MAGNETIC EFFECT OF CURRENT

Transcription

1 PMEC 1 MAGNETIC EFFECT OF CURRENT C1 MAGNETIC FIELD OF UNIFORMLY MOVING CHARGE A moving electric charge creates a magnetic field in the space around it. The law defining the magnetic field B of a point charge moving at a constant non-relativistic velocity v (v r) is written as B q 4 3 r where is the magnetic constant, the coefficient H / m, and r is the radius vector from the point charge q to the point where the field is to be determined as shown in figure. The direction of B can be obtained by applying screw rule on Practice Problem : v r A positive point charge q = 3. C has velocity v (4 1 m / s) î relative to a reference frame. At the instant when the point charge is at the origin in this frame, the magnetic field B that it produces at (.5,, ) m is mt 1 mt.5 mt.5 mt [Answers : (1) a] C C3 MAGNETIC FIELD OF A CURRENT ELEMENT The law of Biot and Savart gives the magnetic field created by a small elements of length dl carrying a dl r current I. According to this law : db I where d l vector points in the direction of current 4 3 r I. The vector r goes from the current element to the point where the field db can be obtained by applying the screw rule (d l r). This law can be used to find the magnetic field created by any configuration of current carrying wire. MAGNETIC FIELD FOR SOME IMPORTANT CONFIGURATION 1. A straight current carrying wire case a : case b : Consider a wire AB carrying current I as shown in figure. For points along the length of the wire the field is always zero. I The magnetic field at point P is given by B (cos cos) 4 r case c : If the wire is of infinite length then the field at P is I. 4 r B Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,

2 PMEC case d : The magnetic field at point P is given by Icos B 4 r case e :. A circular current loop If the wire is of semi infinite length then the field at P is given by Consider a current loop of radius r carrying a current I and having N turns. I. 4 r B Field at centre of the loop : NI B 4 r Field at axis passing through the centre and perpendicular to the plane of the loop : Note that only axial component of the field will exist. 3. A Current Arc I Il Field at centre of the arc is B, 4 r 4 r where l is the length of the arc and is the angle made by the arc at the centre. 4. Solenoid NIr B 4 (r x ) 3/ If several turns of an insulated wires are wounded around the cylinder then the resulting coil is called a solenoid. Field at an axial point of a solenoid is given by B (ni)[cos cos] 4 Here n is then number of turns per unit length. case a : If the solenoid is of infinite length then B = ni case b : Field near the end of an infinite solenoid is case c : Field outside the solenoid is zero. 1 ni Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : , Conventionally the direction of the field perpendicular to the plane of the paper is represented by if into the page and if out of the page. Practice Problems : 1. A square coil of side a carries a current I. The magnetic field at the centre of the coil is I a I a I a I a. Two circular coils have number of turns in the ratio 1 : and radii in the ratio : 1. If the same current flows through them, the magnetic fields at their centres will be in the ratio 1 : 1 1 : : 1 1 : 4

3 PMEC 3 3. The magnetic field inside a current carrying toroidal solenoid is B. If its radius is doubled and the current through it is also doubled, the magnetic field inside the solenoid will be B/ B B 4B 4. Two identical coils have a common centre and their planes are at right angles to each other. They carry equal currents. If the magnitude of the magnetic field at the centre due to one of the coils is B then that due to the combination is B B B/ B 5. A current of 1 A is flowing in the sides of an equilateral triangle of side m. The magnetic field at the centroid of the triangle is 1 5 T T T T 6. Match the following column : Column-I Column-II (p) (i) I I out of the page 4r 4r I (q) (ii) 1 4r out of the page (r) (iii) I r 1 r 4 r1r into the page (p)-(i), (q)-(ii), (r)-(iii) (p)-(ii), (q)-(i), (r)-(iii) (p)-(iii), (q)-(ii), (r)-(i) (p)-(i), (q)-(iii), (r)-(ii) [Answers : (1) d () d (3) c (4) b (5) b (6) a] C4 MAGNETIC FORCE ON MOVING CHARGES A magnetic field exerts a force on a moving charge but not on a stationary charged particle. The force on a charge q moving with velocity v in a magnetic field B is given by F q(v B). F = q v B sin where is the angle between the B and velocity v. The direction of F is given by right hand screw rule. The force on a positive charge is in the direction of v B but the force on a negative charge is opposite to the vector v B. For v B, F = and for F qvb Note that force F is always perpendicular to v, so it cannot change the magnitude of velocity, only its direction. So the magnetic field can never do work on the particle and this is true even if the magnetic field is not uniform. Hence the kinetic energy of the particle is constant. MOTION OF CHARGED PARTICLE IN A MAGNETIC FIELD Case I Motion in Straight Line A charged particle projected into a magnetic field with a velocity parallel or antiparallel to the field, experiences a zero force. Hence the velocity of this particle in this case is constant and hence it travels in a straight line. Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,

