1) At time t = t 1 = 0.032 s, what is I 1 , the induced current in the loop? I 1
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Question
1) At time t = t1 = 0.032 s, what is I1, the induced current in the loop? I1 is defined to be positive if it is in the counterclockwise direction. 2) 2) At time t = t2 = 0.61 s, what is I2, the induced current in the loop? I2 is defined to be positive if it is in the counterclockwise direction 3) What is Fx(t2), the x-component of the force that must be applied to the loop to maintain its constant velocity v = 40 cm/s at t = t2 = 0.61 s? 4) At time t = t3 = 0.497 s, what is I3, the induced current in the loop? I3 is defined to be positive if it is in the counterclockwise direction. Thank you will rate Lifesaver!Explanation / Answer
SOL: total lengths of sides of the conducting loop=s1+s2=3.8=6cm=9.8cm length of the regionL=18.6cm total moving length =L+s1+s2=28.4cm if the conducting loop moves with a 40m/s velocity, then, the entire loop will pass over the magnetic region in time=totaldistance/velocity =28.4/40 =0.71s 1) at atime t=0.035s , distance of the loop will enter the field, distance=velocity.time =40.(0.035) =1.4cm this distance we can take as entering length of the loop. l=1.4cm, where breadth not changes , it is b=3.8cm ( AS square loop given) the area associated by this loop at given time(dA/dt) =l.b =1.4(3.8)cm^2 =5.32cm^2 =5.32*10^-4 m^2 we know that, induced e.m.f in the loop of this area, e=-B.dA/dt =1.2*5.32*10^-4 v (B=1.2 T given) =6.384 *10^-4 v the induced current at this time i=e/R =6.384*10^-4/2.8 (R=2.8 ohm given) = 2.28*10^-4 A =0.228 mA 2) at t=0.61 s the distnce moved by the loop, distance=velocity.time =40.(0.61) cm = 24.4 cm but, the total distasnce =28.4cm loop travelled distance=24.4cm the remaining distance of the loop= 4cm now, this can be taken as length of the loop at this time so, l=4cm breadth remains same b=6 cm area associated with this loop at htis given time(dA/dt)=l.b =4.6 =24cm^2 or 24*10^-4 m^2 the induced e.m.f at this time at this loop will be, e=-B.dA/dt =1.2(24*10^-4) v =28.8 *10^-4 v the induced current in this loop at this time is given by i=e/R i=28.8*10^-4/2.8 i= 1.02 mA 3) the force must be applied to mantain the loop's constant velocity F=ilB =(1.02 *10^-3)(6)(1.2) =7.344*10^-3 N 4) at t=0.497 s , the the distance travelled by the loop dis tnace= velocity.time =40(0.497) =19.88 cm the total distance =28.4 cm travelled distance=19.88 cm remaining distance=8.52 cm this should be taken as length of the loop enteraining l=8.52 cm breadth remains same b=6 cm the area associated with this loop at this given time(dA/dt) =l.b =6(8.52) =51.12 cm^2 = 51.12*10^-4 m^2 the induced e.m.f in the loop at this time is given by, e=-B.dA/dt =1.2(51.12*10^-4) v = 61.344 *10^-4 v the induced current in this loop at this time is given by, i=e/R =61.344*10^-4/2.8 =2.19 mA
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