A infinite solenoid with 3 x 107/m turns passes through the x-y plane. In the x-

Question

A infinite solenoid with 3 x 107/m turns passes through the x-y plane. In the x-y plane, the current is traversing clockwise or counterclockwise (must be determined). The initial amount of that current 458 amps. A square loop with sides of 0.43 meters, wound 3 times is placed inside the solenoid in the x-y plane. The current of the solenoid is then decreased and after 4.7 seconds, there is an induced current of 61 amps running clockwise around the square loop. If the resistance of the square loop is 12 ohms, what is the final amount, in amps, and direction of the current in the solenoid? Indicate direction by including a negative sign for a clockwise flow of current, and do not put a sign for counter-clockwise flow.

Explanation / Answer

The direction should be the same as the square loop, clockwise, because the current decreases in the solenoid, so does the magnetic flux in the square loop, so the induced current in the square loop will opposite the change, thus has has direction as that in the solenoid.

the voltage = 61*12 = 732 V,

B = u0nI,

flux phi = BA = u0nI *A, V = dphi/dt = u0nAdI/dt

u0 n*A * (458-Ie)/4.7 = 732,

458-Ie = 732*4.7/(u0nA) Ie = 458-732*4.7/(u0nA) = 458-493.56 =-35.56 Amps

"-" indicates that it is counterclockwise

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