13.3 Solutions
619
J = ev
(1)
The magnetic field due to a current-carrying circuit is given by the Biot–
Savart law
dB =
μ 0 I
4π
dl × R
R 3
(2)
When dl is the circuit element, R is the vector which points from dl to the
field point. As there is only one proton there is no need to integrate to find
B. Replacing the current by the current density J , the Biot–Savart law is
modified as
B =
μ 0
4π
J × R
R 3
(3)
Substituting (1) into (3)
B =
μ 0 e
4π
v × R
R 3
R =
ˆ
i + 2 ˆ
j
m
v =
ˆ
i + 3 ˆ
j
10
4 m/s
v × R = 10
4
ˆ
i
ˆ
j
ˆ
k
1 3 0
1 2 0
= −10
4 ˆ
k
B =
4π × 10 −7
4π
× 1.6 × 10
−19
×
−10 4 ˆ
k
√
5
3
= 1.43 × 10
−23 ˆ
k T
13.3.3 Magnetic Force
13.52
F
l
=
μ 0 i 1 i 2
2π d
=
4π × 10 −7 × 2 × 3
2π × 0.05
= 2.4 × 10 −5 N
13.53
F
l
=
μ 0 i 1 i 2
2 π d
∴
i 2 =
2π d
μ 0 i 1
F
l
=
2π × 0.2 × 10 −5
4π × 10 −7 × 10
= 1.0 A
The currents are parallel.
619
J = ev
(1)
The magnetic field due to a current-carrying circuit is given by the Biot–
Savart law
dB =
μ 0 I
4π
dl × R
R 3
(2)
When dl is the circuit element, R is the vector which points from dl to the
field point. As there is only one proton there is no need to integrate to find
B. Replacing the current by the current density J , the Biot–Savart law is
modified as
B =
μ 0
4π
J × R
R 3
(3)
Substituting (1) into (3)
B =
μ 0 e
4π
v × R
R 3
R =
ˆ
i + 2 ˆ
j
m
v =
ˆ
i + 3 ˆ
j
10
4 m/s
v × R = 10
4
ˆ
i
ˆ
j
ˆ
k
1 3 0
1 2 0
= −10
4 ˆ
k
B =
4π × 10 −7
4π
× 1.6 × 10
−19
×
−10 4 ˆ
k
√
5
3
= 1.43 × 10
−23 ˆ
k T
13.3.3 Magnetic Force
13.52
F
l
=
μ 0 i 1 i 2
2π d
=
4π × 10 −7 × 2 × 3
2π × 0.05
= 2.4 × 10 −5 N
13.53
F
l
=
μ 0 i 1 i 2
2 π d
∴
i 2 =
2π d
μ 0 i 1
F
l
=
2π × 0.2 × 10 −5
4π × 10 −7 × 10
= 1.0 A
The currents are parallel.