14.3 Solutions
675
14.58 Magnetic energy density
u B =
1
2
B 2
μ 0
=
1
2
4 2
4π × 10 −7 =
2
π
× 10
7 J/m
3
Energy stored in the magnetic field U B = u B × volume = u B πr 2 l
= 2 × 10
7
× 3
2
× 12.5 = 2.25 × 10
9 J
14.59 If P 0 is the power radiated by sun of radius r , then using the results of
prob. (14.51)
P 0 = S 4πr
2
=
1
2μ 0 c
E
2
m 4πr
2
∴ E m =
1
r
P 0 μ 0 c
2π
∴ E m =
1
7 × 10 8
4 × 10 26 × 4π × 10 −7 × 3 × 10 8
2π
= 2.21 × 10
5 V/m
B m =
E m
c
=
2.21 × 10 5
3 × 10 8 = 7.37 × 10
−4 T = 7.37 G
14.60 By prob. (14.51)
S =
E 2
m
2μ 0 c
∴ E m =
2μ 0 c S =
2 × 4π × 10 −7 × 3 × 10 8 × 1300 = 990 V/m
14.61 B m =
E m
c
=
300
3 × 10 8 = 1 × 10
−6 T
S =
E 2
m
2μ 0 c
=
(300) 2
2 × 4π × 10 −7 × 3 × 10 8 = 119.4 W/m
2
14.62
|E|
|H |
=
E μ 0
B
=
(Bc)μ 0
B
=
μ 0
√
μ 0 ε 0
=
μ 0
ε 0
where we have used the equations B = μ 0 H and c =
1
√
μ 0 ε 0
.
μ 0
ε 0
=
4π × 10 −7
8.85 × 10 −12 = 377
Units of
|E|
|H |
are
Volt/metre
Ampere/metre
=
Volt
Ampere
= Resistance
675
14.58 Magnetic energy density
u B =
1
2
B 2
μ 0
=
1
2
4 2
4π × 10 −7 =
2
π
× 10
7 J/m
3
Energy stored in the magnetic field U B = u B × volume = u B πr 2 l
= 2 × 10
7
× 3
2
× 12.5 = 2.25 × 10
9 J
14.59 If P 0 is the power radiated by sun of radius r , then using the results of
prob. (14.51)
P 0 = S 4πr
2
=
1
2μ 0 c
E
2
m 4πr
2
∴ E m =
1
r
P 0 μ 0 c
2π
∴ E m =
1
7 × 10 8
4 × 10 26 × 4π × 10 −7 × 3 × 10 8
2π
= 2.21 × 10
5 V/m
B m =
E m
c
=
2.21 × 10 5
3 × 10 8 = 7.37 × 10
−4 T = 7.37 G
14.60 By prob. (14.51)
S =
E 2
m
2μ 0 c
∴ E m =
2μ 0 c S =
2 × 4π × 10 −7 × 3 × 10 8 × 1300 = 990 V/m
14.61 B m =
E m
c
=
300
3 × 10 8 = 1 × 10
−6 T
S =
E 2
m
2μ 0 c
=
(300) 2
2 × 4π × 10 −7 × 3 × 10 8 = 119.4 W/m
2
14.62
|E|
|H |
=
E μ 0
B
=
(Bc)μ 0
B
=
μ 0
√
μ 0 ε 0
=
μ 0
ε 0
where we have used the equations B = μ 0 H and c =
1
√
μ 0 ε 0
.
μ 0
ε 0
=
4π × 10 −7
8.85 × 10 −12 = 377
Units of
|E|
|H |
are
Volt/metre
Ampere/metre
=
Volt
Ampere
= Resistance