Given:
\(I_1 = 4.50 \, A\)
\(I_2 = 8.00 \, A\)
\(d = 14.0 \, cm = 14 \times 10^{-2} \, m\)
Using the formula \(B_{21} = \frac{\mu_0}{2\pi} \frac{I_1}{d}\), where \(\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A\):
\[B_{21} = \frac{4\pi \times 10^{-7}}{2\pi}
Show more…