2) 3) 4) 5) An infinite straight wire carries a current I that varies with time
ID: 2253521 • Letter: 2
Question
2)
3)
4)
5)
An infinite straight wire carries a current I that varies with time as shown above. It increases from 0 at t = 0 to a maximum value I1 = 3.8 A at t = t1 = 13 s, remains constant at this value until t = t2 when it decreases linearly to a value I4 = -3.8 A at t = t4 = 23 s, passing through zero at t = t3 = 20.5 s. A conducting loop with sides W = 23 cm and L 68 cm is fixed in the x-y plane at a distance d = 55 cm from the wire as shown. What is the magnitude of the magnetic flux phi through the loop at time t = t1 = 13 s?Explanation / Answer
dFi(r) = B(r)*dS
B(r) = (mu*I) / (2*pi*r)
dS =W*dr
Fi =mu*I *W/ (2*pi) integral (r from d to d+L) 1/r*dr = mu*I *W / (2*pi) ln(r) (from d to d+L)
Fi = mu*I *W /(2*pi)*ln*((d+L)/d) =4*pi*10^-7*3.8*0.23/(2*pi)*ln((0.55+0.68)/0.55) = 1.407*10^-7 Wb
Fi = I*3.7*10^-8 Wb if we separate the current I
b) E1 = -dFi/dt = -3.7*10^-8* dI/dt
dI/dt = Delta(I)/Delta(T) =3.8*13 =0.2923 A/s
E1 = -1.08*10^-8 V
(the current in the loop is rotating counterclockwise - see explnation at point d)
c) dI/dt for decreasing current is
dI/dt =-3.8/(t4-t3) =-3.8/(23-20.5) =-3.8/2.5 =-1.52 A/s
I1 =3.8 A is max it means t2-t3 = I1/(dI/dt) =-3.8/1.52 =-2.5 sec
t2 = t3-2.5 =20.5*2.5 =18 sec
It means at t=15 sec the current is constant I = I1 =3.8 A it means dFi/dt =0 it means
E2 =-dFi/dt =0 V
d) at 20.5 sec the current is decreasing means the flux through the loop is decreasing. The induced current in loop will be such a way to compensate for the decreasing flux.
B of current I is into the page and decreasing it means B of the induced current is into the page (to compensate decreasing flux), it means the current in the loop is rotating clockwise.
e) E4 = -dFi/dt = -3.7*10^-8*dI/dt = + 3.7*10^-8*1.52 =+5.624*10^-8 V
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