched thumb. (fig.-c)   (maximum) (xii) The magnetic lines of force due to a current (x) The graph between B and i, B and   or , B carrying element are in the form of concentric circles and r, B and 1/r2 and B and  are as follows. with their centres at the element.  3 APPLICATION OF BIOT AND SAVARTS LAW Magnetic field due to a circular coil carrying current (i) If a circular coil of radius R is made from a wire of length L, then the number of turns in the coil will be Moving Charges and Magnetism (b) when current flows in opposite directions, n B ' 0 (ii) If the current flowing in the coil is i amperes, then from Biot-Savart’s law the magnetic field at the centre (iv) If any two points of a circular coil are connected of current carrying coil is to a battery and then current is passed through the coil. The magnetic field at the centre of the coil will be zero, i.e., B 0 = 0 The direction of magnetic field is along the axis of the coil. If the current flowing in the coil is anticlockwise, then the direction of B0 will be upward along the axis of the coil. If the current is clockwise, then the direction of B0 will be inward along the axis (v) The resultant magnetic field at the centre of two perpendicular concentric circular current carrying two of the coil. perpendicular concentric circular current carrying If a coil of one turn is made from a wire of length coils is L and another coil of n turns is made from a wire of same length and same amount of current flows through both coils, then the relation between the magnetic fields produced at their centres will be B0n = n2B01 B 0n B 01 or (iii) If same amount of current is passed through two concentric and coplanar circular coils of radii R 1 and R 2 and number of turns n 1 and n 2 respectively, then the intensity of magnetic field at the centre will be (a) when current flows in the same direction, If B0 makes an angle with B01, then tan 01 or If the two coils are identical, then    or B0 and 4 (vi) The magnetic field at an axial point situated at a distance x from the centre of a current carrying circular coil: Moving Charges and Magnetism (d) B is not uniform near the coil. The rate of change of magnetic field with distance is different at different points. At the centre of the coil, i.e. at x = 0 B0 Thus the magnetic field at the centre of the coil is maximum. (e) The graph between B and x is found to be of the following type. In this position the observation point P is at the same distance r from each element of the coil where and the angle between the current element  and the position vector r is 90o. Hence . According to Biot-Savart’s law the magnetic induction due to a current element is l The components of magnetic induction due to different elements along the axis are in the same direction but the perpendicular components are symmetrically in different directions  coil at the observation point, along the axis and normal to the axis. l Thus 2 2 or along the axis 2 B (f) The points of inflexion are those points on the curve where curvature becomes zero and the direction of curvature changes sign. The rate of change of magnetic field becomes constant at these points, i.e., dB dx or d2B dx2 . (g) In the (B – x) graph of current carrying coil two points of inflexion are found. They are at a distance x = ± R/2 from the centre. The distance between the two points of inflexion is equal to the radius of the coil. Helmholtz Coils If x >> R, then 2 B B 3 (a) B depends upon the distance x. B decreases as x increases and B = 0 at x = (b) At a distance ± R from the centre of the coil B B0 (c) when x = ± 0.766 R, then B . (i) If two identical coils placed coaxially parallel to each other in such a way that the distance between them is equal to the radius of the coil and same amount of current is passed through them in the same direction, then in the middle region of the coil (near x = ± R/2), on increasing the distance from the first coil field intensity will decrease but on decreasing the distance from the second coil the field intensity will increase by the same amount. Thus in this middle region the resultant magnetic field will be uniform. The two coils arranged in this way are called Helmholtz coils. 5 (ii) These coils are used to produce uniform magnetic field. (iii) The uniform magnetic field produced in the middle region is R Moving Charges and Magnetism (a) If length of the are is one fourth of the coil, then  = /2  B0 (b) If length of the arc is half of the coil, then  =  B0 (ii) Magnetic field due to current carrying conductor shown in the following figure The magnetic field at O B0 = BAC + BCDE + BEF Magnetic Field due to a Current Carrying Arc (i) If an arc of radius R subtends an angle  at its centre, then the lenght of the arc = R The magnetic field due to an element of the arc at the point O (from Biot-Savart’s law) (Inward normal to the plane of the paper) (iii) Mangetic field at the centre O of a current carrying conductor shown in the following figure l 2 but  = 90º  sin 90º = 1 l Thus B0 = BAC + BCDE + BEF + BFGA The magnetic field due to entire arc at the point O l but l l = R where  is in radians  Thus B0  i, B0   and B0  1 R Negative sign shown that the magnetic field is inward normal to the palne of paper. 