Discussion
A long straight conductor carries a current of 5 A. The magnitude of the magnetic field at a point 20 cm from the conductor is:
Answer: B
The field of a long straight current-carrying wire is \(B = \mu_0 I / (2\pi r)\). Substituting \(\mu_0 = 4\pi \times 10^{-7}\) T m/A, I = 5 A and r = 0.20 m gives:
\[ \begin{aligned} B = \frac{2 \times 10^{-7} \times 5}{0.2} \\ &= 5 \times 10^{-6}\ \text{T} \end{aligned} \]This is 5 microtesla. The values 0.5, 50 and 500 microtesla are wrong by factors of 10, 10 and 100 and would result from slips in the power of ten or from using the wrong distance, for example 2 cm instead of 20 cm.
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