36.chapter 36
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CHAPTER – 36
PERMANENT MAGNETS 1.
m = 10 A-m,
2.
m 10 7 10 10 2 –4 B= 0 2 = = = 4 × 10 Tesla 2 2 25 4 r 5 10 m1 =m2 = 10 A-m r = 2 cm = 0.02 m we know
d = 5 cm = 0.05 m
N
Force exerted by tow magnetic poles on each other = 3. 4.
S
0 m1m 2 4 10 7 10 2 –2 = = 2.5 × 10 N 4 4 r 2 4 4 10
dv –3 –3 dv = –B dℓ = – 0.2 × 10 × 0.5 = – 0.1 × 10 T-m d Since the sigh is –ve therefore potential decreases. Here dx = 10 sin 30° cm = 5 cm
B=–
dV 0.1 10 4 T m =B= dx 5 10 2 m
5.
Since B is perpendicular to equipotential surface. –4 Here it is at angle 120° with (+ve) x-axis and B = 2 × 10 T –4 B = 2 × 10 T d = 10 cm = 0.1 m (a) if the point at end-on postion.
V
0.1×10–4 T-m
7
10 2M 0 2M –4 2 × 10 = 4 d3 (10 1 )3
B=
2 10
4
10
3
= M M = 1 Am 10 2 (b) If the point is at broad-on position
6.
0.2×10
T-m
30° 0.4×10
–4
X
T-m
In cm
2
7
0 M 10 7 M –4 2 2 × 10 = M = 2 Am 3 4 d (10 1 )3 Given : 2 2 tan = 2 2 = tan tan tan = 2 cot = cot 2 tan We know = tan 2 Comparing we get, tan = cot or, tan = tan(90 – ) or = 90 – or + = 90 Hence magnetic field due to the dipole is r to the magnetic axis. Magnetic field at the broad side on position : M 2ℓ = 8 cm d = 3 cm B= 0 4 d2 2 3 / 2
= tan
7.
30° –4
0.3×10–4 T-m
–1
4 × 10 m=
–6
=
10 7 m 8 10 2
9 10
4
16 10
4 10 6 125 10 8 8 10 9
4 3 / 2
–6
4 × 10
= 62.5 × 10
–5
=
A-m
36.1
10 9 m 8
10
4 3 / 2
25 3 / 2
P S
N
Permanent Magnets 8.
We know for a magnetic dipole with its north pointing the north, the neutral point in the broadside on position. N 0M Again B in this case = 4d3 M 0 3 = BH due to earth P 4d
d3
d
= 18 T
10 7 1.44
–6 S = 18 × 10 d3 3 –3 d = 8 × 10 –1 d = 2 × 10 m = 20 cm In the plane bisecting the dipole. When the magnet is such that its North faces the geographic south of earth. The neutral point lies along the axial line of the magnet.
9.
10 7 1.44
7 0 2M –6 10 7 2 0.72 = 18 × 10–6 d3 = 2 0.7 10 = 18 × 10 4 d3 18 10 - 6 d3
8 10 9 d= 10 6
1/ 3
= 2 × 10
–1
m = 20 cm
S
N
2
N2
10. Magnetic moment = 0.72 2 A-m = M M B= 0 3 BH = 18 T 4 d
4 10 7 0.72 2 4 d 3
d =
tan
W
–6
= 18 × 10
3
0.72 1.414 10 7 6
= 0.005656
18 10 d ≈ 0.2 m = 20 cm 11. The geomagnetic pole is at the end on position of the earth. B=
0 2M 10 7 2 8 10 22 –6 = ≈ 60 × 10 T = 60 T 3 4 d (6400 10 3 )3
BH
BM
N –1
2
BH
S
E
N
S
N1
d
–5 12. B = 3.4 × 10 T M –5 Given 0 3 = 3.4 × 10 4 R M=
3.4 10 5 R 3 4 7
2
3
= 3.4 × 10 R
4 10 2M –5 B at Poles = 0 3 = = 6.8 × 10 T 4 R 13. (dip) = 60° BH = B cos 60° –6 B = 52 × 10 = 52 T
3 = 44.98 T ≈ 45 T 2 14. If 1 and 2 be the apparent dips shown by the dip circle in the 2r positions, the true dip is given by 2 2 2 Cot = Cot 1 + Cot 2 2 2 2 Cot = Cot 45° + Cot 53° 2 Cot = 1.56 = 38.6 ≈ 39° –6
BV = B sin = 52 × 10
36.2
Permanent Magnets 15. We know
BH =
Give : BH = 3.6 × 10 –2 i = 10 mA = 10 A n=?
