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An infinite straight wire carries a current I that varies with time as shown abo

ID: 2117158 • 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 = 3.4 A at t = t1 = 14 s, remains constant at this value until t = t2 when it decreases linearly to a value I4 = -3.4 A at t = t4 = 29 s, passing through zero at t = t3 = 24 s. A conducting loop with sides W = 30 cm and L = 58 cm is fixed in the x-y plane at a distance d = 46 cm from the wire as shown.

1) What is the magnitude of the magnetic flux %u03A6 through the loop at time t = t1 = 14 s?



2)

What is e1, the induced emf in the loop at time t = 7 s? Define the emf to be positive if the induced current in the loop is clockwise and negative if the current is counter-clockwise.



3)

What is e2, the induced emf in the loop at time t = 16 s? Define the emf to be positive if the induced current in the loop is clockwise and negative if the current is counter-clockwise.



4) What is e4, the induced emf in the loop at time t = 26.5 s? Define the emf to be positive if the induced current in the loop is clockwise and negative if the current is counter-clockwise.



Explanation / Answer

a)

Magnxetic flux = Magnetic field. dA

= (u0*I/2*pi*x).dA

dA = w dx

Integrate from x= d to x = d+l

(u0*I/2*pi*x)*w dx

At t = 14s I = 3.4A w = 0.34m d = 0.46 l = 0.58

Put these values

Flux = 3.08*(10^ -8) * I

put I = 3.4A

and we get the flux equals = 9.08*(10^ -7)


(b)Induced emf = -d(flux)/dt

= -d(5.08*(10^ -8) * I)/dt

= -5.08*(10^ -8) * dI/dt


At t = 5.5 s dI/dt = 5.8/11 = 0.527 from the graph


Induced emf = -5.08*(10^ -8) * 0.527 = -2.678*(10^ -8) V

As the current is counter-clockwise emf is negative



(c)Induced emf = -d(flux)/dt

= -d(5.08*(10^ -8) * I)/dt

= -5.08*(10^ -8) * dI/dt   


At t = 13 s dI/dt =0 from the graph

Induced emf = -5.08*(10^ -8) * 0 = 0 V


(d)Induced emf = -d(flux)/dt

= -d(5.08*(10^ -8) * I)/dt

= -5.08*(10^ -8) * dI/dt


At t = 20.5 s dI/dt = -5.8/4.5 = -1.29 from the graph


Induced emf = -5.08*(10^ -8) * -1.29 = 6.547*(10^ -8) V

As the current is clockwise emf is positive