Question 6 of 8: Magnetic Field Above a Semicircular Loop with Straight Leads
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Phys-A3, Electromagnetics — National Exam, December 2017. 3-hour closed-book exam (Casio or Sharp approved calculators only); any FIVE of the eight questions constitute a complete paper and only the first five as they appear in a candidate's answer book are marked, each of equal value. Aids given on the paper: ε0 = 8.85×10-12 F/m, μ0 = 4π×10-7 H/m. All eight printed questions are solved below as a complete study resource.
Reference texts: Sadiku, Elements of Electromagnetics (7th ed.) — plane waves and boundaries, transmission lines, waveguides, antennas; Hayt & Buck, Engineering Electromagnetics (9th ed.) — magnetic forces and torque, transmission-line transients; Pozar, Microwave Engineering (4th ed.) — transmission-line theory, stub matching, rectangular waveguide cutoff; Balanis, Antenna Theory (4th ed.) — infinitesimal-dipole far field.
Question 6: Magnetic Field Above a Semicircular Loop with Straight Leads (equal value)
Given. Current $I=2$ A in a horizontal circuit: a semicircular arc of diameter 50 cm ($r=25$ cm) whose two ends continue as straight, mutually parallel semi-infinite leads perpendicular to the diameter; viewed from above, current circulates clockwise; field point $P$ is $h=25$ cm directly above the diameter's midpoint $O$.
Given data
Quantity
Symbol
Value
Current
$I$
2 A
Semicircle diameter / radius
$2r/r$
50 cm / 25 cm
Height of $P$ above $O$
$h$
25 cm ($=r$)
Sense
—
clockwise viewed from above
Find. The magnitude and direction of the HORIZONTAL component of $\mathbf B$ at $P$.
Approach. Place the diameter along the $x$-axis with the arc's centre at the origin; superpose Biot–Savart contributions from the semicircular arc and the two semi-infinite straight leads at the axial point $P=(0,0,h)$. By the mirror symmetry of the whole circuit about the vertical plane $x=0$, the $x$-component of $\mathbf B$ at $P$ must vanish, leaving a horizontal ($y$) component plus a vertical ($z$) component; only the horizontal one is asked for.
Arc contribution. Parametrizing the arc from $A=(-r,0,0)$ over the top to $B=(r,0,0)$ and integrating Biot–Savart (constant $|\mathbf R|=\sqrt{r^2+h^2}$ to every arc element) gives, at $h=r$:
$$B_{arc,y}=-\frac{\mu_0 I}{4\sqrt2\,\pi r},\qquad B_{arc,z}=-\frac{\mu_0 I}{8\sqrt2\,r}\qquad(B_{arc,x}=0\text{ by symmetry}).$$
Straight-lead contributions. Each semi-infinite lead (from the arc's end to infinity) contributes, by direct integration of $d\mathbf B=\frac{\mu_0I}{4\pi}\dfrac{d\boldsymbol\ell\times\hat R}{|\mathbf R|^2}$ from its own end out to infinity:
$$\mathbf B_{lead\,1}=\frac{\mu_0 I}{8\pi r}(1,0,-1),\qquad \mathbf B_{lead\,2}=\frac{\mu_0 I}{8\pi r}(-1,0,-1)\quad(\text{at }h=r).$$
The $x$-components cancel (symmetry check) and the $y$-components of the two leads are individually zero; summing:
$$\mathbf B_{leads}=\left(0,\ 0,\ -\frac{\mu_0 I}{4\pi r}\right).$$
Total field and its horizontal component. Adding arc and leads, the ONLY horizontal ($y$) contribution comes from the arc (the leads contribute none):
$$B_y=B_{arc,y}=-\frac{\mu_0 I}{4\sqrt2\,\pi r}.$$
Numerically, with $\mu_0=4\pi\times10^{-7}$ H/m, $I=2$ A, $r=0.25$ m:
$$|B_y|=\frac{(4\pi\times10^{-7})(2)}{4\sqrt2\,\pi(0.25)}=\boxed{5.66\times10^{-7}\ \text{T}=0.566\ \mu\text{T}}.$$
Direction. $B_y<0$ in our coordinates ($y=$ the direction the arc bulges), i.e. the horizontal component points AWAY from the semicircular bulge, along the same horizontal line as the two straight leads — the axis of symmetry of the loop, pointing from the arc side toward the leads' side.
Semicircular loop (radius $r=25$ cm) with two straight leads perpendicular to the diameter; $P$ sits at height $h=r$ above the centre $O$.
