The strength of Earth's magnetic field, as measured on the surface, is approximately \(6.0 \times 10^{-5} \mathrm{T}\) at the poles and $3.0 \times 10^{-5} \mathrm{T}$ at the equator. Suppose an alien from outer space were at the North Pole with a single loop of wire of the same circumference as his space helmet. The diameter of his helmet is \(20.0 \mathrm{cm} .\) The space invader wishes to cancel Earth's magnetic field at his location. (a) What is the current required to produce a magnetic ficld (due to the current alone) at the center of his loop of the same size as that of Earth's field at the North Pole? (b) In what direction does the current circulate in the loop, CW or CCW, as viewed from above, if it is to cancel Earth's field?

A single loop of wire has a diameter of 20.0 cm. The Earth's magnetic field at the North Pole is \(6.0 \times 10^{-5}\,\mathrm{T}\)

First, we need to find the radius of the loop. Since the diameter is 20.0 cm, the radius \(r\) is: \(r = \frac{d}{2} = \frac{20.0 \, \mathrm{cm}}{2} = 10.0 \, \mathrm{cm} = 0.1 \, \mathrm{m}\) Now, we can find the circumference \(C\): \(C = 2\pi r = 2\pi (0.1\,\mathrm{m})= 0.2\pi \, \mathrm{m}\)

We will use Ampere's Law to calculate the magnetic field produced by the loop at its center. \(B_\text{loop}=\frac{\mu_0 I N}{2r}\) Since we only have a single loop, \(N = 1\). We have the radius and Earth's magnetic field but we need to calculate the current \(I\). We want the magnetic field produced by the loop to be equal to Earth's magnetic field at the North Pole: \(B_\text{loop} = B_\text{Earth} = 6.0 \times 10^{-5}\, \mathrm{T}\) Now, we can rearrange the equation and solve for \(I\): \(I = \frac{B_\text{Earth} \cdot 2r}{\mu_0 N} \)

Now, we can use the given values to calculate the current needed. Since we want to cancel the Earth's magnetic field, the magnetic field produced by the loop should be equal to Earth's magnetic field. The permeability of free space is \(\mu_0 = 4\pi\times10^{-7}\,\mathrm{T\,m/A}\) \(I = \frac{(6.0 \times 10^{-5}\, \mathrm{T}) \cdot (2\cdot 0.1 \, \mathrm{m})}{4\pi\times10^{-7}\,\mathrm{T\,m/A}}\) \(I ≈ 9.55 \, \mathrm{A}\) The space invader will need a current of approximately 9.55 A to produce a magnetic field equal to Earth's magnetic field at the North Pole.

Now, we need to find the direction of the current in the loop to cancel the Earth's magnetic field at the North Pole. According to the right-hand rule, if the thumb points in the direction of the magnetic field, the fingers will curl in the direction of the current. At the north pole, the Earth's magnetic field is vertically downward. To produce an opposite magnetic field, the invader must induce a current that produces an upward magnetic field. If we follow the right-hand rule, the circulation of the current in the loop will be counterclockwise (CCW) when viewed from above. In conclusion, a) The current required to cancel Earth's magnetic field at the alien's location is approximately 9.55 A. b) The current should circulate in the counterclockwise (CCW) direction when viewed from above to cancel Earth's magnetic field.