4 Case II Motion on a Circular Path PMEC 4 A charged particle of charge q, projected into a magnetic field B with initial velocity v perpendicular to the field, experiences a force F = qvb. Under the action of such force the particle moves in a circular path with constant speed. The plane of the circular path is perpendicular to the lines of force of the magnetic field. The magnetic force provides the centripetal force. According to Newton s Second Law qvb mv r mv r qb where r is the radius of the circular path. T gives the time period of revolution which is given by r m v qb Note that time period is independent of velocity. Case III Helical Motion. Consider a positive charged q has velocity components both perpendicular ( v ) and parallel (v ) to a magnetic field B then it moves in a path, shown in figure, called helical path. The (v ) and ( v ) component is responsible for linear motion and circular motion respectively of the particle and hence the resulting motion is helical. Radius of helix, mv r qb. Time period of revolution, T m qb. p v m qb T v Pitch of the helix (the displacement parallel to the field in one revolution) is Practice Problems : 1. A charged particle enters a magnetic field such that the direction of initial velocity is different from the direction of the field. Which of the following characteristics of the particle doesn t change with time? momentum kinetic energy acceleration direction of motion. Choose the correct statement for the magnetic force acting on a charged particle The maximum mgnetic force will be when the velocity of the charged particle is perpendicular to the direction of magnetic field. Power acted by the magnetic force is always zero Work done by the magnetic force is always zero All are correct 3. Ions of different momenta (p), having the same charge, enter normally a uniform magnetic field. The radius of the orbit of an ion is proportional to p 1/p p 1/p 4. A proton and an -particle, moving with the same kinetic energy, enter a uniform magnetic field normally. The radii of their circular paths will be in the ratio 1 : 1 : 1 1 : 4 : 1 5. Two particles X and Y having same charge, after being accelerated through the same potential difference enter a region of uniform magnetic field and describe circular paths of radii R 1 and R respectively. The ratio of the mass of X to that of Y is (R 1 / R ) 1/ R /R 1 (R 1 / R ) R 1 /R Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,

5 PMEC 5 6. An electron is injected into a uniform magnetic field with components of velocity parallel to and normal to the field direction. The path of the electron is a helix circle parabola straight line 7. A proton and an -particle enter a uniform magnetic field perpendicularly, with the same speed. If the proton takes 5 microseconds to make 5 revolutions, the periodic time for the -particle would be 5 s 5 s 1 s 5 s 8. If a particle of charge 1 1 C moving along the x-direction with a velocity 1 5 m/s experiences a force of 1 1 N in y-direction, then the minimum magnetic field is T in the positive z-direction 1 15 T in the negative z-direction 1 3 T in the positive z-direction 1 3 T in the negative z-direction [Answers : (1) b () d (3) a (4) a (5) c (6) a (7) c (8) d] C5 LORENTZ FORCE When a charged particle moves through a region of space where both electric and magnetic field are present, both fields exerts forces on the particle. The total force F is the vector sum of the electric and magnetic force on a charged particle of charge q is given by F q(e v B), this force is known as Lorentz force. Practice Problems : 1. A uniform electric field and a uniform magnetic field exist in a region in the same direction. An electron is projected with velocity pointed in the same direction. The electron will follow a straight path circular path Helix path Cycloidal path. A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E and B represent the electric and magnetic fields respectively, this region of space may not have E =, B = E =, B E, B = E, B 3. A charged particle enters a region where a uniform electric field E and a uniform magnetic field B exist. If E and B are perpendicular to each other and also perpendicular to the velocity u of the particle, then the particle will move undeviated if u equals E/B E/B E/B E/3B [Answers : (1) a () c (3) b] C6 C7 Velocity Selector When a charged particle is passed through undeflected in the presence of crossed electric field and magnetic field (perpendicular to each other) then the velocity of charged particle is given by the ratio of electric field to magnetic field i.e., v = E/B. The velocity of charged particle is perpendicular to both electric field and magnetic field. This condition can be used to select charged particles of a particular velocity out of a beam containing charges moving with different speeds (irrespective of their charge and mass). The crossed E and B fields, therefore, serve as a velocity selector. Only particles with speed E/B pass undeflected through the region of crossed fields. This method was employed by J.J. Thomson in 1897 to measure the charge of mass ratio (e/m) of an electron. The principle is also employed in Mass Spectrometer-a device that separates charged particles, usually ions, according to their charge to mass ratio. Cyclotron The cyclotron is use to accelerate the charged particles or ions to very high energy. The key to the operation of the cyclotron is that the frequency f at which the proton circulates in the field (and that does not depend on its speed) must be equal to the fixed frequency f osc of the electrical oscillator. Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,

