Answer 13.6

_psets/04.md

@@ -199,11 +199,11 @@ equal to $$\num{2e-7}$$ newton per metre of length."
 Show that that current at that distance produces that force.
 
 First let's find the magnetic field of an infinitely long straight conductor. Let's use a
-cylindrical coordinate system along this axis. Considering the Biot-Savart Law and the symmetry of
-this problem, the magnetic field must be oriented along $$\hat{\mathrm{d} \theta}$$, with a
-magnitude that depends only on $$r$$. Consider then a circle of radius $$r$$ centered on the wire.
-Ampère's Law tells us that the magnitude of the field at any point on this circle is $$I / (2
-\pi r)$$.
+cylindrical coordinate system along this axis. Considering the [Biot-Savart
+Law](https://en.wikipedia.org/wiki/Biot%E2%80%93Savart_law) and the symmetry of this problem, the
+magnetic field must be oriented along $$\hat{\mathrm{d} \theta}$$, with a magnitude that depends
+only on $$r$$. Consider then a circle of radius $$r$$ centered on the wire. Ampère's Law
+tells us that the magnitude of the field at any point on this circle is $$I / (2 \pi r)$$.
 
 The differential force exerted by this field on a differential piece of current is $$dF = I (dl
 \times B)$$. In this case the direction of the current and the magnetic field are perpendicular, so

_psets/10.md

@@ -136,5 +136,18 @@ coercivity of iron is $$\num{4e3} \si{A/m}$$.
 ## (13.6)
 
 {:.question}
-Approximately what current would be required in a straight wire to be able to erase a $$\gamma \text{-} Fe_2 O_3$$
-recording at a distance of 1 cm?
+Approximately what current would be required in a straight wire to be able to erase a $$\gamma
+\text{-} Fe_2 O_3$$ recording at a distance of 1 cm?
+
+As found in problem 6.4 in [problem set 4](/psets/04.html), the magnitude of the magnetic field a
+distance $$r$$ away from an infinitely long and thin conductor carrying a current $$I$$ is $$I/(2
+\pi r)$$. To erase information stored on $$Fe_2 O_3$$ we need this field to be about as strong as
+the coercivity $$H_C = 300 \si{Oe}$$. Thus the current needed is
+
+$$
+\begin{align*}
+I &= 2 \pi r H_C \\
+&= 2 \pi \cdot 10^{-2} \si{m} \cdot 300 \si{Oe} \cdot \frac{79.6 \si{A/m}}{1 \si{Oe}} \\
+&= 1500 \si{A}
+\end{align*}
+$$