A single rectangular wire loop with a height of 240 cm and a

A single rectangular wire loop with a height of 24.0 cm and a width of 10.0 cm is mounted flat on an office door. The resistance the loop is 3.85 Ohm. The Earth\'s magnetic field of 2.6 times 10^-5 T is uniform and perpendicular to the surface of the closed door. The door is then opened at a constant rate of 4.6 rad/s. Calculate the magnitude of the induced current 0.434 s after the door begins opening.

Solution

area of the rectangular loop is,

   A = height*width = 0.24*0.10 = 0.024 m²

the resitance of the loop is,

R = 3.85 Ohms

Earth magnetic field: B = 2.6x10^-5 T

angular speed: w = 4.6 rad/s

time interval, t = 0.434 s

the induced emf is

   E = NBAw sinwt

      = (1)(2.6x10^-5)(0.024 )(4.6)*sin[4.6*0.434]

      = 2.614e-6 V

Hence, the induced current is,

   I = E/R = 2.614e-6 / 3.85 = 6.79e-7 A = 6.8E-7 A = 0.68 uA