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thinkiit solution hcverma class xi
thinkiit solution hcverma class xi
Short answers of hcv
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.
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.
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