–5
0in 2r = 45° tan = 1 r = 10 cm = 0.1 m
T
BH tan 2r 3.6 10 5 2 1 10 1 3 = = 0.5732 × 10 ≈ 573 turns 0i 4 10 7 10 2 –4 2 16. n = 50 A = 2 cm × 2 cm = 2 × 2 × 10 m –3 i = 20 × 10 A B = 0.5 T –3 –4 –4 = ni A B = niAB Sin 90° = 50 × 20 × 10 × 4 × 10 × 0.5 = 2 × 10 N-M n=
17. Given = 37° We know
d = 10 cm = 0.1 m
M 4 ( d2 2 ) 2 4 d4 = tan = tan [As the magnet is short] BH 0 2d 0 2d
4
=
7
(0.1)3 –3 7 4 3 2 –1 tan 37 = 0.5 × 0.75 × 1 × 10 × 10 = 0.375 × 10 = 3.75 ×10 A-m T 2
4 10 M 3 2 –1 18. (found in the previous problem) = 3.75 ×10 A-m T BH = 37°, d=? M 4 2 = (d 2 )3 / 2 tan BH 0 ℓ << d neglecting ℓ w.r.t.d M 1 4 3 3 = d Tan 3.75 × 103 = × d × 0.75 7 BH 0 10
3.75 10 3 10 7 –4 = 5 × 10 0.75 d = 0.079 m = 7.9 cm M 2 19. Given = 40 A-m /T BH 3
d =
Since the magnet is short ‘ℓ’ can be neglected So,
M 4 d3 = = 40 BH 0 2
S
40 4 10 7 2 –6 = 8 × 10 4 –2 d = 2 × 10 m = 2 cm with the northpole pointing towards south. 20. According to oscillation magnetometer, 3
d =
T = 2
N
MBH
1.2 10 4 = 2 10 M 30 10 6 2
1.2 10 4 1 = M 30 10 6 20 M=
1.2 10 4 400 30 10
6
2
2
= 16 × 10 A-m = 1600 A-m
36.3
2
Permanent Magnets
1 mB H 2 For like poles tied together M = M 1 – M2 For unlike poles M = M1 + M2
21. We know : =
1 = 2
S
N
N
S
N
S
M M2 M M2 M1 M2 10 = 1 25 = 1 M1 M2 M1 M2 M1 M2 2
2M1 M 26 13 = 1 = 24 2M2 M2 12 –6
T1 = 0.1
T –6
B = BH – Bwire = 2.4 × 10
24 10 6 MBH
23. T = 2
B BH
2
0 .1 14 0.01 14 2 = T2 = T2 = 0.076 24 24 T2 Here = 2
1 min 40
T1 = T2
o i 2 10 7 18 –6 –6 –6 = 24 × 10 – = (24 –10) × 10 = 14 × 10 2 r 0 .2
T1 = T2
14 10 6
0 .1 = T2
T1 =
–
MBH
T = 2
S
2
22. BH = 24 × 10
N
T2 = ?
1 = 40T2
1 1 1 1 2 = T2 = T2 = 0.03536 min 2 2 800 1600 T2 2
For 1 oscillation Time taken = 0.03536 min. For 40 Oscillation Time = 4 × 0.03536 = 1.414 = 24. 1 = 40 oscillations/minute BH = 25 T 2 m of second magnet = 1.6 A-m d = 20 cm = 0.2 m (a) For north facing north
1 MBH B 2
1 =
1 MBH 2
B=
0 m 10 7 1.6 = 20 T 4 d3 8 10 3
1 = 2
2 =
40 B = BH B 2
40 25 2 = = 17.88 ≈ 18 osci/min 5 5
(b) For north pole facing south
1 MBH 2
1 = 2
2 =
40 B = BH B 2
2 min
1 MBH B 2
25 2 = 45
40 25 45
= 53.66 ≈ 54 osci/min
36.4
PERMANENT MAGNETS 1.
m = 10 A-m,
2.
m 10 7 10 10 2 –4 B= 0 2 = = = 4 × 10 Tesla 2 2 25 4 r 5 10 m1 =m2 = 10 A-m r = 2 cm = 0.02 m we know
d = 5 cm = 0.05 m
N
Force exerted by tow magnetic poles on each other = 3. 4.