6 PMEC 6 This resonance condition says that, if the energy of the circulating proton is to increase, energy must be fed to it at a frequency f osc that is equal to the natural frequency f at which the proton circulates in the magnetic field. Practice Problems : 1. A cyclotron s oscillator frequency is 1 MHz. The operating magnetic field for accelerating protons equals to (e = C, m p = kg, 1 MeV = J)..33 T.66 T.99 T 1.5 T. If the radius of its dees is 6 cm in the above cyclotrone, the kinetic energy of the proton beam produced by the accelerator equals to.5 MeV 6. MeV 7.4 MeV 8.9 MeV [Answers : (1) b () c] C8 FORCE ON A CURRENT CARRYING CONDUCTOR The force experienced by current element of length dl carrying current I placed in a magnetic field B is given by df Idl B. The total force on a current carrying conductor is given by F I(dl B). If the field B is uniform and current I is steady, then F I ( dl ) B Il B where l is the vector whose magnitude is l and the direction is same as that of current. Hence the force on a wire of arbitrary shape is same as that on a straight wire joining the end points of the wire of arbitrary shape. From the above statement we can conclude that the force acting on a closed loop carrying steady state current placed in a uniform field is zero. FORCE BETWEEN TWO PARALLEL CURRENT CARRYING WIRES Consider two long wires kept parallel to each other, distance d apart, and carrying currents I 1 and I respectively, then (i) (ii) if they carry currents in the same direction then there will be force of attraction between them. if they carry current in the opposite direction then there will force of repulsion between them. The magnitude of force experienced by each wire on a unit length is given as force per unit length = I1I 4 d Practice Problems :. 1. A horizontal wire of length 1 cm and mass.3 g carries a current of 5A. The magnitude of the magnetic field which can support the weight of the wire is (g = 1 m/s ) T T T T. A vertical wire carrying a current in the upward direction is placed in a horizontal magnetic field directed towards north. The wire will experience a force towards North South East West 3. A wire of length l carries a current I along the positive x-axis, placed along the x-axis. A magnetic B B î ĵ kˆ. The force acting on the wire is field exists which is given by IB l (kˆ ĵ) IB l (kˆ ĵ) IB l (ĵ kˆ ) IB l(kˆ ĵ) [Answers : (1) b () d (3) a] Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,

7 PMEC 7 C9 MAGNETIC DIPOLE Consider a small current carrying loop of N turns and area A carrying a current I has similar pattern of lines of force as bar magnet, as shown in figure. That s why a current loop has magnetic dipole moment which is defined as NIA. The direction of coincides with the direction of the area vector A. The direction A is perpendicular to the plane of the loop. If we curl the fingers of the right hand along the current, the direction of thumb gives the direction of dipole moment. Practice Problems : 1. The magnetic moment associated with a coil of 1 turn, area of 1 4 m, carrying a current of A is Am 1 4 Am Am Am. An electron in an atom revolves around the nucleus in an orbit of radius.53 Å. The equivalent magnetic moment if the frequency of revolution of electron is MHz is 1 1 Am 1 Am 1 3 Am 1 4 Am [Answers : (1) b () c] C1 MAGNETIC DIPOLE IN A UNIFORM MAGNETIC FIELD A current loop of magnetic dipole in a uniform magnetic field experiences a zero force but non zero torque and the torque is given by B. Potential energy of the dipole When a dipole is placed in a uniform magnetic field, it posses potential energy given by U.B For = U = B = / U = = U = +B Hence for = i.e. when a dipole is placed along the direction of field, the dipole is in stable equilibrium whereas for = the dipole is in unstable equilibrium. Practice Problems : 1. A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic field B such that B is perpendicular to the plane of the loop. The magnetic force, torque and potential energy of the loop is respectively,, ir B,, Bir,, ir B Bir,, ir B. A circular coil of 1 turns has an effective radius.5 m and carries a current of.1 ampere. The plane of the coil is initially perpendicular to the magnetic field. The work required is to turn it in an external magnetic field of 1.5 Wb/m through 18 about an axis perpendicular to the magnetic field is.436 J.336 J.36 J.136 J Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,

8 PMEC 8 3. A circular coil of 16 turns and radius 1 cm carrying a current of.75 A rests with its plane normal to an external field of magnitude 5. 1 T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of. s 1. The moment of inertia of the coil about its axis of rotation is kg m kg m kg m kg m [Answers : (1) a () c (3) b] C1 The moving coil galvanometer The moving coil galvanometer can be used as a detector to check if a current is flowing in the circuit. Galvanometer can be converted into ammeter and volt meter. Practice Problem : 1. Which of the following will have higher resistance? ammeter milliammeter microammeter all have same resistance. The deflection produced in the pointer attached to the spring of galvanometer depends on number of terms of coil area of coil magnetic field of soft iron core all the above 3. The most convenient way to increase the current sensitivity of galvanometer is to increase the magnetic field of soft iron core area of coil number of terms of coil any of the above 4. The voltage sensitivity of galvanometer depends on torsional constant of spring number of terms and area of coil resistance of galvanometer all of the above [Answers : (1) a () d (3) a (4) d] C11 AMPERE S LAW Ampere s law states that the line integral of B around any closed path equals times the net current (I) through the area enclosed by the path i.e. B.dl I. Using Ampere s law, one finds that the field inside a torodal coil and solenoid are NI B (toroid) r N B I ni (solenoid) l where N is the total number of turns, n is the number of turns per unit length and r is the radius of the toroidal coil. Practice Problems : 1. A hollow cylinder of radius r carries a current I. Let the magnetic field inside the cylinder is B 1 and outside the cylinder is B. Then B 1 =, B B 1, B B 1 =, B B 1, B = Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,

