Solutions of Electricity & Magnetism. Lesson 20th to 25th

140 Pages·2016·2.69 MB·English

Solutions
of
Electricity & Magnetism
Lesson 20 to 25
th th
By DC Pandey

20 Current Electricity
Introductory Exercise 20.1
q 2pr 2.0
1. i= , here q=e, t= =
t v 8.43 ¥1028 ¥1.6¥10-19 ¥3.14
ev
\ i= ¥(0.5¥10-3)2
2pr
1.6¥10-19 ¥2.2¥106 =1.88¥10-6 ms-1
=
2¥3.14¥5¥10-11 3. Yes.
=1.12¥10-3 A As curr ent alw ays flows in the dir ect ion of
elec tric field.
=1.12mA
4. False.
2. No. of at oms in 63.45 g of Cu =6.023 ¥1023
In the abs ence of po tent ial diff er ence,
\ No. of atoms in 1 cm3 (8.89 g) of Cu
elec trons passes ran dom mot ion.
6.023 ¥1023
= ¥8.89 5. Cur rent due to both posi t ive and neg at ive
63.54
ions is from left to right, hence, there is a net
=8.43 ¥1022 curr ent from left to right.
As one conduction electron is present per dq
atoms, 6. i=10+4t fi dt =10+4t
n=8.43 ¥1022 cm-3 or 8.43 ¥1028m-3 fi Úqdq=Ú10(10+4t)dt
As i=neAv 0 0
d i fi q=[10t+2t2]10 =300C
fi v = 0
d neA
Introductory Exercise 20.2
rL 0.49¥3.14¥(0.42¥10-3)2
1. R= =
A 2.75¥10-8
35
=1.72¥10-8 ¥
Ê2.05 ˆ2 =9.87A
3.14¥Á ¥10-3˜ (b) V =EL =0.49¥12=5.88 V
Ë 2 ¯
V 5.88
(c) R= = =0.6W
=0.57W
i 9.87
E
2. (a) J = 3. Let us con sider the con duc tor to be made up
r
of a numb er of ele m ent ary discs. The
EA
fi i=JA= con duc tor is supp osed to be ext ended to form
r a comp lete cone and the ver tex O of the cone
3
rl
is taken as ori g in with the cond uct or placed R=
along x-axis with its two ends at x=r and pr(l+ r)tan2q
x=l+ r. Let q be the semi-vert ic al ang le of
But, rtanq=a
the cone.
(r+ l)tanq=b
Consider an elementary disc of thickness dx
rl
at a distance x from origin. R=
pab
Resistance of this disc,
1
dR=rdx 4. True. r=
s
A
1
\ r¥ s= ¥ s=1
If y be the radius of this disc, then
s
A=py2
5. R = R
Cu Fe
But y=xtanq
4.1(1+ a DT)=3.9(1+ a DT)
dx Cu Fe
dR=r 4.1[1+4.0¥10-3(T -20)]
px2tan2q
=3.9[1+5.0¥10-3(T -20)]
\ Resistance of conductor
R=ÚdR=Úl+r rdx 4.1+16.4¥10-3(T -20)
r px2tan2q =3.9+19.5¥10-3(T -20)
r È 1˘l+r 3.1¥10-3(T -20)=0.2
R= -
ptan2qÎÍ x˚˙r T -20= 0.2
3.1¥10-3
r È1 1 ˘
= Í - ˙ =64.5∞C
ptan2 qÎr l+ r˚
T =84.5∞C
Introductory Exercise 20.3
1. Potential difference across both the res ist ors V =V +2=2V
B A
is 10V. V =V +5=5V
C A
i1 i2 VD =VC+10=15 V
V -V
\ i = C B =3 A
