64 A\)
\(R_1 = 6.60 cm = 0.066 m\)
\(R_2 = 3.33 cm = 0.033 m\)
Substitute the values:
\(I_1 = 8.64 \times \frac{0.066}{0.033}\)
\(I_1 = 8.64 \times 2\)
\(I_1 = 17.28 A\)
Therefore, the current in the smaller loop is 17.28 A in the counterclockwise direction.
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