9 PMEC 9. A solid cylinder of radius r carries a current I. Let the magnetic field from the axis of the cylinder is B. Which of the following graph will represent the variation of B with the distance (x) from the axis of the cylinder? [Answers : (1) a () c] C1 Magnetism : There are two types of magnetic poles known as North and South pole. However, every effort to isolate the poles of a magnet have failed. Thus a magnetic monopole does not exist. Pole Stength : In the study of bar magnets it is sometimes useful to introduce a quantity called magnetic pole strength (q m ) which is analogous to charge in electrostatics. In terms of q m the magnetic moment M of bar magnet of length l can be written as m = l q m. The unit of magnetic moment is Am or J/T. Practice Problems : 1. A bar magnet of magnetic moment M is cut into two parts of equal length. The magnetic moment of either part is M M M/ zero [Answers : (1) c] C13 Properties of electric and magnetic dipoles Property Electric Dipole Magnetic Dipole 1 p m Field at distant point along axis E B x Field at distant point along perpendicular bisector (broad-side-on position) E Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : , x p m B x 4 x Force in an external uniform field zero zero Torque in an external field p E m B potential energy in an external field p. E m.b Work done in rotating the dipole in an p E (1 cos ) mb(1 cos ) external field from the equilibrium position Practice Problems : 1. A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 6. The torque required to maintain the needle in this position is 3 3 W W W W. The magnetic field at a point A on the axis of a small bar magnet is equal to the field at a point B on the equator of the same magnet. The ratio of the distances of A and B from the centre of the magnet is 3 1/3 3 1/3 [Answers : (1) a () d]

10 PMEC 1 C14 Time period of small oscillations of a magnet in a magnetic field : Vibration Magnetometer Suppose a magnet of dipole moment m is suspended in a uniform magnetic field B. If it is given a slight rotation from its position of equilibrium, the restoring torque will be = mb sin = mb Therefore the magnet will oscillate with a time period : T the magnet. Practice Problems : I mb, where I is the moment of inertia of 1. The time period of oscillation of a freely suspended magnet is 4 s. It it is broken in length into two equal parts and one part is suspended in the same way, its time period will be 4 s s.5 s.5 s. The time period of oscillation of a bar magnet suspended horizontally along the magnetic meridian is T. If this magnet is replaced by another magnet of the same size and pole strength, but with double the mass, the new time period will be T / T / T T [Answers : (1) b () c] C15 Gauss s Law of Magnetism The law states that B.dS for all closed surfaces. This is a precise expression of the fact that magnetic C16 monopoles do not exist. The Magnetic Field of the Earth Various observations indicate that there is a magnetic field associated with the earth. The field is approximately like that of a fictious huge bar magnet located deep inside the earth with its north pole nearly towards the geographic south and the south pole nearly towards the geographic north. The actual source of the field appears to be some molten charged metallic fluid giving rise to a current flowing inside the core of the earth. Geographic meridian : It is the vertical plane passing through the axis of rotation of the earth. Magnetic meridian : It is the vertical plane passing through the axis of a freely suspended magnet. Declination : It is the angle between the geographic meridian and the magnetic meridian at a place. Declination varies irregularly from place to place. Dip () : It is the angle between the earth s magnetic field and the horizontal direction at a place. It is at the magnetic equator and 9 at the poles, varying gradually as one goes from equator to poles. We have BV tan, where B B H and B V are the horizontal and vertical components of earth s field B. H Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,

11 Practice Problems : PMEC At a certain place the horizontal component of earth s magnetic field is 3 times the vertical component. The angle of dip at that place is [Answers : (1) d] C17 Magnetic Properties of Materials : On the basis of magnetic behaviour, all the materials can be classified into three categories : 1. diamagnetic,. Paramagnetic, 3. Ferromagnetic. 1. Diamagnetic substances are those which are feebly repelled by a magnet. When placed in a non-uniform magnetic field, they tend to move from stronger to weaker parts of the field. Examples : Bi, Cu, Zn, Hg, Au, Pb, NaCl, H O etc. In fact, most of the materials are diamagnetic.. Paramagnetic substances are those which are feebly attracted by a magnet. When placed in a non-uniform field they tend to move from weaker to stronger parts of the field. Examples : Al, Na liquid O, CuCl, wood etc. 3. Ferromagnetic substances are those which are strongly attracted by a magnet. C18 Examples : Fe, Ni, Co etc. Curie Law The variation of magnetic moment per unit volume, M, also called magnetisation, as a function of B/T, where T is the temperature and B is the magnetic field for low value of B/T is given by, M B/T or CB M, where C is a constant. This is called the Curie law. T Curie Temperature : It is that temperature above which the ferromagnetic material becomes paramagnetic. For iron the Curie temperature is 143 K. C19 Relative Permeability ( r ) and Magnetic Susceptibility ( ) C Relation between Relative Permeability ( r ) and Magnetic Susceptibility is given by r = 1 + Material r Diamagnetic slightly less than unity small, negative Paramagnetic slightly more than unity small, positive Ferromagnetic much greater than unity (~ 1 3 ) large, positive. Hysteresis : When a piece of ferromagnetic material is taken through a cycle of magnetisation, the variation of magnetic induction B with the magnetic intensity H is as shown in the diagram. The value of B at which H (magnetic intensity) is zero is called remanence (B r ). The value of H at which B (magnetic field) is zero is called coercivity. The complete cycle of magnetisation is known as hysteresis loop. The following important points concerning hysteresis loop should be noted : 1. The energy spent per cycle in taking a ferromagnetic material through a cycle of magnetisation is proportional to the area of the hysteresis loop. This area is very small for soft iron and is large for steel. Therefore soft iron is useful for cores of transformers and generators. Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,

