At what distance a long, straight wire carrying a current of 3 A can develop a
magnetic field of 7x10-7 T
Answer
Let this distance is d.
Current through wire I=3A
Magnetic field B=7×10−7T7\times10^{-7}T7×10−7T
Then distance is given by
d=μ02I4πBd=\frac{\mu_0 2 I}{4\pi B}d=4πBμ02I
Putting all values
d=10−7×3×27×10−7=0.86md=\frac{10^{-7}\times 3\times2}{7\times10^{-7}}\\=0.86md=7×10−710−7×3×2=0.86m
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