Calculating Current that Produces a Magnetic Field

Find the current in a long straight wire that would produce a magnetic field twice the strength of the Earth’s at a distance of 5.0 cm from the wire.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
The Earth’s field is about $5 . 0 \times 10^{- 5} T$ , and so here $B$ due to the wire is taken to be $1 . 0 \times 10^{- 4} T$ . The equation $B = \frac{μ_{0} I}{2 πr}$ can be used to find $I$ , since all other quantities are known.
Solution
Solving for $I$ and entering known values gives
$\begin{array}{l} I & = & \frac{2 π rB}{μ_{0}} = \frac{2 π (5.0 \times 10^{- 2} m) (1.0 \times 10^{- 4} T)}{4 π \times 10^{- 7} T \cdot m/A} \\ = & 25 A. \end{array}$
22.25
Discussion
So a moderately large current produces a significant magnetic field at a distance of 5.0 cm from a long straight wire. Note that the answer is stated to only two digits, since the Earth’s field is specified to only two digits in this example.

0 of 3 steps open

How did it go?