6 (iv) The magnetic field at the centre O B0 = BAC + BCDE + BEF + BFGA Moving Charges and Magnetism (viii) The magnetic field at the O B0 = BAC + BCD + BDE + BEA or B 0 Upward normal to the plane of paper. (v) The magnetic field at the centre O Negative sign shows that the magnetic field will be inward normal to the plane of paper. (ix) The magnetic field at the centre O B0 = BCDE + BAF Inward normal to the plane of paper. (vi) The magnetic field at the centre O At point P the wires do not touch each other B0 = BOA + BACD + BDO Inward normal to the plane of paper. (x) The magnetic field at the centre O B0 = BAD + BPQC Upward normal to the plane of paper. (vii) The magnetic field at the centre O will be zero. B0 = BACD + BDE + BEFG + BGA Inward normal to the plane of paper. 7 Magnetic Field due to straight current carrying wire of finite length (i) If a current i is flowing through a conductor XY, then the magnetic field B due to it at a point P can be determined using Biot-Savart’s law. The normal distance of point P from the conductor is R. Moving Charges and Magnetism Due to symmetry the magnetic field due to each conductor will be . The direction of magnetic field in this case will be . The direction of magnetic field in this case will be upward perpendicular to the plane of the coil. The magnetic field due to all the four sides of the square  The direction of magnetic field will be normal to the plane of the coil. (iv) If a current i ampere is passed through the arms of an equilateral triangle of side a, then the magnetic If the lines joining the ends X and Y of the conductor field at its centre will be make angles and respectively with the perpendicular drawn on the conductor from the point P, then resultant magnetic induction at P due to current carrying conductor If the lines joining the ends X and Y of the conductor with the point P make angles and respectively with the conductor, then B0 (v) If a current i ampere is passed in a hexagonal coil of side a. The magnetic field at the centre of the coil (ii) If the current carrying conductor is of infinite length, will be then 2 (iii) The magnetic field at the centre of current carrying square coil of side a : If a current i is passed through a square coil ABCD of side a, then the magnetic field due to straight current carrying conductor AD of finite length B0 Magnetic Field due to Infinitely Long Current carrying Wire (i) Let current i be flowing in a long and straight wire. Wire is placed in the plane of paper. Magnetic field is determined at a distance r from this wire. The length of wire being very large relative to distance r, the wire can be assumed to be of infinite length. The magnetic lines of force due to current carrying wire will be in the form of concentric circles. 8 Moving Charges and Magnetism (a) currents in both wires are in the same direction (ii) If the magnetic induction at a distance r from the wire is B, then the circulation on the circular closed path of radius r.   B or B       (b) currents in the two wires are in opposite directions (iii) The magnetic field B due to long current carrying wire at a point near it (a) is proportional to the    current i and (b) is inversely proportiional to the distance r. This result can also be obtained from (ii) If the point P is outside the wire and Bio-Savart’s law. (a) currents in both wires are in the same direction (iv) The curve between B and r is or following type (v) The direction of magnetic field is perpendicular to the plane made by the wire and the position vector of the point. Magnetic filed due to two parallel and straight current carrying wires In the following figure, two long current carrying parallel wires are shown. The magnitude of current flowing in both wires is same. The distance between the wires is d. The magnetic field due to these wires at a point P will be.    (b) current in the two wires are in opposite directions    (I) If the point P is in between the wires and 9 (vii) In a cubical structure made by wires if current is passed from one of its corner, then the magnetic field at its centre will be zero. Because AB and HG; AE and CG, AD and FG, BF and DH, FE and CD, EH and BC are 6 identical current carrying condutor pairs and each pair produces equal and opposite magnetic field at the centre. Thus the resultant field is zero. (viii) If a square ABCD is made by joining four identical wires and current is passed from a battery, then the magnetic field will be zero at the centre of the square. Moving Charges and Magnetism Sol. In region B, field due to 1 is into the page and field due to 2 is out of page. So at some points we expect, these fields to cancel and add up to give zero. Similarly in region D, field due to 1 is out of page, while field due to 2 is into the page. Thus, again, some where fields A and C, fields due to 1 and 2 are in the same directions and can not give zero on addition. Thus the correct answer is (2) B and D. Ex.2 A straight horizontal stretch of copper wire carries a