An infinite straight wire carries a current I that varies with time as shown abo
ID: 2252540 • Letter: A
Question
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 = 3.0 A at t = t1 = 13.0 s, remains constant at this value until t = t2 when it decreases linearly to a value I4 = -3.0 A at t = t4 = 28.0 s, passing through zero at t = t3 = 23.0 s. A conducting loop with sides W = 28.0 cm and L = 54.0 cm is fixed in the x-y plane at a distance d = 45.0 cm from the wire as shown.
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 = 3.0 A at t = t1 = 13.0 s, remains constant at this value until t = t2 when it decreases linearly to a value I4 = -3.0 A at t = t4 = 28.0 s, passing through zero at t = t3 = 23.0 s. A conducting loop with sides W = 28.0 cm and L = 54.0 cm is fixed in the x-y plane at a distance d = 45.0 cm from the wire as shown. What is the magnitude of the magnetic flux ? through the loop at time t = t1= 13.0 s? What is ?1, the induced emf in the loop at time t = 6.5 s? Define the emf to be positive if the induced current in the loop is clockwise and negative if the current is counter-clockwise. What is ?2, the induced emf in the loop at time t = 15.0 s? Define the emf to be positive if the induced current in the loop is clockwise and negative if the current is counter-clockwise. What is ?4, the induced emf in the loop at time t = 25.5 s? Define the emf to be positive if the induced current in the loop is clockwise and negative if the current is counter-clockwiseExplanation / Answer
a) consider at distance x a small length dx.
B = uI/(2pi*x)
dF = uI/(2pi*x) * Wdx
F = integrating from d to d+L
F = (uIW/2pi) * ln (d+L/d) = 4.415*10^-8 * I {Putting all the values except I}
a) I = 3 A at t1.
So, F = 1.324*10^-7 Weber
b) EMF = -dF/dt = -4.415*10^-8 * dI/dt {dI/dt = slope of line}
At t = 6.5, dI/dt = 3/13
EMF = -1.02*10^-8 volts {curent is counter clockwise}
c) At t=15 s, dI/dt = 0.; Slope = 0.
So, EMF = 0.
d) Slope = -3/5
EMF = -4.415*10^-8 * -3/5 = 2.65*10^-8 volts. {clockwise current}