The figure shows a side view and overhead view of two concentric copper loops. T

Question

The figure shows a side view and overhead view of two concentric copper loops. The outer bop has a radius of 50.0 cm, a resistance of 0.70 mOhm, and is connected by ideal wires to a switch and a 12 V battery. The inner bop has a radius of 1.5 cm, a resistance of 0.20 mOhm, and is not connected to a battery. When the switch is cbsed, it takes 3.0 ms for the current in the outer bop to reach its final value. What is the average current in the inner bop in the lime between when the switch is cbsed and when the outer bop's current reaches its final value? (Because the inner bop is much smaller than the outer bop, assume that any magnetic fields generated by the outer bop are constant over the inner bop's area.) 8.1 A, countercbckwise 8.1 A, clockwise 5.1 mA, cbckwise 25 A, cbckwise 25 A, countercbckwise

Explanation / Answer

Current will flow clockwise in outer loop
It will generate a magnetic field into the page into the inner loop
So, induced current in inner loop will be such that it generates magnetic field pointing out of the page
So, current in inner loop should be counter clockwise

Let us 1st calculet induced emf in inner loop
E = A* dB/dt
B = miuo*I/2*r1

dB/dt = miuo/2Pi*r1 * dI/dt
= (1.2566*10^-6) / (2*0.5) * 5.71*10^6
{di/dt = imax/t = V/R*t = 12 / (0.7*10^-3*3*10^-3) = 5.71*10^6 A/s}
dB/dt = 7.18 T/s

E = A* dB/dt
= pi*r2^2 * dB/dt
= pi*(0.015)^2 * 7.18
= 5.08*10^-3 V

I = E/R
= (5.08*10^-3) /(0.2*10^-3)
= 25 A

Answer: E

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