current i = 30 A. The linear mass density of the wire is 45 g/m. What is the magnitude of the magnetic field needed to “float” the wire, that is to be balance its weight ? (1) 147 G (2) 441 G (3) 14.7 G (4) zero G Sol. For L length of wire, to balance Fmagnetic = mg ILB = mg Therefore B = mg/IL = (m/L) g/I  45  10 3  9.8 30 = 1.47 × 10–2 tesla The correct answer is therefore (1) 147 Gauss Ex.3 You are given a closed circuit with radii a and b carrying current i. The magnetic dipole moment of the circuit is Sol. m = current × area = i  1 a 2  1 b 2  = 2 2 1 i(a 2  b 2 ) 2  Ex.1 As shown in the diagram two mutually perpendicular Ex.4 If i1 and i2 currents are flowing in two wires mutually wires are placed very close to each other, but are perpendicular to each other and placed very near, not touching each other. The regions where the then find the equation of locus of the points of zero intensity of magnetic field is zero is (are) magnetic field. Sol. BpQ  BRs For zero magnetic field (BPQ)  = (BRS) (1) A only (2) B, D (3) A, B (4) B only or or This locus is straight line. 10 Ex.5 A current i is flowing in a square coil of side a, find the value of magnetic field at centre. Sol.   Moving Charges and Magnetism    Sol. PQRS is current carrying square of side a due to which magnetic field at its centre O   B = 4 × BPQ Ex.8 Find the magnitude field at O due to curved and linear portions of conducting wire as shown in figure. here 1 = 2 = 45º and r = Sol. a 2    BPSR BRQP B or B [ OQ = r cos ] Ex.6 In an equilateral triangle of side 4.5 × 10–2 m, 1 ampere current is flowing. Find magnitude of B at its centroid. Sol. Due to AB value of magnetic induction B1 at O here i = 1 A, 1 = 2 = 60º by figure or a /2 Ex.9 There are two concentric coils having same number of turns and radius 10 cm and 30 cm respectively. A current is made to flow in them (i) in same direction, (ii) in opposite direction. In both the cases the ratio of resultant magnetic fields at the centre of the coils will be : Sol. (i) When the current flows in same direction     ni  ni | B0 || B0  B 02 | 0  0 2R1 2R 2  a so  0 ni  1 1     2  0.1 0.3   ni 40  0  2 3  2 3 At the centre of triangle O, B = 3B1 and this will be upward perpendicular. (ii) When the current flows in opposite direction    B 0 '  B01  B 02  0ni  0 ni  0ni  1 1     2R1 2R 2 2  0.1 0.3   ni 20  0  2 3 B0 '  4 = 4 × 10–5 Wb/m2 Ex.7 In the body shown in diagram i current is flowing. Radius of circular part is r and linear parts are very long. Find magnitude of magnetic field at O. B0 2  B '  1  2 :1 0  (2) is correct answer.. 11 Ex.10 Two wire ABC and DEF are arranged as shown in figure. Current i is flowing through them. value of magnetic field at point O will be :  i BA   0i 2 r , in the direction PQ and due to B, Sol. B0 = BAB + BBC + BDE + DEF1  i Moving Charges and Magnetism Sol. Magnetic field at P due to wire A is BB   i = 0  4 0a  0  40a  20a Ex.11 In the given figure, find the magnetic field at the centre O : Sol. B0 = BAB + BBCD + BDE  0i 2r , in the direction PR. On resolving the magnetic field into their components, resultant of components parallel to AB will be zero whereas components perpendicular to AB will be added. Resultant magnetic field 2 i a  ia 0 0 B = 2B sin   2r r  2 r   0 i 1  0 i  0i   4x 4 2r 4x 2   0 2i 2  0  i  .    4 r 4   2r    0i  0  i   .  4 x 4  2r   0  2i      2  4  r   4 FORCE ON A CURRENT CARRYING CONDUCTOR DUE TO MAGNETIC FIELD (i) If a current i is passed through a conductor of length l and it is placed in a magnetic field B, then the force acting on it will be Ex.12 A current of 2 ampere is made to flow through a coil which has only one turn. The magnetic field produced at the centre is 4  10 6 Wb/m2, then radius of the coil will be : Sol. B 0  B0   0ni 2R 0 i 2R (but n = 1) or or R R  0i 2B0 4   10 7  2 2  4  10 6  0.1m Ex.13 Two parallel wires situated at a distance 2a are there in which current i is flowing in opposite direction as shown in the figure. Value of magnetic field at point P which is situated at a distance r from both the wires will be : or (ii) If  l   l   , that is, B || l , then F = 0 (minimum) (iii) If , that is, B l , then F = Bil (maximum) (iv) The direction of magnetic force is perpendicular to the plane of l and B according to right hand screw rule. Following two rules are used in determining the direction of the magnetic force. (a) Right hand palm rule : If the right hand and the palm are stretched such that the thumb points in the direction of current and the stretched fingers in the direction of the magnetic field, then the force on the conductor will be perpendicular to the palm in the outward direction. 