Question

A 1.60 cm × 1.60 cm square loop of wire with resistance 1.10×10^?2 ? has one edge parallel to a long straight wire. The near edge of the loop is 1.00 cm from the wire. The current in the wire is increasing at the rate of 100 A/s .

A. What is the current in the loop?

Answer 1

let x = 1 cm = 0.01 m
L = 1.6 cm = 0.016 m
dI/dt = 100 A/s

magnetic field due to staright wire, B = mue*I/(2*pi*r)

magnetic flux through the loop = integral B*L*dr ( r = x to r = L+x)

= integral (mue*I/(2*pi*r))*L*dr

= (mue*I*L/(2*pi))*ln(r)

= (mue*I*L/(2*pi))*(ln((L+x)) - ln(L))

= (mue*I*L/(2*pi))*ln((L+x)/x)

induced emf in the loop = the rate of change of magnetic flux = (mue*L/(2*pi))*ln((L+x)/x)*(dI/dt)

= (4*pi*10^-7*0.016/(2*pi))*ln( (1.6+1)/1)*100

= 3.06*10^-7 V

induced current = induced emf/R

= 3.06*10^-7/(1.1*10^-2)

= 2.78*10^-5 A or 27.8 micro A <<<<<<---------------Answer