12 PMEC 1. For steel, coercivity is very large and remanence is fairly large.therefore, steel is used for making parmanent magnets. The area of the loop is large for steel, but it does not matter because a permanent magnet has not to be taken through a cycle of magnetisation. For soft iron coercivity is small and area of the loop is also small. Therefore it is a suitable materials for making electromagnets. Practice Problems : 1. Electromagnets are made of soft iron because soft iron has low susceptibility and low retentivity high susceptibility and high retentivity high susceptibility and low retentivity low permeability and high retentivity. The area of the B-H hysteresis loop is an indication of the permeability of the material susceptibility of the material retentivity of the material energy dissipated per cycle 3. The material of a permanent magnet has high retentivity, low coercivity low retentivity, high coercivity low retentivity, low coercivity high retentivity, high coercivity [Answers : (1) c () d (3) d] Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,

13 PMEC 13 SINGLE CORRECT CHOICE TYPE 1. A potential difference of 6 V is applied across the plates of a parallel plate capacitor. The separation between the plates is 3 mm. A magnetic field also exists between the plates. An electron projected parallel to the plates as shown with a speed of 1 6 m/s moves undeflected between the plates. The magnitude and direction of the magnetic field is 3. A long horizontally fixed wire carries a current of 1 A. Directly above and parallel to it is another wire carrying a current of A and weighing.4 N/m. What should be the separation between the two wires so that the upper wire is just supported by magnetic repulsion? 1 cm cm 3 cm 4 cm 4. A charge q coulomb is circulating in an orbit of radius r metres making n revolutions per second. The magnetic field produced at the centre of the circle in N/Am is q 7 nr 1 q 7 r 1. T, into the page. T, out of the page.1 T, into the page.1 T, out of the page. A particle of charge +q and mass m moving under the influence of a uniform electric field E î and a uniform magnetic field B kˆ follows a trajectory from P to Q as shown in the figure. The velocities at P and Q are vî and vĵ. Which of the following statements is incorrect? nq 7 r 1 rn In the figure AB is a long straight wire carrying a current of A and CDFG is a rectangular loop of size cm 9 cm carrying a current of 1 A. The edge CG is parallel to AB, at a distance of 1 cm from it. The force exerted on the loop by the magnetic field of the wire is q 3 mv E 4 qa N towards left N towards right towards right towards left Rate of work done by the electric field at P is 3 mv 4 a 3 Rate of work done by the electric field at P is zero. Rate of work done by both the fields at Q is zero. Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,

14 PMEC In the figure x, y and z are long straight wires. The magnetic force on 5 cm length of y is me qd me qd me qd me qd 9. A neutral particle is at rest in a uniform magnetic field B. At time t = it decays into two charged particles, each of mass m. The two particles move off in separate paths, both of which lie in the plane 1 4 N towards right N towards right 1 4 N towards left towards left 7. A circular loop of mass m and radius r lies on a horizontal table (xy-plane). A uniform magnetic field is applied parallel to the x-axis. The current I that should flow in the loop so that it just tilts about one point on the table is mg rb, clockwise mg, clockwise rb mg rb, anticlockwise rb, anticlockwise mg 8. A charged particle having kinetic energy E enters normally a region of uniform magnetic field between two plates P 1 and P as shown in the figure. If the particle just misses hitting the plate P, then the magnetic field B in the region between the plates is perpendicular to B. After some time t the particle collide. The value of t is m/qb m/qb m/qb m/qb 1. A wire of 6. cm length and 13. g mass is suspended by a pair of flexible leads in a uniform magnetic field of magnitude.44 T. The magnitude of the current required to remove the tension in the supporting leads 467 ma 367 ma 67 ma 167 ma 11. A 1. kg copper rod rests on two horizontal rails 1. m apart and carries a current of 5 A from one rail to the other. The coefficient of static friction between rod and rails is.6. The smallest magnetic field (not necessarily vertical) that would cause the rod to slide is.5 T.1 T. T.3 T 1. A particle of charge q moves in a circle of radius a with speed v. Treating the circular path as a current loop with constant current equal to its average current, the maximum torque exerted on the loop by a uniform magnetic field of magnetic B is qvab qvab qvab 3 qvab A current loop, carrying a current of 5. A, is in the shape of a right triangle with sides 3, 4 and 5 cm. The loop is in a uniform magnetic field of magnitude 8 mt whose direction is parallel to the current in the 5 cm side of the loop. The torque on the loop is.4 N-m.8 N-m.14 N-m.4 N-m Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,

