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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,m

Explanation / 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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