Question

A 20 cm times 20 cm square loop of wire lies in the xy-plane with its bottom edge on the x-axis. The resistance of the loop is 0.50 ohm. A magnetic field parallel to the z-axis is given by B = 0.80y^2 t, where B is in tesla, y in meters, and t in seconds. What is the size of the induced current in the loop at t = 0.50 s?

Answer 1

Let

L = 2 cm = 0.02m

a = 1 cm = 0.01 m

b = 1 + 2 = 3 cm = 0.03 m

When I is the current through the wire,

magnetic flux through the loop = (mue*I*L/(2*pi))*ln(b/a)

induced emf in the loop = change in megnetic flux/time

= (mue*L/(2*pi))*ln(b/a)*(dI/dt)

= (4*pi*10^-7*0.02/(2*pi))*ln(0.03/0.01)*100

= 4.39*10^-7 volts

induced current in thr loop, I = induced emf/R

= 4.39*10^-7/0.01

= 4.39*10^-5 A