612
13 Electromagnetism I
dB z =
μ 0 i R 2 dφ
4π r 3
(3)
Integrating
B z = B =
dB z =
μ 0 i R 2
4π r 3
2π
0
dφ =
μ 0 i R 2
2 (R 2 + z 2 ) 3/2
(4)
13.33 B =
μ 0 Ni
l
=
4π × 10 −7 × 500 × 5
1.0
= 3.14 × 10 −3 T
13.34 B =
μ 0 I
2π
1
x
+
1
d − x
13.35 Apply Ampere’s law inside the hollow cylindrical conductor
(B) (2π r ) =
μ 0 i π (r 2 − a 2 )
π (b 2 − a 2 )
(1)
where the right-hand side includes only the fraction of the current that passes
through the surface enclosed by the path of integration. Solving for B,
B =
μ 0 i
2 π r
(r 2 − a 2 )
(b 2 − a 2 )
.
13.36 Magnetic field at P, due to loop A is
B A =
μ 0 INR 2
2(R 2 + x 2 ) 3/2
(1)
where x = AP =
R
2
. Similarly the magnetic field B c due to the second loop
is given by an identical expression with x = CP =
R
2
. As the currents are in
the opposite direction these two fields are added:
∴
B = B A + B C =
μ 0 INR 2
(R 2 + x 2 ) 3/2
(2)
Put x = R/2 to find
B =
8 N μ 0 I
5 3/2 R
(3)
13.37 Consider a ring of radius r , width dr , concentric with the disc. The charge
on the ring, Fig. 13.27