15 PMEC Figure shows a current loop ABCDEFA carrying a current i = 5. A. The sides of the loop are parallel to the coordinate axes, with AB =. cm, BC = 3. cm, and FA = 1. cm. The magnitude and direction of the magnetic dipole moment of the loop is qbd tan 1 qbd cos 1 mv mv qbd sec 1 qbd sin 1 mv mv 19. A particle having mass m and charge e is released from the origin in a region in which electric field and magnetic field are given by B B ĵ and E E kˆ. The speed of the particle as a function of its z-coordinate is given by ( 15ĵ 3kˆ )A m ( 15ĵ 3kˆ )A m ( 15ĵ 3kˆ )A m ( 15ĵ 3kˆ )A m 15. A wire of given length is folded into a circular coils of constant radius. A constant current is flowing through the loop. The magnetic field produced at the centre of the coil is directly proportional to the number of turns as n. The value of is In the above problem the magnetic dipole moment of the loop depends on number of turns as n. The value of is Two infinitely long straight wires lie in the x y plane along the x and y axes respectively. Each wire carries a current I, respectively in the positive x direction and the positive y direction. The locus of a point in this plane where the magnetic field is zero Straight Line Circle Parabola Ellipse 18. A particle of mass m and charge q is projected into a region having a perpendicular uniform magnetic field B of width d. The angle of deviation of the particle as it comes out of the magnetic field is ee z m 6eE z m 4eE z m 8eE z m. A rotating rod is pivoted about one end and carries a total charge q uniformly distributed along its length l. If the rod rotates with angular velocity, then its magnetic moment is 1 q l 3 1 ql 1 q l ql 6 1. Two perpendicular straight wires join the ends of a semicircular loop of radius a, as shown in the figure. If the current is I, then the resultant field at the centre of the circular section is I a I a I a I a. A wire PQ, carrying a current I from P to Q, is kept x y in the x-y plane along the curve 1 in a b the first quadrant with P at x-axis and Q at y-axis. A uniform magnetic field B exists in the negative z-direction. The x component of the force on the wire is Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,

16 PMEC 16 IbB IaB I a b B I a b B 3. Two very long wires are placed parallel to each other. One of the wire is carrying a current I, which is constant with time. The separation between the wires is d and the current carrying wire is fixed whereas other wire is free to move which has mass per unit length. Suddenly a charge Q is passed through the movable wire. The instant speed of the movable wire when the charge is passing through this I 4 I 8 Q d Q d I 4 Q 16 d I Q 4. A solenoid of length.4 m and having 5 turns of wire carries a current of 3 ampere. A thin coil having 1 turns of wire and of radius.1 m carries a current of.4 ampere. The torque required to hold the coil in the middle of the solenoid with its axis perpendicular to the axis of the solenoid N-m N-m N-m 5. A loop of flexible conducting wire of length.5 m lies in a magnetic field of 1. T going inwards normal to the plane of the loop. The tension developed in the wire if the current is 1.57 amp in clockwise direction.375 N.75 N.15 N.175 N 6. A current I flows along a thin wire shaped as a regular polygon with n sides which can be inscribed into a circle of radius R. The magnetic induction at the centre of the polygon 3n I tan (/n)/r n I tan (/n)/r 5n I tan (/n)/r 7n I tan (/n)/r 7. An electron is projected horizontally with a kinetic energy of 1 kev. A magnetic field of strength T exists in the vertically upward direction. The approximate sideways deflection of the electron in travelling through 1 m.5 cm 1. cm 1.5 cm cm d 8. A hypothetical magnetic field existing in a region is given by B B e, where e r r denotes the unit vector along the radial direction. A circular loop of radius a, carrying a current i, is placed with its plane parallel to the X-Y plane and the centre at (,, d). The magnitude of the magnetic force acting on the loop is a ib a d 4a ib a d 6a ib a d a ib a d 9. Consider a solid sphere of radius r and mass m which has a charge q distributed uniformly over its volume. The sphere is rotated about a diameter with an angular speed. The magnetic moment of the sphere is 1 q 5 r 1 q r 4 1 q 3 r 1 q r 3. A narrow beam of singly-charged ions, moving at a constant velocity of m/s, is sent perpendicularly in a rectangular region having uniform magnetic field B =.5 T. It is found that two beams emerge from the field in the backward direction, the separations from the incident beam being 3. cm and 3.5 cm. Take the mass of an ion = A( )kg, where A is the mass number. The isotopes present in the ion beam are 1 C and 14 C 1 H and H 1 C and H none Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,

