13.3 Solutions
613
dq = q
2πr dr
π R 2 =
2qrdr
R 2
(1)
The elementary current due to rotation of charge with frequency f is
di = f dq =
ω
2π
2qrdr
R 2 =
ω qr dr
π R 2
(2)
The induction at the centre due to the current in the ring is
dB =
μ 0 di
2r
=
μ 0
2r
ω qrdr
π R 2 =
μ 0 ω qdr
2π R 2
(3)
Total induction from the rotating disc
B =
dB =
R
0
μ 0 ω qdr
2 π R 2 =
μ 0 ω q
2 π R
(4)
Fig. 13.27 Magnetic
induction at the centre of a
charged rotating disc
13.38 The field at any point P 1 at distance x from P, the middle point will be
B =
μ 0 iNR 2
2
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎩
1
R 2 +
1
2
R + x
2
3/2 +
1
R 2 +
1
2
R − x
2
3/2
⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎭
(1)
by prob. (13.36).
613
dq = q
2πr dr
π R 2 =
2qrdr
R 2
(1)
The elementary current due to rotation of charge with frequency f is
di = f dq =
ω
2π
2qrdr
R 2 =
ω qr dr
π R 2
(2)
The induction at the centre due to the current in the ring is
dB =
μ 0 di
2r
=
μ 0
2r
ω qrdr
π R 2 =
μ 0 ω qdr
2π R 2
(3)
Total induction from the rotating disc
B =
dB =
R
0
μ 0 ω qdr
2 π R 2 =
μ 0 ω q
2 π R
(4)
Fig. 13.27 Magnetic
induction at the centre of a
charged rotating disc
13.38 The field at any point P 1 at distance x from P, the middle point will be
B =
μ 0 iNR 2
2
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎩
1
R 2 +
1
2
R + x
2
3/2 +
1
R 2 +
1
2
R − x
2
3/2
⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎭
(1)
by prob. (13.36).