A 12.5 cm x 12.5 cm square loop of wire having 50 turns and with a resistance of

Question

A 12.5 cm x 12.5 cm square loop of wire having 50 turns and with a resistance of 3.5*10^-2 ohms has one edge parallel to a long straight wire. The near edge of the loop is 2.5 cm from the wire. The current in the wire is increasing at the rate of 2.5 A/s.

A. What is the direction of induced current in the loop?

B. How much bigger is the magnetic field at point 2.5 cm than 15 cm from the wire?

C Solve for the magnetic flux symbolically (as a step to get the current)

D What is the current in the loop?

Explanation / Answer

A) in opposite direction as of wire (In the wire that is closest to wire).

B) magnetic field due to current carrying wire, B = u0 I / 2 pi d

Delta B = (4pi x 10^-7 x I ) / (2pi ) [ (1/0.025) - (1/0.15)]

= 6.67 I x 10^-7 T

Ratio= [(u0 I ) / (2pi 0.025)] / [ (u0 I ) / (2 pi 0.15)]

     = 6 times

C) Magnetic flux at position x = N B A

     d (flux) = 50 (( u0 I ) /( 2 pi x ) ) (0.125 dx )

d (flux) = 1.25I x 10^-6 ( dx/x)

integrating,

flux = 1.25I x 10^-6 ( ln x)

x is from 2,5 cm to 15 cm

flux = 1.25I x 10^-6 x (ln(15/2.5)) = 2.24I x 10^-6

d) induced emf = rate of change of flux = d(flux) / dt

e = 2.24 x 10^-6 dI / dt = 2.24 x 10^-6 x 2.5 =5.6 x 10^-6 V

I = e / R = 1.60 x 10^-4 A

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