Question 4 of 7: Magnetic Field and Force Near a Finite-Width Current Sheet

Question 4: Magnetic Field and Force Near a Finite-Width Current Sheet (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) $B$ at $P_1$, $P_2$ with $I_2=0$. (b) how $B$ changes from the origin to $P_2$. (c) force per unit length on the $I_2=1$A wire at $P_2$. (d) $I_2$ needed to zero $B$ at $P_1$.

Approach. Model the finite sheet as a continuum of infinite line currents $dI=K\,dx'$ and integrate the infinite-wire Biot–Savart result across the sheet's width; all elements give the same field direction at a point outside the strip, so the integral reduces to a single logarithm.

1. Set up the sheet integral. For a field point at $x_0$ to the left of the strip ($x_0 \lt x_1 \lt x_2$), every element contributes in the $+z$ direction (right-hand rule), and $$B(x_0)=\int_{x_1}^{x_2}\frac{\mu_0K\,dx'}{2\pi(x'-x_0)}=\frac{\mu_0K}{2\pi}\ln\!\left(\frac{x_2-x_0}{x_1-x_0}\right)$$
2. Part (a) — evaluate at $P_1$ ($x_0=-1$) and $P_2$ ($x_0=-2$). $$B(P_1)=\frac{\mu_0(1000)}{2\pi}\ln\!\left(\frac{2-(-1)}{1-(-1)}\right)=\frac{\mu_0(1000)}{2\pi}\ln(1.5)$$ $$\boxed{B(P_1)=8.109\times10^{-5}\ \text{T, }+z}$$ $$B(P_2)=\frac{\mu_0(1000)}{2\pi}\ln\!\left(\frac{2-(-2)}{1-(-2)}\right)=\frac{\mu_0(1000)}{2\pi}\ln(1.333)$$ $$\boxed{B(P_2)=5.754\times10^{-5}\ \text{T, }+z}$$
3. Part (b) — from the origin to $P_2$. At the origin ($x_0=0$): $$B(0)=\frac{\mu_0(1000)}{2\pi}\ln(2)=1.386\times10^{-4}\ \text{T}$$ So the field decreases monotonically from $\boxed{1.386\times10^{-4}\ \text{T at the origin to }5.754\times10^{-5}\ \text{T at }P_2}$ as the point moves farther from the sheet.
4. Part (c) — force per unit length on the $I_2=1$A wire at $P_2$. The wire sits in the sheet's own field $B(P_2)=5.754\times10^{-5}$T ($+z$), current $+y$: $$\frac{\vec F}{L}=I_2\hat y\times B\hat z=I_2B\,\hat x$$ $$\boxed{\frac{F}{L}=5.754\times10^{-5}\ \text{N/m, in }+x\text{ (toward the sheet, attractive)}}$$ consistent with parallel currents in the same direction attracting.
5. Part (d) — $I_2$ to zero $B$ at $P_1$. The wire at $P_2$ (distance $d=1$m to the left of $P_1$) must supply $B=-8.109\times10^{-5}$T ($-z$) at $P_1$ to cancel the sheet's field there; a current in $+y$ at $P_2$ gives exactly a $-z$ contribution at points to its right: $$B_{\text{wire}}(P_1)=\frac{\mu_0I_2}{2\pi d}=B(P_1)\ \Rightarrow\ I_2=\frac{2\pi\,d\,B(P_1)}{\mu_0}$$ $$\boxed{I_2=405.5\ \text{A, in the }+y\text{ direction}}$$