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as shown in the figure, a current I passes through a straight conducting wire located...

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handsofgod | (Level 2) eNoter

Posted May 30, 2013 at 6:45 PM via web

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as shown in the figure, a current I passes through a straight conducting wire located within this page.Rectangular circuit ABCD on the page is moving away from the straight conducting wire with speed v, while side AB remains parallel with current I. The length of side AB is 2a,and the length of side AD is 2b. The resistace of the circuit is R

The magnetic permeability of vacuum is μ0, and the magnetic flux density created by circuit's current is negligible. How much current flows through circuit ABCD when the centre of the circuit (O) is distance r from the straight wire?

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted September 13, 2013 at 10:30 AM (Answer #1)

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See the figure below.

The magnetic field B generated by the current I is into the page.

`B(r) = miu*I/(2*pi*r)`   

where `miu =4*pi*10^-7 H/m` is the magnetic permeability of free space.

The induced voltage into the loop ABCD is (see Faraday induction law)

`U_("ind") = -(dPhi)/dt = -(d(B*S))/dt = - (S*dB + B*dS)/dt`

Now by differentiating we have

`dB = -miu*I/(2*pi*r^2)*dr`

`dS = 2a*dr`  (see the attached figure)

Therefore

`U_("ind") = + 4ab*miu*I/(2*pi*r^2)* (dr)/dt -2*a*miu*I/(2*pi*r)*(dr)/dt`

`U_("ind") = miu*I/(pi*r)*((2ab)/r - a)*v`

Since Ohm law states that

`I = U/R`

the induced current in the loop ABCD is

`I_("ind")(r) = miu*I/(pi*r*R)*((2ab)/r -a)*v`

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