(D)-30% A long straight wire carries a current given by I Imax sin(wt). The wire
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Question
(D)-30% A long straight wire carries a current given by I Imax sin(wt). The wire lies in the plane of a rectangular coil of 'N' turns at a distance h' as shown in the Figure. The coil has length 'L' and width W. The quantities Imax, u, h, w and L are constants with values lmax-50A, ?-200.? s-1, N 100, h- w 5 cm and L 20 cm. Determine the time-dependent EMF induced in the coil by the magnetic field created by the current in the straight wire (Use the space below to display your answers: the constant ?.-4. TT-10" T-m/A ls provided; hint use the following expression for the magnetic field of the straight wire at a distance t' from its center: B o(2 IT r) and then use Faraday's law to determine the EMF: the following integral is provided: 1 d. In(x)) tu)Explanation / Answer
magnetic field at a distance d from the wire is given by:
B=mu*I/(2*pi*d)
consider a small area of length dx at a distance of x parallel to the wire where x varies from h and h+w.
magnetic flux linkage=magnetic field at distance x * area of the small segment
=(mu*I/(2*pi*x))*L*dx
total flux=integration of (mu*I/(2*pi*x))*L*dx
=(mu*I*L/(2*pi))*ln(x)
using limits for x=w to x=w+h
total flux=(mu*I*L/(2*pi))*(ln((w+h)/w)
emf induced=time derivative of total flux
=(mu*L/(2*pi))*ln((w+h)/w)*(dI/dt)
I=I_max*sin(w*t)
==>dI/dt=I_max*w*cos(w*t)
then emf induced=(mu*L/(2*pi))*ln((w+h)/w)*I_max*w*cos(w*t)
taking emf induced by all the turns,
total time-dependent EMF=N*(mu*L/(2*pi))*ln((w+h)/w)*I_max*w*cos(w*t)
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