12 (b) Fleming left hand rule : If the thumb, fore finger and central finger of the left hand are stretched such that first finger points in the direction of magnetic field and the central finger in the direction of current, then the thumb will point in the direction of force acting on the conductor. Moving Charges and Magnetism This force will act away from the current-carrying wire. Similarly, the (attractive) force on the side CD of the loop, due to current i 1 is (now R = d + b = 10 cm = 0.10 m) 7  F2  ( 2  10 )  20  16  0.15  0.96  10  4 N 0.10 This force will act towards the current carrying wire. The force on the sides AD and BC of the loop will be equal and acting downward and upward respectively. Hence they will neutralize each other. Ex.14 The distance between the wires of electric mains is 12 cm. These wires experience 4 mg wt. per unit length. The value of current flowing in each wire will be – Sol. F  0i 2  l 2d × 10–4 = 1.44 × 10–4 N acting away from the current - carrying wire. MAGNETIC FORCE BETWEEN 9.8  4  10 6  i  net force on the loop = F1 – F2 = (2.40 – 0.96) 4   10 7  i2 TWO CURRENT CARRYING 2    12  10  2 PARALLEL CONDUCTORS 4  10 6  9.8  0.12 2  10 7  4.85 A Ex.15 In the adjoining fig the force on current carrying loop ABCD due to a long current carrying (i1) wire will be– (i) In the figure two parallel conductors are shown. The distance between them is d. Let i1 and i2 be the currents flowing in these conductors. When current flowing in these conductors are in the same direction, they attract each other. When currents flowing in them are in opposite direction, they repel each other. The force of attraction or repulsion acting on unit length of each conductor will be Sol. The (repulsive) force on the side AB of the loop, due to current i1 is F1  given 0 2  0 i1i2L 2 R = 2 × 10–7 N/A2, i1 = 20A, i2 = 16A, R = d = 4 cm = 0.04 m and L = a = 15 cm = 0.15 m & b = 6 cm  F1  ( 2  10 7 )  20  16  0.15  2.40  10  4 N . 0.04 (ii) Force acting on a conductor of length    newton 13 (iii) Attractive or repulsive force  and    i.e., they are equal in magnitude but opposite in  directions, where F12 is the force on the conductor  1 due to conductor 2 and F21 is the force on second due to first conductor. Moving Charges and Magnetism (vii) As shown in figure current i is flowing through a rectangular loop. It is placed near a long straight wire in such a way that the wire is parallel to one arm of the loop and is in the plane of the loop. If constant current is passed through the wire in a direction shown in the figure, then the loop will move towards the wire because there will be more attraction due to current flowing through nearby arm of the loop in the same direction. (iv) Definition of Ampere : If i1 = i2 = 1 ampere and d = 1 metre, then F = 2 × 10–7 N/m If two conductors of infinite length and negligible cross-sectional area are placed parallel to each other at a distance 1m apart each carrying same current in the same direction produce a force of attraction of 2 × 10–7 N/m between them, then the current flowing in each wire will be 1 ampere. (viii) Current carrying spring : From the above definition of ampere, the new unit of is 2 Dimensions of (a) If current is passed through a spring then it will contract because current will flow through all the turns in the same direction. (b) In the figure shown, when key K is pressed, then the spring will contract. As a result curent will stop flowing and the spring will region its original position. Hence spring will again touch mercury and current will start flowing and again it will contract. In this ay the spring will oscillate up and down. (v) Two parallel wires are placed one over the other. One wire is stationary and other is free. If free wire is above the stationary wire, then in order to stop the free wire falling currents will have to be passed in the two wires in opposite directions. The force of repulsion due to them will balance the weight of free wire. If free wire is below the stationary wire, then currents will have to be passed in the same Ex.16 Two parallel wires are situated at a distance 10 cm and if 1 ampere current is flowing through each direction. In this case force of attraction due to them wire. The calculate the force per unit length on any will balance the weight of free wire. one of them. (vi) Some weight is suspended from a spring. If current is passed through it, then there will be Sol. We know that attraction between the turns due to flow of current in the turns in the same direction and spring will contract.  = 2 × 10–4 N/m 14 Ex.17 As given in the figure X, Y and Z are three current carrying wires. What will be the direction of resultant magnetic force on wire Y ? Sol. There will be attraction between X and Y and repulsion between X and Z. Both forces will act towards left     B 2 Resultant force,    0 x  FTotal  2r1 or y  0 y z 2r2 Ex.18 A flexible conducting loop of wire of length 0.5 meter is kept in a magnetic field perpendicular to the plane of loop of value 1 tesla. Show that when the current flows in the loop then it acquires circular shape. Also calculate the tension in the wire if 1.57 amperes current is flowing through it. Sol. Force on each small element dl of loop placed in magnetic field will be iBdl and it will be perpendicular to the element d  Moving Charges and Magnetism (ii) If charge is not stationary, in that case magnetic field also acts in addition to the electric field. (iii) Each charge q moving with velocity v produces a magnetic field whose intensity at a point having the position vector r will be Loop