Question

College Physics-there are two questions below, please show all your works and explain your answers!!!! MT13-CT1: MOVING RECTANGULAR LOOPS IN UNIFORM MAGNETIC FIELDS-CURRENT In each case below, a rectangular loop of wire is moving to the right at 4 m/s into a region with a uniform magnetic field pointing into the page. In Case B, the field is four times as strong as in Case A. The rectangular loops have different widths and heights, but they are made from identical wires. At the instant shown, both loops are halfway into the uniform magnetic field regions. There is an induced current in the left loop of 12 mA. 18 cm 24 cm 8 cm 2 cm 100 4 m/s 4 m/s Case A Case B Is the induced current in the loop in Case A greater than, less than, or equal to the induced current in the loop in Case B at the instant shown? Explain. MT13-CT2: MOVING RECTANGULAR LOOPS IN UNIFORM MAGNETIC FIELDS-CURRENT In each case below, a rectangular loop of wire is moving to the right at the same velocity into a region with a uniform external magnetic field pointing into the page. The magnetic field is the same in both cases. The rectangular loops have different widths but have the same height. The loops are made from copper wires that have the same cross-section. At the instant shown, the leading edge of both loops is the same distance into the uniform magnetic field region. 24 cm +18 cm + 8 cm 8 cm 1 Case A Case B Is the induced current in the loop in Case A greater than, less than or equal to the induced current in the loop in Case B at the instant shown? Explain.

Answer 1

CTA Totel length Total length - length of of of are made wires. صلى of wire in care Az 2x (8+18) 2 52cm wie in Care B2 2x(2+24) = 52cm. wires are some and they from type Therefore, resistances both the loop will be same. Induced voltage Induced E 틈 R R 1 2 € (Ris name) {a Blo → EA Box 0.08% 4. 20.32 Bo EB2 4 Box 0.0244 = 0.32 Bo Induced current in care A and Care B are the same. cament, 12 CT2 Total length wire in cese Az 2/24+8) 2 64 cm Total length of of 2 52cm. - lenghts of wire in care A is wire in Care B2 2(18+8) greater than length of RA > RB Kel RA=0.64k and RB 20.52k will in Care B I a Blo 0.125 Bole R In Box0.08x u 0.64k IB a Box 0.08x к 0.154 Bol 0.52K An Bo & and k are name, IB > IA ( IA <I8) ie, Induced current in care A is less than that of case B.