南京大学大学物理答案17-19prob

南京大学大学物理电磁学习题参考答案

Problem 17.2 For a single ring2µ B= 4π ( R 2 + x 2 ) 3 2

µ0

For Helmholtz coils z apart µ0 1 1 B= × 2µ + 32 32 2 2 4π 2 z z 2 R + x R + + x 2 2

南京大学大学物理电磁学习题参考答案

z z +x x 3µ 0 dB 2 2 = ×µ 32 32 2 2 dx 2π 2 z z 2 R + + x R + x 2 2

=0

(odd function of x)2 2 R 2 4( z 2 ) 2 3µ 0 R 4( z 2 ) = + ×µ 72 72 2 2 2 π R 2 + (z 2 )2 R + (z 2 )

d B 2 dx

2

x =0

[

]

[

]

南京大学大学物理电磁学习题参考答案

Plot for

z = 0.4,0.5,0.6 2R

南京大学大学物理电磁学习题参考答案

Problem 17.5 One side contributes:

π π sin sin B= 4π R cos π n n n µ 0 2I π = tan 4π R n I

µ0

南京大学大学物理电磁学习题参考答案

Discuss:

π µ0 I n π tan = nB = tan 4π R n 2R π nn π lim tan = 1 leads to a loop solution n→∞ π nB=

µ 0 2nI

µ0

2 z2 + R

(

IR 2

2 32

)

k

南京大学大学物理电磁学习题参考答案

Problem 17.9 2keV<<0.511MeV non-relativistic

V// = V0 cos 89°; V⊥ = V0 cos 1°(a) a translation of V // and a circular motion of V⊥ . So the trajectory is a helix.2 V⊥

V⊥ qB (b) qV⊥ B = m ω= = r m r 2π 10 T= = 3.57 × 10 s ω pitch p = V// T = 0.17mmradius r = 1.5 × 10 3 m

南京大学大学物理电磁学习题参考答案

*Problem 17.10 The cyclotron radius is a = mv/qB.

The trajectories are symmetric about x = – a .

南京大学大学物理电磁学习题参考答案

Problem 18.1

B( r ) =

µ0 2 I4π r

;

dΦ dA (r ) = B = B dr v dt dt

b ε = B (r ) v dr = 2 Iv ln 4π a a

∫

b

µ0

= 10 × 2 × 100 × 5.0 × ln 20V

7

= 3.0 × 10 V

4

南京大学大学物理电磁学习题参考答案

Problem 18.2

dB 2πrE = πr dt2

1 dB E = r 2 dtThe tangential component is

1 dB 1 dB E t = r cos = d 2 dt 2 dtwhich is independent and points left. The emf in the rod is then

南京大学大学物理电磁学习题参考答案

dB 2 ε= R dt 2 2

2

南京大学大学物理电磁学习题参考答案

Problem 18.5 Under approximation r << d, dipole expressions can be used.

B(r ) =

µ0 14π r3

[3( µ er )er µ ]

The flux can be found as follows (Bk)

Φ = πr 2

µ0 14π d3

(

πr 2 Iπr = 3 µ 0 I = MI 2d4 2

)

南京大学大学物理电磁学习题参考答案

πr M = 3 µ0 2d4

I = k 0t

2

πr 4 ε 1 3 µ 0 k 0t I′ = = R R d Because of

(e )

er ~

dk + re ρ d +r2 2

Only ρ component of B contributes to the force:

南京大学大学物理电磁学习题参考答案

µ0 1 2 r F = 2πrI ′ 3Iπr k 3 4π d d4

Bρ

sin θ

µ 0 1 2πr πr 2 2 r 3 µ 0 k 0t = 3k 0t πr k d 4π d 3 R d

µ0 = 4π

2

4 8 2 3 24π r k 0 t 7 d R

k

南京大学大学物理电磁学习题参考答案

Problem 18.6 Known

t B = B0 ; (0 ≤ t ≤ T ) Tt Φ = πr B0 T2

The flux is

Faraday’s law reads:

πr B0 rB0 E 2 πr = ; or E = T 2T2

南京大学大学物理电磁学习题参考答案

Tangential velocity is

v=

E q m

t

The centripetal force on the bead is (–eρ)

v m = r

4mT

2

2 2 2 rB0 q t 2

南京大学大学物理电磁学习题参考答案

The magnetic force isF = qv × B = q E q m tBeρ

2 rB0 q 2t 2 = ( eρ ) 2 2mT

The track contributes2 2 rB0 q 2 t 2 rB0 q 2 t 2 v m ( eρ ) F = ( eρ ) ( eρ ) 2 2 r 4mT 2mT 2

=

2 2 2 rB0 q t 2

4mT

eρ