1 1
2W 10 V 4W and i = VD -VA =15 =7.5 A
2 2 2
3. Curr ent in the given loop is
E+15
10 i=
Hence, i = =5A 8
1 2
ÊE+15ˆ
10 V =E-2i=E-2◊Á ˜ =0
and i2 = 4 =2.5A AB Ë 8 ¯
fi E=5V
2. As A is grounded, V =0
A
1W i 4. Ef fec tive emf,
C 1 B
E=8¥1-2¥1=6V
Effective resistance of circuit
5 V
10 V 2 V R1 i1
i2 i3
D i2 2W A 10 V R2 R3 10 V
4
R= Rexternal +10r =2+10¥1=12W 6. i= E
E 6 R+ r
\ i= = =0.5A
R 12 Also, V =E-ir
5. As R2 = R3 and V1 =V2 i= E-V
Potential difference across R is zero. r
1
Hence, current through R1fi i1 =0 (a) I (b) I
and current through R
2
V E E
fi i = 1 r r
2 R
2
10
= =1A
10 O R O E V
Introductory Exercise 20.4
1. 6=1-E
12 V r E=-5 V
i
P Q And from
V =V
E 1W 1A ST QP
U R 6=-ir+12
12-6 6
r= = =2W
3W i 3
T S
2A 2. Power de liv ered by the 12 V power supp ly,
Applying KCL at junction R P =Vi =12¥3 =36W
1
i=1+2=3 A and power dissipated in 3W resister,
V =V =V P =i2R =22 ¥3 =12W
ST RU QP 3 3 3
Taking V =V
ST RU
Introductory Exercise 20.5
E1 + E2 + E3 10 + 4 + 6
1. E= r1 r2 r3 = 1 2 2 E r
1 1 1 1 1 1
+ + + +
r r r 1 2 2
1 2 3
10+2+3
=
2
R
=7.5 V
1 1 1 1 Rate of dissipation of energy
and = + +
r r r r E2R
1 2 3 P=i2R =
1 1 1 (R+ r)2
= + + =2
1 2 2
For maximum or minimum power
1
fi r= dP
=0
2
dR
=0.5W È(R+ r)2 -2R(R+ r)˘
E fi E2Í ˙ =0
2. i= Î (R+ r)4 ˚
R+ r
5
(R+ r)(r- R)
fi E2 =0 5. G=100W, ig =50mA, i=5mA
(R+ r)4 i G 50¥10-6 ¥100
\ S= g =
E2(r- R) i-i 5¥10-3 -50¥10-6
fi =0 g
(R+ r)3 1 1
= =
fi R=r 1-0.01 0.99
ddR2P2 =E2ÈÎÍ(R+ r)3(-1()R-3+(rr)-6 R)(R+ r)2˘˚˙ =19090W
100
-E2(4r-2R) By connecting a shunt resistance of W.
= 99
(R+ r)4 V
d2P 6. ig =G
Clearly is negative at R=r.
dR2 and R= nV -G=(n-1)G
i
Hence, P is maximum at R=r g
E2r E2 15
and P = = 7. V = E
max (r+ r)2 4r AB 16
Potential gradient
3. When the batt er ies are con nected in se ries V 15E
k= AB =
E =2E=4V, r =2r=2W
eff eff L 16¥600
For maximum power
E
= V/cm
R=reff =2W 640
E2 (4)2 E E
and P = eff = =2W (a) =kL fi L= =320 cm
max 4r 4¥2 2 2k
eff
E 7E
4. Ig =5mA, G=1W, V =5 V (b) V =kl =640 ¥560= 8
V 5
R= -G= -1 Also, V = E-ir
I 5¥10-3
g 7E
\ E-ir=
=999W 8
A 999 W resistance must be connected in i= E
series with the galvanometer. 8r
AIEEE Corner
Subjective Questions (Level 1)
q ne 337.5
1. i= = = =2.11¥1021
t t 1.6¥10-19
Given, 2pr 1 v
