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for sample answer but values change I want 1st problem.. The figure below is a top...

The figure below is a top view of a rectangular thin wire loop that moves with velocity vo on a horizontal table. The frictio

The figure below is a top view of a rectangular thin wire loop that moves with velocity vo on a horizontal table. The frictiofor sample answer but values change

solutions BE 21206 0.212T As it enters it in which in F: IIB xl e B) going to experience a Jone Roy B2lv ma 2 R du dt B²12 MRImax 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

I want 1st problem..

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=3074 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 - у dt т" 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 cm, b=29.5 cm, as well as its total resistance R=0.428 I and mass m=5 g. b ХХХХХХХХХХ ХХХХ ХХХХ Ххх в ХХХХ ХХХХХХХХХХХХХХХХ ХХХХХХХХХХХХХХХХ xxxx ХХХХХ ХХХХ xxx XXX ХХХХ ХХХХХ а R M 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 V min 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.
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 V 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. 늘 b xxxxxxxxx XXX ХХХХХХХХХХ В XX ХХХХХХХХХХХХХХХХ ХХХХХХХХХХХХХХХХ ХХХХХХХХ XXX XX ХХХХХХ XXX ХХХХХХХХХ XXX a R 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.
solutions BE 21206 0.212T As it enters it in which in F: IIB xl e B) going to experience a Jone Roy B2lv ma 2 R du dt B²12 MR V v du atx 82v MR re { che la de la du B²1² do duz mR Va -B?l?x MR tc at VE Vnin x=0 • B² 1² MR + Vnin at x=b V=0 Pasin V min - B²1²b + (0-212] (0.102) 70-178 1za del comune mR 5.2 x 103 x 0.3 0.052 US mily 5.25cmia Umin
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
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Please refer to the below images for detailed explanation:

= 0.0001 T solution : - Given B= 3074 G ;{IG= 3) B- 0.3074 T - 29.5cm maso こ 5gm T locm dr. = - 1 dt L R=0.428 vo =) Force exby Imax Il BI Vmin R on putting Imax Putting values :- 0.3074 * 0.10 to 0.1302 O. 4 28 Amps 2) Imax = 9.351 * 10 3 Amps o V m

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