3 The wire is now replaced by a conducting rectangular loop

3) The wire is now replaced by a conducting rectangular loop as shown. The loop has length L-43 cm and width W-18 cm. At timet-0 the loop moves with velocity v 15 cm/s with its left end located a distance d 52 cm from the y axis. The resistance of the loop is R - 2.7 , what is i(0), the induced current in the loop at time t-T Define the current to be positive if it flows in the counter clockwise direction. Submit

Solution

3.

Net Magnetic field

B=B_ab+B_ad = (u_oI₁/2pi)[(1/d) - 1/(d+L)]

B=((4pi*10^-7)*2/2pi)[1/0.52 -1/(0.52+0.43)]

B=3.48*10^-7 A

Induced current in the loop at t=0 is

I=BvW/R =(3.48*10^-7)*0.15*0.18/2.7

I=3.48*10^-9 A

---------------------------------------------------------------------------------------------------------------------------------

Net magnetic field

B=u_oI/2pid =(4pi*10^-7)*2/2pi*0.52

B=7.69*10^-7 T

Induced emf

E=BvW =7.69*10^-7 *0.15*0.18

E=2.08*10^-8 Volts