2. TRABAJO DE CAMPO
2.1 DESCRIPCIÓN GENERAL DEL PROCESO
1. Answer the following questions ;
(a) What happens if a bar magnet is cut into two pieces (i) transverse to its length (ii) along its length?
(b) What happens if an iron bar magnet is melted? Does it retain its magnetism?
(c) A magnetized needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition a toroid. Why?
(d) Must every magnetic field configuration have a north pole and a south pole? What about the field due to a toroid?
(e) Can you think of a magnetic field configuration with three poles?
(f) Two identical looking iron bars A and B are given , one of which is definitely known to be magnetized.(
We do not know which one.) How would one ascertain whether or not both ate magnetized? If only one is magnetized, how does one ascertain which one? [Use nothing else but the two bars A and B]
2. Answer the following questions regarding earth’s magnetism :
(a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.
(b) The angle of dip at a location in southern India is about 18º. W ould you expect a greater of smaller dip angle in Britain?
(c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground?
3. A short bar magnet placed with its axis at 30º with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 x 10–2 J. What is the magnitude of magnetic moment of the magnet?
4. A short bar magnet of magnetic moment m = 0.32 JT –1 is placed in a uniform external magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientations would correspond to its (i) stable and (ii) unstable equilibrium? What is the potential energy of the magnet in each case?
5. A closely wound solenoid of 800 turns and area of cross-section 2.5 x 10–4 m2 carries a current of 3.0 A Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment ?
6. A bar magnet of magnetic moment 1.5 JT–1 lies aligned with the direction of a uniform magnetic field of 0.22 T.
(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment, (i) normal to the field direction (ii) opposite to the field direction?
(b) What is the torque on the magnet in cases (i) and (ii)?
7. A closely wound solenoid of 2000 turns and area of cross-section 1.6 x 10–4 m2 , carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.
(a) What is the magnetic moment associated with the solenoid?
(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 x10–2 T is set up at an angle of 30º with the axis of the solenoid ?
8. At a certain location in Africa, a compass points 12º west of the geographic north. the north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60º above the horizontal.
The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.
9. A monoenergetic ( 18 keV ) electron beam initially in the horizontal direction is subject to a horizontal magnetic field of 0.40 G. normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm,( me= 9.11 x 10–31 kg, e = 1.60 x 10–19 C ).
[Note : Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set]
Exercise # 1
PART-I
A-1.* (ABC) A-2. (B) A-3. (A) B-1. (A) B-2. (B) B-3. (C) C-1. (A)
C-2. (B) C-3. (D) C-4. (C) C-5. (A) C-6. (D) C-7. (B) D-1. (B)
D-2. (D) D-3. (C) D-4. (B) D-5. (C) D-6. (B) D-7.* (AC) D 8.* (CD)
E-1. (D) E-2.* (AD) E-3.* (BD) E 4*. (BD) E-5.* (AB) F-1. (C) F-2. (C)
F-3. (B) F-4. (B) F-5. (D) G-1. (B) G-2. (A) H-1. (C) H-2. (A)
-1. (C)
PART-II
1. (C) 2. (A) 3. (A) 4. (A) – p,q, r ; (B) – p, q, r, s ; (C) – r ; (D) – p, q, r, s 5. (A) – p, q ; (B) – p, r ; (C) – p ; (D) – p, q, s 6. (i) – R ; (ii) – Q, V ; (iii) – V ; (iv) – U
Exercise # 2
PART-I
1. (C) 2. (C) 3. (A) 4. (A) 5. (D) 6. (C) 7. (A)
8. (D) 9. (B) 10. (A) 11. (D) 12. (C) 13. (C) 14.* (CD)
15.* (AD) 16.* (BCD) 17.* (BCD) 18.* (ABD) 19.* (ABC)
PART-II
1.. 13
2
10 -4 wb/m2
2. (i) 0, (ii) 10–13 kˆ (iii) –10–13 ˆj (iv) 27
4
× 10–13 (–2ˆj + kˆ) (v) yes, no (vi) yes, no.