will acquire circular shape At equilibrium,  = iBdl 2 2T sin B 2 Position vector r is with respect to source charge. (iv) If charges q1 and q2 are moving with velocities v1 and v2 respectively and at any instant the distance between them is r, then the force between them will be Fm newton 2 (a) If both charges are of same nature and moving in the same direction, then the magnetic force acting between them will be attractive in nature. (b) If the two charges are moving in opposite directions or their nature are opposite, then the magnetic force acting between them will be repulsive in nature. (v) If both charges are moving with the same velocity, then the magnetic force acting between them will be 2 If  is small sin  2 Fm 2  or 2T 2 = iBdl   B T dl    r but    or T  Bid l = Bir = T i iBl 2 1.57  1 0.5  0.125 N 2  3.14 FORCE BETWEEN TWO MOVING CHARGES (i) Electric force acts between two stationary charges which is determined by Coulomb’s law 2 newton That is, the magnetic force acting between the charges depends upon their velocities. (vi) If an observer moving with velocity v observes a charge moving with the same velocity v, then it will appear stationary. In that case electric force will appear to be acting between the charges and not the magnetic force. (vii) If charges are stationary and they are seen by an observer moving with velocity v, then both electric and the magnetic force both will appear to be acting between them. (viii) The existence of the magnetic force acting between two charges depends upon the motion of charges relative to the observer. Thus for the observers moving with different velocities the magnetic force acting between two charges will appear different. 15 Moving Charges and Magnetism (ix) The ratio of the magnetic force and the electric force acting between two moving charges is (1) (2) 2 Fm Fe 2 but 2 where c is velocity of light in vacuum. (4) If v << c, then Fm << Fe. (x) The force acting between the charges is represented in the following figure. (3) Sol. According to the figure, at point P due to the motion of proton current i+ will produce field B+ whose direction will be inward ( B    0i / 2r ). Ex.19 A proton is moving in circular path perpendicular  to some magnetic field of induction B . If value of B is doubled and radius of circular path is kept constant then what will be the value of kinetic energy of the particle ? In the same way at point P, due to the motion of electrons a current i– produces field B–. Whose Sol. Value of kinetic energy of proton direction will be outward ( B O   0i / 2r ). Ans. (3) Ex.21 In hydrogen atom electron revolves around proton Value of kinetic energy if value of B is doubled in an orbit of radius 0.5Å. If at the position of proton 2 2 2 2 2 2 magnetic field of magnitude 14 tesla is produced, R q (2B) 4R q B (KE )new   then what will be the value of angular frequency of 2m 2m motion of electron: Which is four times the initial kinetic energy. Ex.20 Two beams of electrons and protons are equivalent Sol. Magnitude of magnetic induction on proton will be  ef to same value of current are at a distance d. Electron B0  0 2R and proton are moving in opposite direction. There is a 14  2  0.5  10 10 B  2R point P situated between these two beams at a distance   f 0 0 e 12.57  10 7  1.6  10 19 x from one of the beams having magnetic flux density B. If curve is drawn between B and x, then the resultant 14  10 10 14  1015 = 7 × 1015 = or =  26 2 curve will be : 20.112  10 KE  1 R 2 q2B 2 mv 2  2 2m 16 Moving Charges and Magnetism (vi) For maximum torque or , i.e., the plane or the loop should be in the direction of magnetic field or the vector area of the loop should be normal to the magnetic field. FORCE AND TORQUE ON A CURRENT CARRYING LOOP PLACED IN A UNIFORM MAGNETIC FIELD (i) A rectangular conducting loop PQRS is placed in a uniform magnetic field. Its length is l and breadth is b. The rotational axis of the loop is perpendicular to the magnetic field B. When current i is passed through the loop, then the resultant force acting on the loop is zero but a torque acts on it. (vii) For minimum torque or , i.e., the plane of the loop should be normal to the magnetic field or vector area should be in the direction of magnetic field. (viii) The torque acting on a current - carrying loop placed in a magnetic field depends on the current. Moving coil galvanometer is based on this principle. (ix) The work done in deflecting a coil from angle to The work done in deflecting a coil from an angle 0 to in a magnetic field (ii) If plane of the loop makes an angle with the direction of magnetic field, then the torque acting on the loop  but b = A = area of the rectangular loop . Ex.22 If i current is flowing in wire PQR. It is given the shape according to the figure and kept in a magnetic field B. If PQ =  and PQR  45 o , then ratio of force on QR and PQ is : If the loop has N turns, then (iii) If the vector area of the loop makes an angle with the direction of magnetic field, then Torque For N turns (iv) NiA = M = magnetic moment of the loop Sol. FQR FPQ o  Ex.23 The length of a solenoid is 0.4 m and the