17 PMEC 17 EXCERCISE BASED ON NEW PATTERN COMPREHENSION TYPE Comprehension-1 Two long staright parallel wires are m apart, perpendicular to the plane of the paper. The wire A carries a current of 9.6 ampere, directed into the plane of the paper. The wire B carries a current such that the magnetic field of induction at the point P at a distance (1/11) m from the wire B is zero. 1. The magnitude and direction of the current in B 1 ampere, in the page ampere, in the page 3 ampere, out of the page ampere, out of the page. The magnitude of the magnetic field of induction at the point S 1.3 T.6 T 3.9 T 5. T 3. The force per unit length on the wire B 1.44 N/m.88 N/m 3.6 N/m 4.8 N/m Comprehension- A top view of the region of a cyclotron in which the particles (protons, say) circulate. The two hollow D-shaped objects (open on their straight edges) are made of sheet copper. The radius of the dee is.5 m These dees, as they are called, are part of an electrical oscillator that alternates the electric potential difference across the gap between the dees. The electrical signs of the dees are alternated so that the electric field in the gap alternates in direction, first toward one dee and then the other dee, back and forth. The dees are immersed in a magnetic field (B = 3. T) whose direction is out of the plane of the page and that is set up by a large electromagnet. Suppose that a proton, injected by source S at the center of the cyclotron in figure, initially moves towards a negatively charged etc. It will accelerate towards this dee and enter it. Once inside, it is shielded from electric fields by the copper walls of the dee; that is, the electric field does not enter the dee. The magnetic field, however, is not screened by the (nonmagnetic) copper dee, so the proton moves in a circular path whose radius, which depends on its speed. Let us assume that the instant the proton emerges into the center gap from the first dee, the potential difference between the dees is reversed. Thus, the proton again faces a negatively charged dee and is again accelerated. This process continues, until the proton has spiraled out to the edge of the dee system. There a deflector plate sends it out through a portal. The key to the operation of the cyclotron is that the frequency f at which the proton circulates in the field (and that does not depend on its speed) must be equal to the fixed frequency f osc of the electrical oscillator. This resonance condition says that, if the energy of the circulating proton is to increase, energy must be fed to it at a frequency f osc that is equal to the natural frequency f at which the proton circulates in the magnetic field. 4. The maximum velocity of the proton m/s m/s m/s m/s 5. The kinetic energy of the ejected proton is 75 MeV 15 MeV 18 MeV 15 MeV 6. The period for a half cycle of the oscillator is s s s s 7. Cyclotrone is not suitable for the acceleration of proton He ++ ion deutron electron Comprehension-3 Two long parallel wires carrying currents.5 ampere and I ampere in the same direction (directed into the plane of the paper) are held at P and Q respectively such that they are perpendicular to the plane of the paper. The points P and Q are located at a distance of 5m and m respectively from a collinear point R Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,

18 PMEC An electron moving with velocity of m/s along the positive x-direction experiences a force of magnitude 3. 1 N at the point R. The value of I is 1 A A 3 A 4 A 9. The null point along the x-axis is 15 at m at m at m at m 15 from P in the right from P in the left from P in the right from P in the left 1. The positions at which a third long parallel wire carrying a current of magnitude.5 ampere may be placed so that the magnetic induction at R is zero. 1 m left and right from R.5 m left and right from R.5 m left and right from R none Comprehension-4 This Hall effect allows us to find out whether the charge carriers in a conductor are positively or negatively charged. Beyond that, we can measure the number of such carriers per unit volume of the conductor. Consider a copper strip of width d and thickness l, carrying a current i. The charge carriers are electrons and, as we know, they drift (with drift speed v d ) in the opposite direction of the direction of current. An external magnetic field B, pointing into the plane of the strip, has just been turned on. 11. Choose the correct statement : Due to the magnetic field, the electron in the conductor will experience a force. An electric field will be developed in the strip. After some time an equilibrium will acheive in which the electric force on the electron is balanced by magnetic force. All the above statemens are correct. 1. The potential difference (known as Hall potential difference) across the strip is Bdv d Bdv d ½ Bdv d can t be calculated 13. The number of electrons per unit volume in the conductor is 3i dv i dv d l d l e e Comprehension-5 i dv d l 4i dv Moving coil galvanometer is an instrument to detect and measure the current. It is based on the principle : the torque exerted by a magnetic field on a current-carrying coil. The reading is displayed by means of the deflection of a pointer over a scale. It consists of a coil having large number (N) of turns of insulated copper wire so that it can rotate about an axis (into the page) in a uniform radial magnetic field (B). The coil is wound over a non-magnetic metallic frame (usually brass) which may be rectangular or circular in shape. For any orientation of the coil, the net magnetic field through the coil is perpendicular to the normal vector of the coil (and thus parallel to the plane of the coil). A spring Sp provides a countertorque that balances the magnetic torque, so that a given steady current i in the coil results in a steady angular deflection ( = k) where k is known as torsional constant of the spring. 14. To make the field radial in a moving coil galvanometer d l the number of turns in the form of increased magnet is taken in the form of horse-shoe poles are cylindrically cut coil is wound on aluminium frame e e Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,

19 PMEC In a moving coil galvanometer the current i is related to the deflection as i = n, where n is a constant. The value of n is 1 1 ½ ½ 16. If the coil is.1 cm high and 1. cm wide, has 5 turns, and is mounted so that it can rotate about an axis (into the page) in a uniform radial magnetic field with B =.3 T. If a current of 1 A produces an angular deflection of 8, the torsional constant k of the spring is N m/degree N m/degree N m/degree N m/degree 17. The current sensitivity of the moving coil galvanometer is defined as the ratio of angular deflection to the current. The current sensitivity of a moving coil galvanometer does not depend on Comprehension-6 Rail Gun the number of turns in the coil moment of inertia of the coil strength of the magnetic field torsional constant of the the spring A rail gun is a device in which a magnetic force can accelerate a projectile to a high speed in a short time. Figure shows an idealized schematic drawing of a rain gun. Projectile P sits between two wide rails of circular cross section; a source of current sends current through the rail and through the (conducting) projectile itself. Let w be the distance between the rails, R the radius of the rails and i the current. 18. The direction of force on the projectile is upward downward right left 19. The force on the projectile is given approximately. Assume that i = 45 ka, w = 1 mm, R = 6.7 cm, L = 4. m, and mass of the projectile is m = 1 g. If the projectile is initially at rest at the left end then the speed attained by the projectile is 5 km/s 1 km/s 15 km/s none MATRIX-MATCH TYPE Matching-1 Column - A Column - B (A) A charge particle enters (P) straight line at in the uniform magnetic field (B) A charge particle enters (Q) parabolic at 9 in the uniform electric field (C) A charge particle enters (R) circular at 9 in the uniform magnetic field (D) A charge particle released (S) helical from rest in a uniform electric & magnetic field parallel to each other Matching- Column - A (T) Column - B cycloid (A) Moving coil (P) wheatstone galvanometer bridge (B) Cyclotron (Q) magnetic effect of current (C) Post office box (R) particle accelerator (D) Mass spectrometer (S) particle detector Matching-3 (T) none of these Match the following column for the magnetic field produced by different configuration at O : Column - A Column - B I (A) (P) 4 a b c i w R n l R i w R n l R i w R l n R 4i w R ln R I (B) (Q) a b c Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,

