Question

Currents in Field The diagram below depicts a wire carrying a current 1 = 1.60 A. The wire splits into two channels, of resistance R2 = 7.75 ? and R1 = 4.75 ?, and re-joins, forming a current loop in the shape of an isosceles triangle with base distance d 9.00 cm and height L-22.0 cm. The loop is entered into the space between the two poles of a magnet with a uniform magnetic field, B-8.60 x 10? T, that runs from one pole to the other. The loop is placed such that the field lies in the plane of the loop. What is the torque on the circuit about the wire's axis? I S Submit Answer Tries 0/99

Answer 1

Current through each wire I_1 = I times [R_2/R_1 + R_1] = 1.6[7.75/4.75 + 7.75] = 0.992 I_2 = I - I_2 = 1.6 - 0.992 = 0.608A magnetic force on I_1: F_1 = BI_1 squarerootL^2 + (d/2) ^2 times sin theta since, sin theta = L/squarerootL^2 + (d/2) ^2 magnetic force on I_2: F_2 = BI_2L since F_1 and F_2 are out of pase, torque is given by: r = F_1 times d/4 - F_2 times d/4 = BLd/4[I_1 - I_2] = 8.6 times 10^-2 times 22 times 10^-2 times 9 times 10^-2/4[0.992 - 0.608] r = 1.63 times 10^-4 N-m