number   turns in it is 500. A current of 3 amp. is flowing in In vector form,  it. In a small coil of radius 0.01 m and number of (v) The torque acting on a current carrying loop turns 10, a current of 0.4 amp. is flowing. The torque placed in a magnetic field depends on its magnetic necessary to keep the axis of this coil perpendicular moment but not the shape of the loop. to the axis of solenoid will be – 17 Sol. B solenoid   0 ns is or   B s .iNA Moving Charges and Magnetism Ex.26 The magnetic force on segment PQ, due to a current of 5 amp. flowing in it, if it is placed in a magnetic field of 0.25 Tesla, will be – s    0NsisiNr 2 Ls 4  10 7  500  3  0.4  10    ( 0.01) 2 0.4 F = Bilsin  = 0.25 × 5 × 0.25 sin 65° Sol. –6 = 5.92 × 10 N-m. = 0.3125 sin 65° Ex.24 The effective radius of a coil of 100 turns is 0.05 m and a current of 0.1 amp is flowing in it. The work required to turn this coil in an external magnetic field MOTION OF A CHARGED of 1.5 Tesla through 180° will be, if initially the plane PARTICLE IN A MAGNETIC FIELD (i) If a particle of charge q moves in a magnetic field of the coil is normal to the magnetic field.   B with a velocity v , then the force acting on the Sol. W = 2MB particle will be W = 2i Na2B    or W = 3.14 × 2 × 0.1 × 100 × (0.05)2 × 1.5 or = 0.236 Joule where is the Ex.25 A conducting wire of length l is turned in the form angle between of a circular coil and a current i is passed through   B and v . it. For torque due to magnetic field produced at its (ii) If , i.e., the charged particle moves in the centre, to be maximum, the number of turns in the direction of magnetic field, then coil will be– Sol. max  MB or max  nia 2B Let number of turns in length  is n  a or  2 iB  Thus no force will act on the particle. In this case the particle will remain in motion without any deflection and the energy and momentum of the particle will remain constant. (iii) If , i.e. the charged particle moves in a direction normal to the magnetic field, then (a) F = qvB sin 90 = qvB, this force will be maximum. (b) Force will act in a direction perpendicular to the motion of particle. (c) Particle will move in a circular path.  min nmin = 1 (d) The kinetic energy of the particle will remain constant. 18 Moving Charges and Magnetism (e) The magnitude of momentum of the particle will remain constant but its direction will change continuously. (f) If R is the radius of the circular path of the particle, then mv 2 R (v) Moving charged particle produces both electric and magnetic fields. R where p is the momentum of the particle. (g) Kinetic energy of the particle – (vi) If a proton and -particle having same kinetic energy move perpendicular to the magnetic field B , then the ratio of the radii of their paths will be RP (h) If E is the energy of the particle, then the radius of the circular path will be but and p R and Rp R (i) The work done by the charged particle in completing a full circle will be zero. (j) Period of the particle T and frequency n Thus the period or the frequency does not depend Ex.27 A point charge of q coulomb is moving along a upon the speed of the particle. As the speed of the circular path of radius r meter at the rate of n rotations particle increases, the radius of circular path increases per second. Magnitude of magnetic induction at the in the same ratio so that the time taken in completing centre of circle (in tesla) will be : a full circle remains same. This frequency for electron is called cyclotron frequency. Sol. From any point on the circle charge flowing or circular  (iv) When the velocity v of moving charge (+q0) is not current per second will be i = nq. perpendicular to the magnetic field, i.e., , Magnetic induction produce at the centre due to then the velocity component v cos parallel to the circular current magnetic field remains uneffected, as a result particle  i moves in the forward direction and due to velocity B 0 2r moves in the forward direction and due to velocity component v sin normal to the magnetic field the or 2r particle moves in a circular path. Thus the particle has both types of motion simultaneously so the 2nq   10 7 T resultant motion is in the form of helix. r 19 5 Ex.28 An electron is projected with a velocity of 10 m/s at right angles to a magnetic field of 0.019 G. Calculate the radius of the circular path described by the electron, if e = 1.6 × 10 –19C, m = 9.1 × 10–31 kg. Sol.  v = 105 m/s; e = 1.6 × 10–19 C; m = 9.1 × 10–31 kg; B = 0.019 G = 0.019 × 10–4 T Ex.29 Electron moving at right angles to the uniform magnetic field completes a circular orbit in 10–6 s. Calculate the value of magnetic field. or B  2 m Be velocity v  (2 î  3 ĵ ) m/s in a magnetic field  B  (2 î  3 ĵ ) Wb/m2. Find the force acting on the test charge.   