Question

(a) How large a current would a very long, straight wire has to carry so that the magnetic field 2 cm from the wire is equal to (comparable to the earth’s northward-pointing magnetic field)? (b) If the wire is horizontal with the current running from east to west, at what locations would the magnetic field of the wire point in the same direction as the horizontal component of the earth’s magnetic field? (c) Repeat part (b) except the wire is vertical with the current going upward.

Short answer

a) l = 10A

b) At all points directly above the wire the field of the wire would point northward.

c) At all points directly east of the wire, its field would point northward.

Step-by-step solution

The formula to calculate the current

The magnetic field due to a very long wire is given by$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}\mathbf{\pi r}}$ .

The current can be found by$I=\frac{2\mathrm{\pi rB}}{{\mu }_0}$ .

Calculate the current

Given that r = 2 cm and$B={10}^{-4}T$ .

$\begin{array}{r}I=\frac{2\pi rB}{{\mu }_0}\\ I=\frac{2\pi \times 0.02\times {10}^{-4}}{4\pi \times {10}^{-7}}\\ I=10\mathrm{A}\end{array}$

Find the locations of magnetic field

We know that earth’s horizontal field points northwards. Therefore, at all points directly above the wire the field of the wire would point northward as shown in the figure

Now, the wire is vertical with the current going upward, at all points directly east of the wire, its field would point northward as shown in the figure