Chapter 28: Problem 25

Suppose that the magnetic field of the Earth were due to a single current moving in a circle of radius \(2.00 \cdot 10^{3} \mathrm{~km}\) through the Earth's molten core. The strength of the Earth's magnetic field on the surface near a magnetic pole is about \(6.00 \cdot 10^{-5} \mathrm{~T}\). About how large a current would be required to produce such a field?

Short Answer

Expert verified

Answer: The required current to produce the Earth's magnetic field is approximately 3 x 10^7 A.

Step by step solution

Ampere's law states that the magnetic field (B) produced by a current (I) in a circular loop of radius (r) is given by \[B = \frac{\mu_0 I}{2\pi r}\] Where: - B is the magnetic field strength - I is the current - r is the radius of the loop - \(\mu_0\) is the permeability of free space, which is \(4\pi \times 10^{-7} \frac{T \cdot m}{A}\) #Step 2: Rearrange the equation to solve for the current (I)#

To determine the current needed to produce the given magnetic field strength, rearrange the equation to solve for I: \[I = \frac{2\pi r B}{\mu_0}\] #Step 3: Substitute the given values into the equation#

We are given the radius of the loop (r) as \(2.00 \cdot 10^{3} km\) and the magnetic field strength (B) as \(6.00 \cdot 10^{-5} T\). Convert the radius from kilometers to meters: \[r = 2.00 \cdot 10^{3} km \times \frac{10^3 m}{1 km} = 2.00 \cdot 10^{6} m\] Now, substitute the given values and the permeability of free space into the equation: \[I = \frac{2\pi (2.00 \cdot 10^{6} m)(6.00 \cdot 10^{-5} T)}{4\pi \times 10^{-7} \frac{T \cdot m}{A}}\] #Step 4: Calculate the current (I) required to produce the given magnetic field strength#

Calculate the current needed to produce Earth's magnetic field: \[I = \frac{2\pi (2.00 \cdot 10^{6} m)(6.00 \cdot 10^{-5} T)}{4\pi \times 10^{-7} \frac{T \cdot m}{A}} = 3 \times 10^7 A\] Hence, a current of about \(3 \times 10^7 A\) would be required to produce the Earth's magnetic field strength of \(6.00 \times 10^{-5} T\).