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4.23.2 Toys other than magnetic/electrical experimental sets) i* j4 b9 n1 C/ e1 \
a) Any loose as-received magnet(s) and magnetic component(s) shall either have a magnetic flux index less than 50 kG2mm2 (0,5 T2mm2) when tested according to 8.35 (magnetic flux index), or shall not fit entirely in the cylinder when tested according to 8.2 (small parts cylinder)., S1 k& @; o& V9 I# E
b) Any magnet(s) and magnetic component(s) that become(s) released from a toy, or from a loose asreceived magnetic component, when tested according to 8.3 (torque test), 8.4.2.1 (tension test, general), 8.4.2.2 (tension test, seams and materials), 8.5 (drop test), 8.7 (impact test), 8.8 (compression test), and finally, for magnets that are accessible but not grippable (as specified in 8.4.2.1), 8.34 (tension test for magnets), shall either have a magnetic flux index less than 50 kG2mm2 (0,5 T2mm2) when tested according to 8.35 (magnetic flux index), or shall not fit entirely in the cylinder when tested according to 8.2 (small parts cylinder).! b X1 B! X ?3 ~: a, M3 F0 J
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8.35.4 Calculation of magnetic flux index
- L- [# R* V1 o5 M- G1 n( V! uThe flux index (kG2 mm2) is calculated by multiplying the calculated area of the pole surface (mm2) of the magnet by the square of the maximum flux density (kG2).0 `& V2 v1 b. I4 g( b
以上是摘自EN 71-1:2005+A8:2009 (E)的二段,看到许多网友问到magnetic flux index
3 }0 z! }" G' J# A3 a4 M: d(磁通指数)和单位(kGs)2*mm2的意义是什么?+ `- W: n( L* X/ E4 N# ]! d' @
我不做玩具检测,也不做玩具,查了一下以前学的电磁学方面书籍,试说一下自己的看法:磁通指数从量纲上看,相当于(磁通B)与(磁通密度Φ=B×S)的乘积,从磁学方面分析推算,它应当是一个表征磁石吸力的量:F=(SB2)/(2μ0),即磁石吸力(F)的大小正比于磁通指数)F∝(SB2)。下面仅从单位换算说明一下(kG2 mm2)的意义:% E0 c6 {) u% Z6 U' h
3 K8 d; r4 h( z1【特斯拉】=1【韦伯】【米】-2=(1【牛顿】【米】)/(1【安培】【米】2)2 `+ ]2 N( V# p7 o
=(105【达因】×102【厘米】)/(10-1【电磁系电流单位】×104【厘米】2)
! @3 R- _; s: V8 p% b7 R: Z4 O9 Y=104【高斯】 ——————(1)
& v* V1 `! C7 }# _4 O. B* G! g ?由电磁系电流单位定义,得:
5 v- y/ }2 c A$ O, D$ A- Z1【电磁系电流单位】={(【达因】【厘米】)/【厘米】}1/2=【达因】1/2
1 T) B7 {. n) \$ W ——————(2)
- |. l6 c) q5 D' t6 K(2)带入(1):9 i' |6 Z3 t$ l9 ?
1 【千高斯】=103【高斯】
" j- }* @6 i3 o* P2 h$ c; C=(104【达因】×102【厘米】)/(10-1【达因】1/2×104【厘米】2)1 ]" U- W5 D; K3 j( g) w5 i
=(104【达因】×103【毫米】)/(10-1【达因】1/2×106【毫米】2)
$ U3 a8 @& r+ k1 s+ P+ X: N=104【达因】/(【达因】1/2×102【毫米】) ——————(3)
4 Y% }7 g% ^. n. f则:
9 N" Z9 m9 n8 ^" ^% Y1 (kG2 mm2)= 1 【千高斯】×1 【千高斯】×1【毫米】2- f0 s) }1 T0 e. [6 @+ { l: B2 C
=104【达因】=10-1【牛顿】
5 ^! `- `' C0 t6 c: b
/ o- M2 x7 {( k# Q" E8 `( K. J以上推导和看法如有错误,请批评指正。
8 B u- r; ~7 M
6 Z; w- Q; B1 L3 D% |1 j5 ~5 k注:以上推导中数字的第3位和单位后的数字应当是指数上标,这里显示不出来,请注意,谅解。 |
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