A long vertical wire carries a steady 9 A current. A pair of rails are horizonta
ID: 1367943 • Letter: A
Question
A long vertical wire carries a steady 9 A current. A pair of rails are horizontal and are 0.2 m apart. A 13.8 ? resistor connects points a and b, at the end of the rails. A bar is in contact with the rails, and is moved by an external force with a constant velocity of 0.7 m/s as shown. The bar and the rails have negligible resistance. At a given instant t1, the bar is 0.13 m from the wire, as shown. In Figure, at time t1, the induced current in µA is:
Express the answer with three decimal places.
2.00 m I, A V, m/s R, L,mExplanation / Answer
first find the magnetic field as B = uoi/2pi R
where uo is 4pi e-7
i is current
r is ths distance at which magnetic field is produced
so
B = (4*3.14 e-7 * 9) /(2*3.14 * 0.13)
B = 13.84 uT
so now use the formula for Induced emf e = L v B
where L is length
v is speed
B is magnetic field
so emf e = 0.2* 0.7 * 13.84 e -6
emf e = 1.937 e -6 V
induced current = emf/R
R is Resisatnce
I = 1.937 e -6/13.8
I = 1.404 e -7 Amps or 0.14 uA
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