20 PMEC Column - A Column - B I (C) (R) 4 a b c I (D) (S) 4 a b c Matching-4 Match the initial direction of the deflection of charged particles as they enter the magnetic field as shown in figures Column - A Column - B (A) (P) no deflection (B) (Q) up (C) ` (R) into the plane of the paper (A) They are projected (P) 1 : 1 : 1 : : 1837 with the same speed (B) They are projected (Q) 1 :1 : : 4 : 1837 with the same momentum (C) They are projected (R) :1 : : : with the same kinetic energy (D) They are accelerated (S) 1 1 :1 : :1: 1 with the same potential difference and then projected MULTIPLE CORRECT CHOICE TYPE 1. The path of a charged particle moving in a magnetic field can be a Straight line Circle Parabola Helix. In case of motion of a charged particle in a steady magnetic field speed remains constant momentum remains constant kinetic energy remains constant work done is always zero 3. When a current loop is placed in a non-uniform magnetic field and F R but F R but F R (D) (S) towards you Matching-5 out of the plane of the paper Electron, proton, -particle, He + ion and deutron are projected in the same magnetic field normally. Let the radius of their circular path in the ratio R e : R p : R : R : R He d and F R 4. H +, He + and O + all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. The masses of H +, He + and O + are 1 amu, 4 amu and 16 amu respectively, then : H + will be deflected most O + will be deflected most He + and O + will be deflected equally All will be deflected equally Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,

21 5. Let E and B denote the electric and magnetic fields in a certain region of space. A proton moving with a velocity v along a straight line enters the region and is found to pass through is undeflected. Indicate which of the following statements are consistent with the observations : E and B E and B E and B and both E and B are parallel to v E is parallel to v but B is perpendicular to v 6. A proton moving with constant velocity passes through a region of space without any change in its velocity. If E and B represent electric and magnetic fields respectively, this region of space may have E = and B = E = and B E and B = E and B 7. A long straight wire of radius r carries a current distributed uniformly over its cross-section. The magnitude of the magnetic field is maximum at the axis of the wire minimum at the axis of the wire maximum at the surface of the wire minimum at the surface of the wire 8. A charged particle of unit mass and unit charge moves with velocity of v (8î 6ĵ) m/s in a magnetic field of B kˆ T. Choose the correct alternatives. The path of the particle may be x + y 4x 1 = The path of the particle may be x + y = 5. The path of the particle may be y + z = 5 The time period of the particle will be 3.14 s 9. A current-carrying ring is placed in a magnetic field. The direction of the field is perpendicular to the plane of the ring. There is no net force on the ring The ring may tend to expand The ring may tend to contract None of these PMEC 1 1. Suppose a uniform electric field and a uniform magnetic field exist along mutually perpendicular directions in a gravity free space. If a charged particle is released from rest, at a point in the space particle can not remain in static equilibrium. first particle will move along a curved path but after some time its velocity will become constant. particle will come to rest at regular interval of time. acceleration of the particle will never become equal to zero. 11. A beam of proton with a velocity m/s enters a uniform magnetic field of.3 tesla at an angle of 6 to the magnetic field. Let the radius of the helical path taken by the proton beam is r and the pitch of the helix is p then r = 1. cm r = 4.3 cm p = 1. cm p = 4.3 cm Assertion-Reason Type Each question contains STATEMENT-1 (Assertion) and STATEMENT- (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. (A) Statement-1 is True, Statement- is True; Statement- is a correct explanation for Statement-1 (B) Statement-1 is True, Statement- is True; Statement- is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement- is False (D) Statement-1 is False, Statement- is True 1. STATEMENT-1 : A hanging spring is attached to a powerful battery and a switch. When the switch is closed so that current suddenly exists in the spring, the spring compresses STATEMENT- : Each coil of the spring becomes a magnet because a coil acts as a current loop. The sense of rotation of the current is the same in all coils, each coil becomes a magnet with the same orientation of poles.. STATEMENT-1 : An observer move along a line parallel to a metal wire at a speed equals to the drift speed of the electron in the current. He measures a zero magnetic field. STATEMENT- : With respect to the observer there is a magnetic field due to the moving positive crystal lattice ions. 3. STATEMENT-1 : In a lightning stroke, negative charge rapidly moves from the cloud to the ground. The lightning stroke deflected by the earth magnetic field to the west. Einstein Classes, Unit No. 1, 13, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi 11 18, Ph. : ,