S
0 m1m 2 4 10 7 10 2 –2 = = 2.5 × 10 N 4 4 r 2 4 4 10
dv –3 –3 dv = –B dℓ = – 0.2 × 10 × 0.5 = – 0.1 × 10 T-m d Since the sigh is –ve therefore potential decreases. Here dx = 10 sin 30° cm = 5 cm
B=–
dV 0.1 10 4 T m =B= dx 5 10 2 m
5.
Since B is perpendicular to equipotential surface. –4 Here it is at angle 120° with (+ve) x-axis and B = 2 × 10 T –4 B = 2 × 10 T d = 10 cm = 0.1 m (a) if the point at end-on postion.
V
0.1×10–4 T-m
7
10 2M 0 2M –4 2 × 10 = 4 d3 (10 1 )3
B=
2 10
4
10
3
= M M = 1 Am 10 2 (b) If the point is at broad-on position
6.
0.2×10
T-m
30° 0.4×10
–4
X
T-m
In cm
2
7
0 M 10 7 M –4 2 2 × 10 = M = 2 Am 3 4 d (10 1 )3 Given : 2 2 tan = 2 2 = tan tan tan = 2 cot = cot 2 tan We know = tan 2 Comparing we get, tan = cot or, tan = tan(90 – ) or = 90 – or + = 90 Hence magnetic field due to the dipole is r to the magnetic axis. Magnetic field at the broad side on position : M 2ℓ = 8 cm d = 3 cm B= 0 4 d2 2 3 / 2
= tan
7.
30° –4
0.3×10–4 T-m
–1
4 × 10 m=
–6
=
10 7 m 8 10 2
9 10
4
16 10
4 10 6 125 10 8 8 10 9
4 3 / 2
–6
4 × 10
= 62.5 × 10
–5
=
A-m
36.1
10 9 m 8
10
4 3 / 2
25 3 / 2
P S
N
Permanent Magnets 8.
We know for a magnetic dipole with its north pointing the north, the neutral point in the broadside on position. N 0M Again B in this case = 4d3 M 0 3 = BH due to earth P 4d
d3
d
= 18 T
10 7 1.44
–6 S = 18 × 10 d3 3 –3 d = 8 × 10 –1 d = 2 × 10 m = 20 cm In the plane bisecting the dipole. When the magnet is such that its North faces the geographic south of earth. The neutral point lies along the axial line of the magnet.
9.
10 7 1.44
7 0 2M –6 10 7 2 0.72 = 18 × 10–6 d3 = 2 0.7 10 = 18 × 10 4 d3 18 10 - 6 d3
8 10 9 d= 10 6
1/ 3
= 2 × 10
–1
m = 20 cm
S
N
2
N2
10. Magnetic moment = 0.72 2 A-m = M M B= 0 3 BH = 18 T 4 d
4 10 7 0.72 2 4 d 3
d =
tan
W
–6
= 18 × 10
3
0.72 1.414 10 7 6
= 0.005656
18 10 d ≈ 0.2 m = 20 cm 11. The geomagnetic pole is at the end on position of the earth. B=
0 2M 10 7 2 8 10 22 –6 = ≈ 60 × 10 T = 60 T 3 4 d (6400 10 3 )3
BH
BM
N –1
2
BH
S
E
N
S
N1
d
–5 12. B = 3.4 × 10 T M –5 Given 0 3 = 3.4 × 10 4 R M=
3.4 10 5 R 3 4 7
2
3
= 3.4 × 10 R
4 10 2M –5 B at Poles = 0 3 = = 6.8 × 10 T 4 R 13. (dip) = 60° BH = B cos 60° –6 B = 52 × 10 = 52 T
3 = 44.98 T ≈ 45 T 2 14. If 1 and 2 be the apparent dips shown by the dip circle in the 2r positions, the true dip is given by 2 2 2 Cot = Cot 1 + Cot 2 2 2 2 Cot = Cot 45° + Cot 53° 2 Cot = 1.56 = 38.6 ≈ 39° –6
BV = B sin = 52 × 10
36.2
Permanent Magnets 15. We know
BH =
Give : BH = 3.6 × 10 –2 i = 10 mA = 10 A n=?