i=0.7 , t=1 s, e=1.6¥10-19C 4. T = v fi f = T =2pr
it 0.7¥1
\ n= = 2.2¥106
e 1.6¥10-19 =
2¥3.14¥5.3 ¥10-11
=4.375¥108
=6.6¥1019 s-1
2. q=it=3.6¥3 ¥3600 q
I = =ef
=38880 C T
3. (a) q=it =7.5¥45=337.5C =1.6¥10-19 ¥6.6¥1019
q
(b) q=ne fi n= =10.56A
e
6
5. (a) I =55-0.65t2 =1.7¥10-8 W-m
I = dq l=24.0 m
dt Êdˆ2 Ê2.05 ˆ2
A=pÁ ˜ =3.14¥Á ¥10-3˜
fi dq=Idt Ë2¯ Ë 2 ¯
fi q=ÚIdt =3.29¥10-6 m2
\ q=Ú8 Idt=Ú8(55-0.65t2)dt R=1.7¥10-8 ¥ 24.0
0 0 3.29¥10-6
Èt2˘8
=55[t]08 -0.65ÎÍ2˚˙ =0L.12W
0 10. R=r
=440-20.8=419.2C A
rL
(b) If current is constant A=
q 419.2 R
I = = =52.4A
t 8 If D is density, then
DrL2
6. iµv m=DV =DAL=
d
v i R
\ vd2 = i2 8.9¥103 ¥1.72¥10-8 ¥(3.5)2
d1 1 = 0.125
i 6.00
fi vd2 = i2vd1 =1.20 ¥1.20¥10-4 =1.5¥10-2 kg = 15 g
1
=6.00¥10-4 ms-1 11. At 20∞C,
i R1 =600W, R2 =300W
7. vd = neA At 50∞C,
1 R¢ = R(1+ a Dt)
= 1 1 1
8.5¥1028 ¥1.6¥10-19 ¥1¥10-4 =600(1+0.001¥30)=600¥1.03
=0.735¥10-6 ms-1 =618W
=0.735mm/s R¢ = R (1+ a Dt)
2 2 2
l 103 =300(1+0.004¥30)=336
t= =
v 0.735¥10-6 \ R¢= R ¢+ R ¢=618+336
d 1 2
=1.36¥109 s =43 yr =954W
R¢- R 954-900
8. Dist ance cov ered by one elec tron in 1 s a = =
R¥Dt 900¥30
=1¥0.05=0.05 cm
R=600+300=900W
Number of electrons in 1 cm of wire
=2¥1021 =0.002∞C-1
\ Number of electrons crossing a given area 12. As both the wires are conn ected in par all el,
per second V =V
Al Cu
=Number of electrons in 0.05 cm of wire i R =i R
Al Al Cu Cu
=0.05¥2¥1021 =1020 L L
i r Al =i r Cu
i= q = ne Al Al pd2 Cu Cu pd2
Al Cu
t t
i r L
1020 ¥1.6¥10-19 fi d =d Cu Cu Cu
= =1.6¥10 =16A Cu Al i r L
1 Al Al Al
L 2¥0.017¥6
9. R=r =1¥10-3
A 3 ¥0.028¥7.5
Given, =0.569¥10-3 m
r=0.017mW-m =0.569 mm.
7
V 0.938 20
13. (a) E= L = 75¥10-2 =1.25 V/m fi R2 =11W =1.82W
E 1.25 and R =20- R =20-1.82W
(b) J = fi r= 1 2
r 4.4¥107 =18.2W
r=2.84¥10-8 W-m 17. The circ uit can be red rawn as
E V
14. (a) J = =
r rL
Current density is maximum when L is 24 V 12W 8W
minimum, ie, L=d, potential difference
should be applied to faces with dimensions
2d¥3d.
V 8¥12
Jmin. =rd. Reff =12+ 8 =4.8W
V VA V 24
(b) i= = I = = =5 A
R rL R 4.8
eff
Current is maximum when L is minimum 18. Here, A and C are at same po ten tial and B
and A is maximum. and D are at same po ten tial,
Hence, in this case also, V should be applied 8W
to faces with dimensions 2d¥3d A B
V(2d¥3d) 6Vd
and i = = .
