The diagram below depicts a wire carrying a current I = 4.75 A. The wire splits
ID: 1373197 • Letter: T
Question
The diagram below depicts a wire carrying a current I = 4.75 A. The wire splits into two channels; of resistance R2 = 6.30 ? and R1 = 4.25 ?, and re-joins, forming a current loop in the shape of an isosceles triangle with base distance d = 5.80 cm and height L = 20.5 cm. The loop is entered into the space between the two poles of a magnet with a uniform magnetic field, B = 1.95
The diagram below depicts a wire carrying a current I = 4.75 A. The wire splits into two channels; of resistance R2 = 6.30 ? and R1 = 4.25 ?, and re-joins, forming a current loop in the shape of an isosceles triangle with base distance d = 5.80 cm and height L = 20.5 cm. The loop is entered into the space between the two poles of a magnet with a uniform magnetic field, B = 1.95 A?10?2 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?Explanation / Answer
torque =I ABsintheta where A = area of triangle
i2= It(total current)*R1/(R1+R2)= 2.83A
And other other current is I1=It-I2= 4.75-2.83= 1.91A
using F=I*l*B*sin(theta)
for finding lenth of one leg, using pythagoras theorem 2.9^2 +5.8^2= 428.66cm= 4.28m
theta = arc cos (0.2/4.2) = 87.7degree
l*sin(theta) would give you an arm perpendicular to the B field = 4.48 * sin(87.7) = 4.43
F1 =I1*l*sin(theta)*B= 8.17N
f2=I2*l*sin(theta)*B =10.4N
torque=I1*l*sin(theta)*B*d/3-I1*l*sin(theta)*B=|I1-I2|*l*B*sin(theta)*d/3