Sol. e = 1.6 × 10–19 C; v  (2 î  3 ĵ) m/s; B  ( 2 î  3 ĵ) Wb m–2    F  e(v  B) = 1.6 × 10–19 [(2 î  3 ĵ )  (2 î  3 ĵ )]  0 mv 9.1 10 31  10 5 r   0.299 m Be 0.019  10  4  1.6  10 19 Sol. Period of revolution T  Moving Charges and Magnetism Ex.31 A test charge of 1.6 × 10–19 C is moving with [ T = 10–6 s] 2m 2  9  1031   3.534  105 Tesla eT 1.6  1019  106 AMPERES LAW (i) If an imaginary path is considered in a magnetic field, then the integral of the scalar product of the magnetic  field B at the position of an element with the  displacement  along the path is called line integral. Thus the line integral of the magnetic field for the Q    B . d path PQ  Ex.30 A beam of protons with velocity 4 × 105 m/s enters P a uniform magnetic field of 0.3 tesla at an angle of 60° to the magnetic field. Find the radius of the helical (ii) If the path is closed, then the line integral along the path taken by the proton beam. Also find the pitch of helix. Mass of proton = 1.67 × 10–27 kg. claose path is called circulation.  Circulation = line integral along the closed path    B . d  (iii) The ratio of the magnetic induction due to current or moving charge to the permebaility  of the medium,  i.e. B /  is called magnetizing field H. The  mv sin  ( component of velocity  to field qB is v sin  ) Sol. r   (1.67  1027 )(4 105 ) 3 / 2 (1.6  10 19  2  102 m  1.2cm 3 )0.3   Megnetomotive force Fm   H. d  Again, pitch p  v cos   T Where 1   B . d +   Unit of magnetomotive force is ampere. This quantity is analogous to electromotive force which is 2r T v sin    o  p magnetizing field H depends upon the current flowing in the conductor and its geometry. (iv) The circulation or the the integral of the magnetizing field along the closed path is called magnetomotive force. v cos   2r cos 60  2  (1.2  10  v sin  sin 60 = 4.35 × 10–2 m = 4.35 cm 2 ) equivalent to the circulation  E . d , where E is the intensity of electric field. Unit of electromotive force is volt. 20 (v) According to Ampere’s law, the integral of magnetic field along a closed path, i.e., circulation is equal to the product of 0 and the algebraic sum of currents crossing the area bound by the closed path, where m0 is the permeability of the medium.  Moving Charges and Magnetism (iii) Magnetic field at the surface of the cylinder (r = R) Bsureface (iv) Magnetic field inside the conductor (r < R)  Circulation   B . d   0  i c (vi) If no current passes through the area bound by the closed path, then the circulation is zero, i.e.   B . d  0 c  (vii) From Ampere’s law the magnetic field B can be determined but it can be used only in those circumstance when the distribution of current is uniform and symmetrical, that is, current should be completely defined and amperian path should be symmetrical. (viii) If magnetic field B is defined in terms of magnetising field H, i.e., H = B/0, then Ampere’s law can be written in the following form    H . d   i (ix) Magnetomotive force on the closed path is equivalent to the algebric sum of the electric currents crossing the area bound by the closed path. (a) Magnetic Field due to a Long Cylindreical Current Carrying Conductor (i) Current i is flowing symmetrically through a straigh long solid cylindrical conductor. Magnetic field is produced due to current flowing in the conductor around the conductor whose lines of force are concentric circles outside and inside the conductor. Thus the magnetic field inside the cylindrical conductor is proportional to the distance r from the axis. (v) In the figure the dependence of magnetic field due to cylindrical conductor with the distance r from the axis is shown. (vi) The magnetic field inside a current current pipe (hollow cylinder) : If cylindrical condcutor is hollow, i.e., it is in the form of pipe then the current flows through the surface only. In this case the current crossing an area r2 of a circular path inside the hollow part is zero. Thus from Ampere’s law     B=0 That is, the magnetic field inside the current carrying pipe is zero. For this case the graph between B and the distance r is given below. (b) Megnetic Field Inside a Solenoid (ii) Magnetic filed outside the conductor (r > R) B ext Thus the magnetic field due to current carrying cylinder outside it is equivalent to the magnetic field due to thin current carrying wire. (i) Solenoid is a long cylindrical tube on which electically insulated wire is wound symmetrically along its length. (ii) The diameter of a solenoid is small in comparsion of its length and the plane of each turn of wire of an ideal solenoid can be considered to be normal to its axis. 21 (iii) When current is passed through the soleniod, the magnetic behaviour at the points near each turn of the wire is similar to the straight current carrrying wire and magnetic lines of force are in the form of concentric circles, as shown in figure. Moving Charges and Magnetism (viii) If a magnetic material of permeability  is placed inside the solenoid, then B = ni or B = (N/L) i (ix) The magnetic field at one end of the solenoid is half of the magnetic field at its centre i.e., Bend (x) The magnetising field inside the solenoid is  (iv) The magnetic field B at any point of the solenoid and at its end Hend is equal to the vector sum of the magnetic fields (xii) B is uniform over the cross-sectional area of produced by the different turns. the solenoid and it does not depend upon the length (v) The magnetic field inside a solenoid is almost or diameter of the solenoid. uniform