1 \begin{problem*}{31.45}
2 A circular coil enclosing an area $A=0.0100\U{m$^2$}$ is made of $200$
3 turns of copper wire as shown in Figure~P31.45. Initially, a uniform
4 magnetic field of magnitude $B=1.10\U{T}$ points upward in a direction
5 perpendicular to the plane of the coil. The direction of the field
6 then reverses in a time interval $\Delta t$. Determine how much
7 charge enters one end of the resistor during this time interval if
11 % | | | (from back of coil)
16 % | | | (from font of coil)
24 int n = 10; // number of coil loops
25 real tense = 8; // increasing tension flattens the coil loops
27 MultiTerminal R = resistor(dir=90, Label("$R$", align=E));
28 real rlen = R.terminal[1].y - R.terminal[0].y;
29 real xr = R.center.x - dx;
32 for (int i = 0; i < n; i += 1) { // back sides of coil
37 real y0 = R.terminal[0].y + i * dy;
38 draw((xl, y0 + dy/2){N}..tension tense ..{dout}(xr, y0 + dy));
40 Vector B = BField(dir=90);
41 B.outline += linewidth(0.5mm);
42 vector_field(((xl+xr)/2, R.center.y+rlen/3), width=2*r, height=2*rlen, v=B);
43 for (int i = 0; i < n; i += 1) { // front sides of coil
48 real y0 = R.terminal[0].y + i * dy;
49 draw((xr, y0){din}..tension tense ..{N}(xl, y0 + dy/2));
51 draw((xr, R.terminal[0].y) -- R.terminal[0]);
52 draw((xr, R.terminal[1].y) -- R.terminal[1]);
58 Picking up as the positive flux direction, the induced \EMF\ is
60 \EMF = -\deriv{t}{\Phi_B} = -\deriv{t}{NAB} = -NA\deriv{t}{B}
61 = -NA\frac{-2B}{\Delta t} = \frac{2NAB}{\Delta t} \;.
63 This \EMF\ drives a current through the resistor
66 I &= \frac{\EMF}{R} = \frac{-2NAB}{R\Delta t} \;.
68 Current is defined as the charge passing through a cross section of
69 your circuit per unit time, so the charge entering one end of the
72 \Delta q = \frac{\Delta q}{\Delta t} \Delta t = I \Delta t
73 = \frac{2NAB}{R} = \ans{0.880\U{C}} \;.