Question

Electricity and Magnetism: Will rate high for correct solutions!

The figure below is a top view of a rectangular thin wire loop that moves with velocity vo on a horizontal table. The friction against the table is negligible so that this motion would continue practically with the same velocity if not for crossing the boundary of the region of the uniform magnetic field B=2120 G, also shown in the figure. From our discussions in class, you know that this crossing is accompanied by the slowing down of the loop motion facilitated by the phenomenon of induction. In fact, while crossing the boundary, the loop's velocity v(t) would be changing with time t according to the equation dv dt just as if the loop were experiencing the friction force with some effective parameter t. The geometrical dimensions of the loop are given: a=10.2 cm, b=17.5 cm, as well as its total resistance R=0.3 12 and mass m=5.2 g. T b ХХХХХХХХХХХХХХХХ XXXXXXXXXX ВХХХХХ ХХХХХХХХХХХХХХХХ ХХХХХХХХ ХХХХХХХХ ХХХХХХХХ ХХХХХХХХ X X X X X X X a R ХХХХ ХХХХХ Х XXX XX Х XX VO Understandably, if the velocity vo is "too low", the loop would stop before completely crossing the boundary of the magnetic field region. Find the minimum value Vmin of the velocity vo, at which the complete crossing is still possible: Vmin= cm/s. What would be the maximum magnitude Imax of the current induced in the loop corresponding to this limiting velocity?: Imax= mA.

Answer 1

solutions B: 2 1 2061 0 212 T As enteres it in going to experience a Jone Ray x28 which in F-I l B -maz B²l² R du dt B212 - V MR - B²l²v V du do du du de do z du at mR - B²1² do dva mR B² l ²x vo tc MR at V = Vnin X=0 -B² e ²x MR + Vnin @ at x=b V=0 Darin V mina B²12b (0-2123 (0.102) ČO:128) lza 5-2 X 10-3 x 0:3 mR 0.052 US min V min 5.25cmia

Imax B l Vimin = 0.212 x 0.102 * 5.25X10-2 li R 0.3 11 3.78X10-3 A Imax 3.78 m A

1) 5.25 cm/s

2) 3.78 mA