–5
0in 2r = 45° tan = 1 r = 10 cm = 0.1 m
T
BH tan 2r 3.6 10 5 2 1 10 1 3 = = 0.5732 × 10 ≈ 573 turns 0i 4 10 7 10 2 –4 2 16. n = 50 A = 2 cm × 2 cm = 2 × 2 × 10 m –3 i = 20 × 10 A B = 0.5 T –3 –4 –4 = ni A B = niAB Sin 90° = 50 × 20 × 10 × 4 × 10 × 0.5 = 2 × 10 N-M n=
17. Given = 37° We know
d = 10 cm = 0.1 m
M 4 ( d2 2 ) 2 4 d4 = tan = tan [As the magnet is short] BH 0 2d 0 2d
4
=
7
(0.1)3 –3 7 4 3 2 –1 tan 37 = 0.5 × 0.75 × 1 × 10 × 10 = 0.375 × 10 = 3.75 ×10 A-m T 2
4 10 M 3 2 –1 18. (found in the previous problem) = 3.75 ×10 A-m T BH = 37°, d=? M 4 2 = (d 2 )3 / 2 tan BH 0 ℓ << d neglecting ℓ w.r.t.d M 1 4 3 3 = d Tan 3.75 × 103 = × d × 0.75 7 BH 0 10
3.75 10 3 10 7 –4 = 5 × 10 0.75 d = 0.079 m = 7.9 cm M 2 19. Given = 40 A-m /T BH 3
d =
Since the magnet is short ‘ℓ’ can be neglected So,
M 4 d3 = = 40 BH 0 2
S
40 4 10 7 2 –6 = 8 × 10 4 –2 d = 2 × 10 m = 2 cm with the northpole pointing towards south. 20. According to oscillation magnetometer, 3
d =
T = 2
N
MBH
1.2 10 4 = 2 10 M 30 10 6 2
1.2 10 4 1 = M 30 10 6 20 M=
1.2 10 4 400 30 10
6
2
2
= 16 × 10 A-m = 1600 A-m
36.3
2
Permanent Magnets
1 mB H 2 For like poles tied together M = M 1 – M2 For unlike poles M = M1 + M2
21. We know : =
1 = 2
S
N
N
S
N
S
M M2 M M2 M1 M2 10 = 1 25 = 1 M1 M2 M1 M2 M1 M2 2
2M1 M 26 13 = 1 = 24 2M2 M2 12 –6
T1 = 0.1
T –6
B = BH – Bwire = 2.4 × 10
24 10 6 MBH
23. T = 2
B BH
2
0 .1 14 0.01 14 2 = T2 = T2 = 0.076 24 24 T2 Here = 2
1 min 40
T1 = T2
o i 2 10 7 18 –6 –6 –6 = 24 × 10 – = (24 –10) × 10 = 14 × 10 2 r 0 .2
T1 = T2
14 10 6
0 .1 = T2
T1 =
–
MBH
T = 2
S
2
22. BH = 24 × 10
N
T2 = ?
1 = 40T2
1 1 1 1 2 = T2 = T2 = 0.03536 min 2 2 800 1600 T2 2
For 1 oscillation Time taken = 0.03536 min. For 40 Oscillation Time = 4 × 0.03536 = 1.414 = 24. 1 = 40 oscillations/minute BH = 25 T 2 m of second magnet = 1.6 A-m d = 20 cm = 0.2 m (a) For north facing north
1 MBH B 2
1 =
1 MBH 2
B=
0 m 10 7 1.6 = 20 T 4 d3 8 10 3
1 = 2
2 =
40 B = BH B 2
40 25 2 = = 17.88 ≈ 18 osci/min 5 5
(b) For north pole facing south
1 MBH 2
1 = 2
2 =
40 B = BH B 2
2 min
1 MBH B 2
25 2 = 45
40 25 45
= 53.66 ≈ 54 osci/min
36.4