3. (i) 0 2
R 4
qv
, inwards (ii)
) R x ( 4
qv
2 2 0
, No
4. 1 × 10-4 wb/m2, towards the reader 5. 4 × 10-5 wb/m2
6. (i) 2 3 × 10–5 tesla (ii) 2 × 10–5 T 7. 0
8. 4 d
0i
, 2 4 d
0i ð ì ,
2 1
9. 2 x
0i
10. a
i 2
2 0
11. 0
12. (i) (a) 4 × 10–4 (b) zero (ii) 2 2x10–4 T 13. (a) (ii) current directed out of from paper, 1m from R on RQ (between R and Q)
27. 30º 28. (–75ˆi + 100 jˆ) m/s 29. 3.0 40. 20 cm, 5 103s1
60. 8 × 10–2 N 61. 2B0i 62. 2 N towards the inside of the circuit
83. (a) 1.0 A , (b) 2.0 V perpendicular to the magnetic meridian
84. 6× 10–4 T, 10 4
n 1 a , where x is the perpendicular distance from the wire i0. It will try to become antiparallel to i0.
92. i = mg
BN
2
= 2.5A
Exercise # 3
PART-I
1. (A) 2. (D) 3. RP : R
= 1 : 2 4. (B)
5. (i) K = NAB (ii)
0
NAB
2 (iii)
C NABQ
6.* (AC) 7. (A)
8. (A) – q, r ; (B) – p ; (C) – q, r ; (D) – q,s 9. (ACD) 10. (A) – p, r, s; (B) – r, s; (C) – p, q, t ; (D) – r, s 11. 7
12. (C) 13.* (BD) 14. (A) 15.* (CD) 16. 5 17. (B)
18. (D) 19. (AC) 20. (AD)
PART-II
1. (1) 2. (4) 3. (2) 4. (2) 5. (1) 6. (3) 7. (2)
8. (3) 9. (1) 10. (3) 11. (1) 12. (4) 13. (3) 14. (3)
15. (3) 16. (2) 17. (3) 18. (3) 19. (1) 20. (1) 21. (2)
22. (1) 23. (2) 24. (1) 25. (1) 26. (1) 27. (1) 28. (2)
Exercise # 4
1. (a) In either case, one gets two magnets, each with a north and south pole,
(b) Molten iron is above the Curie temperature (770 ºC) and is, therefore, not ferromagnetic. An iron bar magnet when melted does not does not retain its magnetism.
(c) No force if the field is uniform. The iron nail experiences a non-uniform magnetic field due to the bar magnet. The induced magnetic moment in the nail, experiences both force and torque. The net force is attractive because the induced (say) south pole in the nail is closer to the north pole of the magnet than the induced north pole.
(d) Not necessarily. True only if the source of the field has a net non-zero magnetic moment. This is not so for a toroid of even for a straight infinite conductor.
(e) Depends on what one means by three poles. Poles must always occur in pairs. But one can think to two bar magnets with (say) their north ends glued together as providing a three-pole field configuration.
(f) Try to bring different ends of the magnets closer. A repulsive force in some situation establishes that both are magnetized. If it is always attractive, then one of them is not magnetized. To see which one, pick up one, say , A and lower one of the middle of B is magnetized. If you do not notice any change from the end to the middle of B, then A is magnetized.
2. (a) Magnetic declination, angle of dip, horizontal component of earth’s magnetic field.
(b) Greater in Britain (it is about 70º) , because Britain is closer to the magnetic north pole.
(c) Field lines of B due to the earth’s magnetism would seen to come out of the ground.
3. 0.36 JT–1
4. (a) m parallel to B ; U = –mB = –4.8 x 10–2 J ; stable
(b) m anti-parallel to B ; U’ = +mB = +4.8 x 10–2 J ; unstable
5. 0.60 JT–1 along the axis of the solenoid ; the direction determined by the sense of flow of current.
6. (a) (i) 0.33 J (ii) 0.66 J
(b) (i) torque of magnitude 0.33 J in a direction that tends to align the magnetic moment vector B. (ii) Zero.
7. (a) 1.28 A m2 along the axis in the direction related to the sense of current via the right-handed screw rule.
(b) force is zero in uniform field; torque= 0.048 Nm in a direction that tends to align the axis of the solenoid ( i.e., its magnetic moment vector) along. B.
8. The earth’s field lines in a vertical 12º west of the geographic meridian making an angle of 60º ( upwards ) with the horizontal (magnetic south to magnetic north) direction. Magnitude = 0.32 G.
9. R =
eB meV
eB
energy kinetic x m 2 e
= 11.3 m
Up or down deflection R (1 –cos ) where sin = 0.3 / 11.3. We get deflection ~ 4 mm.