max r(d) r 24V 4W 6W
L
15. (a) R=r
RAA D 12W C
r =
L Hence, the circuit can be redrawn as
d
[r= =1.25 mm =1.25¥10-3 m] A,C
2
0.104¥3.14¥(1.25¥10-3)2
=
14.0 24V 4W W8 W12 6W
=3.64¥10-7W-m
V EL 1.28¥14
(b) i= = = =172.3 A
R R 0.104
B,D
(c) i=neAv
d 1 1 1 1 1
i \ = + + +
v = R 4 8 12 6
d neA
6+3 +2+4
172.3 =
= 24
8.5¥1028 ¥1.6¥10-19 ¥3.14¥(1.25¥10-3)2 15 5
= =
=2.58¥10-3 ms-1 24 8
8
16. For zero ther mal coe f fic ient of resistance, R= W
DR=0 5
R a DT + R a DT =0 =1.6W
C C Fe Fe
R1 = -aFe = -5.0¥10-3 =10 i= V = 24
R a -0.5¥10-3 R 1.6
2 C
=15A
R =10R
1 2 19. Given cir cuit is sim i lar to that in pre vi ous
Also, R + R =20
1 2 ques tion but 4W re sist or is rem oved. So the
fi 10R2 + R2 =20 ef fec tive cir cuit is given by
8
V 12
A,C i= =
R 36/13
13
= A
24 V 8W 2W 6W 3
1 12+6
21. (a) i= =3A
1+2+3
B,D i = 3A
A
1 =1 + 1 + 1 12 V 1W
R 8 12 6
1 3 +2+4 9 3 B
R = 24 =24 = 8 G 2W
8
R= W=2.67W C
3 6 V 3W
V 24
i= = =9A
R 2.67 D
V =0
20. 6W G
A B V =V +12=12 V
A G
4W V -V =3 V
A B
12V 12W 6W V =12-3 =9 V
B
3W V -V =6 V
B C
D 2W C VC=9-6=3 V
fl
V -V =6 V, V =-6V
A G D D
6W 4W (b) If 6V battery is reversed
4W 12-6
12 B C 12W i= =1 A
1+2+3
3W 2W
i = 1A
D
A
Wheatstone bridge is bal anced, hence 4W
12 V
re sis tance con nected be tween B and C be
B
re moved and the eff ec tive cir cuit becomes
G
A
6W 4W C
12V B C 12W 6 V
3W 2W
D
D
fl
A V =0,
G
V -v =12 V, V =12 V
A G A
12V 9W 6W 12W V -V =1 V
A B
fi V =11 V
B
fl V -V =2 V
B C
fi V =9 V
C
36
12V 13 W VD -VG =6 V
fi V =6 V
D
9
22. i= 200 =5A =42.26W
5+10+25 E
i= =0.102W
R
1 eff
5W Reading of voltmeter
200 V
V =E-ir
2
=4.3 -0.102¥1
10W
ª4.2W
0
Reading of Ammeter,
25W V 4.2
i = = =0.08A
1 R+ R 42
3 a
(i) V -V =5¥25 24. Con sider the dir ec tions of cur rent as shown
3 0
in fig ure.
fi V =125 V
3
(ii) V0 -V2 =5¥10 42V 5W A 4W
V =-50 V
2
(iii) V -V =5¥5
2 1 V =-75 V I1 6W I2 10V
(iv) V =5¥35=1175 V 8W
3-2 B
(v) V1-2 =-5¥5=-25 V 1W C
(vi) V1-3 =-200 V I3
23. (a) 6W
i E r D
4V
Applying KVL in loop 1, 2 and 3, we
i R
1
A respectively get,
i2 S I1 +6(I1 -I2)+5I1 =42
V fi 12I -6I =42
1 2
Reff = R||Rv + Ra + r fi 2I1 -I2 =7 …(i)
50¥200 4I +6(I -I )+ 8(I + I )=10
= +2+1 2 2 1 2 3
50+200 fi 9I -3I +4I =5 …(ii)
2 1 3
8(I + I )+16I =4
=43W 2 3 3
E 4.3 2I +6I =1 …(iii)
i= = =0.1A 2 3
R 43 On solving, we get,
eff
\ Reading of ammeter, i=0.1A I1 =4.7A, I2 =2.4A, I3 =0.5A
and reading of voltmeter =i(R||R ) Resistor 5W 1W 4W 6W 8W 16W
v
=0.1¥40=4V Current 4.7A 4.7A 2.4A 2.3A 2.9A 0.5A
(b) i E r 25. R = 400W
V400W 200W
i R
1
A
100W 100W 200W fiA100W 100WD
i C
2 i2 i2 B 200W
V 100W 100W
i i
R =(R + R)||R + r 1 i 1
eff a v
52¥200
= +1 i 10 V i 10 V
52+200