and along the axis and the magnetic field outside the solenoid at large distance is negligible (c) Magnetic Field in a Toroid in comparison to the magnetic field inside the solenoid (i) Toroid is like an endless cylindrical solenoid, i.e. thus the resultant magnetic field outside the solenoid if a long solenoid is bent round in the form of a closed at large distances can be considered to be zero. ring, then it becomes a toroid. (vi) A solenoid of finite length : The magnetic field (ii) Electrically insulated wire is wound uniformly over at a point P inside a current carrying solenoid of finite the toroid as shown in the firgure. length If L is the length of the solenoid and R its radius, then the magnetic field at its centre B oc where n is number of turns per unit length of solenoid. (vii) A Solenoid of infinite length : Magnetic field inside a solenoid B0 = 0ni where n is number of turns per unit length of solenoid. If N is number of turns in a solenoid of length L, then n  B0 = 0(N/L) i (iii) The thickness of toroid is kept small in comparison to its radius and the number of turns is kept very large. (iv) When a current i is passed through the toroid, each turn of the toroid produces a magnetic field along the axis at its centre. Due to uniform distribution of turns this magnetic filed has same magnitude at their centres. Thus the magnetic lines of force inside the toroid are circular. 22 (v) The magnetic field inside a toroid at all points is same but outside the toroid it is zero. (vi) If total number of turns in a toroid is N and R is its radius, then number of turns per unit length of the toroid will be Moving Charges and Magnetism (vi) Magnetic moment of a current carrying coil is equal to the product of the current flowing in the coil and its effective area i.e., M = IA (vii) For a coil of one turn n (vii) The magnetic field due to toroid is determined by Ampere’s law. (viii) The magnetic field due to toroid is B0 = 0ni or (ix) If a substance of permeability  is placed inside the toroid, then B = ni If r is relative magnetic permeability of the substance, then B = r0 ni MAGNETIC BEHAVIOUR OF A CURRENT CARRYING COIL AND ITS MAGNETIC MOMENT (i) If current is passed in a small coil (loop), then the magnetic field due to it at a point on its axis is equivalent to the magnetic field due to a small magnet or magnetic dipole at a point on its axis. (ii) A current carrying small coil behaves like a small magnet. (iii) If the current in the coil is observed anticlockwise, then that plane of the coil behaves like the north pole. (iv) If the current in the coil is observed clockwise, then that plane of the coil behaves like the south pole. (v) Current carrying coil is equivalent to a magnetic strip. where R is the radius of a coil and I is the current flowing in the coil. For a coil of N turns (viii) Unit of magnetic moment ampere – m2 (ix) The magnetic field B0 at the centre of a current carrying circular coil and its magnetic moment M have the following relation. B0 (x) If the magnetic moment of a current carrying small coil is M, then the magnetic field at a distance x(x >> R) on its axis from the centre will be BP CURRENT AND MAGNETIC FIELD DUE TO CIRCULAR MOTION OF A CHARGE (i) According to the theory of atomic structure every atom is made of electrons, protons and neutrons. protons and neutrons are in the nucleus of each atom and electrons are assumed to be moving in different orbits around the nucleus. (ii) An electron and a proton present in the atom constitute an electric dipole at every moment but the direction of this dipole changes continuously and hence at any time the average dipole moment is zero. As a result static electric field is not observed. (iii) Moving charge produces magnetic field and the average value of this field in the atom is not zero. (iv) In an atom an electron moving in a circular path around the nucleus. Due to this motion current appears to be flowing in the electronic orbit and the orbit behaves like a current carrying coil. If e is the electron charge, R is the radius of the orbit and f is the frequency of motion of electron in the orbit, then 23 (a) current in the orbit = charge × frequency = ef If T is the period, then Moving Charges and Magnetism Writing M in terms of L f M According to Bohr’s second postulate i In ground state n = 1 L M (b) Magnetic field at the nucleus (centre) B0 (c) If the angular velocity of the electron is , then and f (vi) If a charge q (or a charged ring of charge q) is moving in a circular path of radius R with a frequency f or angular velocity , then (a) current due to moving charge (b) magnetic field at the centre of ring B0 B0 (d) If the linear velocity of the electron is v, then or f or B0 (c) magnetic moment B0 (v) Magnetic moment due to motion of electron in an orbit (vii) If a charge q is distributed uniformly over the surface of plastic disc of radius R and it is rotated about its axis with an angular velocity , then (a) the magnetic field produced at its centre will be or M or M B0 (b) the magnetic moment of the disc will be If the angular momentum of the electron is L, then M 24