ENGG2520D Assignment 2 Li Man Hin 1155047511 1. (a) ∮ B · ds

ENGG2520D Assignment 2
Li Man Hin
1155047511
I
B · ds = µ0 ienc = 4π × 10−7 × (30 − 10) = 8π × 10−6 = 2.51 × 10−5 Tm
I
B · ds = µ0 ienc = 4π × 10−7 × (2 × 30 − 10) = 2π × 10−5 = 6.28 × 10−5 Tm
1. (a)
(b)
mv 2
= |qv × B| = qvB
r
qrB
1.6 × 10−19 × 210 × 1
⇒v=
=
= 8.56 × 107 ms−1
m
235 × 1.67 × 10−27
1
KE = mv 2 = 21 (235 × 1.67 × 10−27 )(8.56 × 107 )2 = 1.44 × 10−9 J
2
8.56 × 107
v=
c = 0.285c
3 × 108
2. At maximum radius,
3. (a) µ = N iA = iA ⇒ i =
(b) B =
8 × 1022
µ
=
= 1.86 × 109 A
A
π(3700000)2
µ0 i
4π × 10−7 × 1.86 × 109
=
= 3.16 × 10−4 T
2R
2 × 3700000
(c)
µ0 irds
cos θ
4π r2
µ0 ids
R
=
2 3/2
2
4π (R + RE
)
Z 2πR
µ0 i
R
ds
Bz =
2 3/2
4π (R2 + RE
)
0
R2
µ0 i
=
2 3/2
2 (R2 + RE
)
−5
= 4.00 × 10 T
z-component = dB cos θ =
1
4. (a) By Kirchoff’s voltage law,
3 × 10−2
q
=
V =
= 300 V
C
1 × 10−4
(b)
di
q
+L =0
C
dt
q(t) = q0 sin(ωt + φ0 )
q(0) = q0 sin(φ0 ) = q0
π
⇒ φ0 =
2
1
t
1
π
⇒ q(t) = 0.03 sin( √
ω=√
=√
+ )
2 × 10−7 2
2 × 10−7
LC
0.03
π
q(t)
t
=
+ )
V (t) =
sin( √
−4
C
1 × 10
2 × 10−7 2
π
= 300 sin(ωt + )
2
π
t
+ )
i(t) = q 0 (t) = 0.03ω cos( √
2 × 10−7 2
(c)
1
EC (t) = CV (t)2
2
π
1
= × 1 × 10−4 × (300 sin(ωt + ))2
2
2
t
π
9 2
+ )
= sin ( √
2
2 × 10−7 2
1
EL (t) = Li(t)2
2
1
1
π
= × 2 × 10−3 × (0.03 × √
cos(ωt + ))2
2
2
LC
π
t
cos(ωt + ))2
= 10−3 × (0.03 × √
−7
2
2 × 10
9
t
π
= cos2 ( √
+ )
2
2 × 10−7 2
π
π
9
t
9
t
+ ) + cos2 ( √
+ )
EC (t) + EL (t) = sin2 ( √
2
2
2 × 10−7 2
2 × 10−7 2
9
= J
2
µ0 i
5. (a) B-field at x = B(x) =
2π(x + 1)
Z HZ W
Z Z
µ0 ln 4
µ0 i 2 3 1
ΦB =
B(x)dxdy =
dxdy =
i
2π 0 0 x + 1
π
0
0
2
(b) ΦB (t) = (4 × 10−7 ln 4)(2t3 × 106 ) = (0.8 ln 4)t3
ΦB (1) = 0.8 ln 4 = 1.11 Wb
dΦB
d
(c) E = −
= − (0.8 ln 4)t3 = −(2.4 ln 4)t2
dt
dt
E(1) = −2.4 ln 4 = −3.33 V
(d) q(t) = C|V (t)| = C|E(t)|
q(1) = C|E(1)| = 1 × 10−3 (3.33) = 3.33 × 10−3 C
d
(e) i(t) = q 0 (t) = (2.4 ln 4 × 10−3 )t2 = (4.8 ln 4 × 10−3 )t = 6.65 × 10−3 t
dt
By Lenz’s Law, an induced electromotive force always gives rise to a current
whose magnetic field opposes the original change in magnetic flux. Therefore
in order to generate magnetic field out of paper, with right-hand grip rule, the
current will be anticlockwise.
3